Mixing
jQuery set value of td cell depending on values of other cells using Array and if/else if
0
I've Run into another problem trying to show different SLA statuses in the slaClass td.
I was previously using console.log("red") etc to show the statuses for testing. This worked perfectly, however when then trying to set the text for the slaClass field it just shows Warning and not the correct values.
$(document).ready(function() {
var daysLeft = $('.DaysLeft');
var sla = $('.slaClass');
$(daysLeft).each(function() {
var daysLeft = $(this).text(); // assign a current value to increase its performance.
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Demo code - http://jsfiddle.net/ZmesE/294/
jquery html arrays
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
asked Nov 21 '18 at 15:54
WHowardWHoward
518
add a comment |
0
I've Run into another problem trying to show different SLA statuses in the slaClass td.
I was previously using console.log("red") etc to show the statuses for testing. This worked perfectly, however when then trying to set the text for the slaClass field it just shows Warning and not the correct values.
$(document).ready(function() {
var daysLeft = $('.DaysLeft');
var sla = $('.slaClass');
$(daysLeft).each(function() {
var daysLeft = $(this).text(); // assign a current value to increase its performance.
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Demo code - http://jsfiddle.net/ZmesE/294/
jquery html arrays
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
asked Nov 21 '18 at 15:54
WHowardWHoward
518
1
Does your demo code not work in a StackOverflow snippet?
– connexo
Nov 21 '18 at 15:57
add a comment |
0
0
0
I've Run into another problem trying to show different SLA statuses in the slaClass td.
I was previously using console.log("red") etc to show the statuses for testing. This worked perfectly, however when then trying to set the text for the slaClass field it just shows Warning and not the correct values.
$(document).ready(function() {
var daysLeft = $('.DaysLeft');
var sla = $('.slaClass');
$(daysLeft).each(function() {
var daysLeft = $(this).text(); // assign a current value to increase its performance.
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Demo code - http://jsfiddle.net/ZmesE/294/
jquery html arrays
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
asked Nov 21 '18 at 15:54
WHowardWHoward
518
I've Run into another problem trying to show different SLA statuses in the slaClass td.
I was previously using console.log("red") etc to show the statuses for testing. This worked perfectly, however when then trying to set the text for the slaClass field it just shows Warning and not the correct values.
$(document).ready(function() {
var daysLeft = $('.DaysLeft');
var sla = $('.slaClass');
$(daysLeft).each(function() {
var daysLeft = $(this).text(); // assign a current value to increase its performance.
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Demo code - http://jsfiddle.net/ZmesE/294/
$(document).ready(function() {
var daysLeft = $('.DaysLeft');
var sla = $('.slaClass');
$(daysLeft).each(function() {
var daysLeft = $(this).text(); // assign a current value to increase its performance.
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>$(document).ready(function() {
var daysLeft = $('.DaysLeft');
var sla = $('.slaClass');
$(daysLeft).each(function() {
var daysLeft = $(this).text(); // assign a current value to increase its performance.
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>jquery html arrays
jquery html arrays
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
asked Nov 21 '18 at 15:54
WHowardWHoward
518
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
asked Nov 21 '18 at 15:54
WHowardWHoward
518
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
edited Nov 21 '18 at 16:00
Rory McCrossan
245k29212248
245k29212248
asked Nov 21 '18 at 15:54
WHowardWHoward
518
asked Nov 21 '18 at 15:54
WHowardWHoward
518
asked Nov 21 '18 at 15:54
WHowardWHoward
518
518
1
Does your demo code not work in a StackOverflow snippet?
– connexo
Nov 21 '18 at 15:57
add a comment |
1
Does your demo code not work in a StackOverflow snippet?
– connexo
Nov 21 '18 at 15:57
1
1
Does your demo code not work in a StackOverflow snippet?
– connexo
Nov 21 '18 at 15:57
Does your demo code not work in a StackOverflow snippet?
– connexo
Nov 21 '18 at 15:57
add a comment |
2 Answers
2
active
oldest
votes
2
The problem is that you are not grabbing the sibling ".slaClass" element from your current ".daysLeft" element.
Try this:
$(document).ready(function(){
var daysLeft = $('.DaysLeft');
$(daysLeft).each(function() {
var daysLeft=$(this).text(); // assign a current value to increase its performance.
var sla = $(this).siblings('.slaClass');
if(daysLeft !="" && daysLeft!=undefined)
{
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
}else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
}else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
}
else(alert("Error"));
})
});
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
add a comment |
2
The issue is because you're updating all the .slaClass elements instead of the one related to the current .DaysLeft element in the each iteration. To fix this, use closest() to get the parent tr of the current element. Try this:
$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Also note that in your original logic DaysLeft and slaClass are already jQuery objects, so you don't need to put them in jQuery objects again.
It's also not a good idea to re-use the same variable name, even if it is in a different scope. That could very easily lead to some unnecessary confusion.
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
May I be allowed to add my opinion to your answer? I suggest to use.next()
– Foo
Nov 21 '18 at 16:05
1
@TânNguyễn that's an alternative, but I prefer the use ofclosest()andfind()as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Usingnext()would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.
– Rory McCrossan
Nov 21 '18 at 16:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
The problem is that you are not grabbing the sibling ".slaClass" element from your current ".daysLeft" element.
Try this:
$(document).ready(function(){
var daysLeft = $('.DaysLeft');
$(daysLeft).each(function() {
var daysLeft=$(this).text(); // assign a current value to increase its performance.
var sla = $(this).siblings('.slaClass');
if(daysLeft !="" && daysLeft!=undefined)
{
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
}else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
}else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
}
else(alert("Error"));
})
});
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
add a comment |
2
The problem is that you are not grabbing the sibling ".slaClass" element from your current ".daysLeft" element.
Try this:
$(document).ready(function(){
var daysLeft = $('.DaysLeft');
$(daysLeft).each(function() {
var daysLeft=$(this).text(); // assign a current value to increase its performance.
var sla = $(this).siblings('.slaClass');
if(daysLeft !="" && daysLeft!=undefined)
{
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
}else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
}else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
}
else(alert("Error"));
})
});
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
add a comment |
2
2
2
The problem is that you are not grabbing the sibling ".slaClass" element from your current ".daysLeft" element.
Try this:
$(document).ready(function(){
var daysLeft = $('.DaysLeft');
$(daysLeft).each(function() {
var daysLeft=$(this).text(); // assign a current value to increase its performance.
var sla = $(this).siblings('.slaClass');
if(daysLeft !="" && daysLeft!=undefined)
{
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
}else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
}else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
}
else(alert("Error"));
})
});
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
The problem is that you are not grabbing the sibling ".slaClass" element from your current ".daysLeft" element.
Try this:
$(document).ready(function(){
var daysLeft = $('.DaysLeft');
$(daysLeft).each(function() {
var daysLeft=$(this).text(); // assign a current value to increase its performance.
var sla = $(this).siblings('.slaClass');
if(daysLeft !="" && daysLeft!=undefined)
{
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$(sla).text('Red');
}else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$(sla).text('Warning');
}else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$(sla).text('Green');
}
}
else(alert("Error"));
})
});
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
answered Nov 21 '18 at 16:00
MikeSMikeS
1,5441510
1,5441510
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
add a comment |
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
Looks good to me
– Jamie_D
Nov 21 '18 at 16:05
add a comment |
2
The issue is because you're updating all the .slaClass elements instead of the one related to the current .DaysLeft element in the each iteration. To fix this, use closest() to get the parent tr of the current element. Try this:
$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Also note that in your original logic DaysLeft and slaClass are already jQuery objects, so you don't need to put them in jQuery objects again.
It's also not a good idea to re-use the same variable name, even if it is in a different scope. That could very easily lead to some unnecessary confusion.
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
May I be allowed to add my opinion to your answer? I suggest to use.next()
– Foo
Nov 21 '18 at 16:05
1
@TânNguyễn that's an alternative, but I prefer the use ofclosest()andfind()as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Usingnext()would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.
– Rory McCrossan
Nov 21 '18 at 16:07
add a comment |
2
The issue is because you're updating all the .slaClass elements instead of the one related to the current .DaysLeft element in the each iteration. To fix this, use closest() to get the parent tr of the current element. Try this:
$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Also note that in your original logic DaysLeft and slaClass are already jQuery objects, so you don't need to put them in jQuery objects again.
It's also not a good idea to re-use the same variable name, even if it is in a different scope. That could very easily lead to some unnecessary confusion.
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
May I be allowed to add my opinion to your answer? I suggest to use.next()
– Foo
Nov 21 '18 at 16:05
1
@TânNguyễn that's an alternative, but I prefer the use ofclosest()andfind()as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Usingnext()would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.
– Rory McCrossan
Nov 21 '18 at 16:07
add a comment |
2
2
2
The issue is because you're updating all the .slaClass elements instead of the one related to the current .DaysLeft element in the each iteration. To fix this, use closest() to get the parent tr of the current element. Try this:
$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Also note that in your original logic DaysLeft and slaClass are already jQuery objects, so you don't need to put them in jQuery objects again.
It's also not a good idea to re-use the same variable name, even if it is in a different scope. That could very easily lead to some unnecessary confusion.
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
The issue is because you're updating all the .slaClass elements instead of the one related to the current .DaysLeft element in the each iteration. To fix this, use closest() to get the parent tr of the current element. Try this:
$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>Also note that in your original logic DaysLeft and slaClass are already jQuery objects, so you don't need to put them in jQuery objects again.
It's also not a good idea to re-use the same variable name, even if it is in a different scope. That could very easily lead to some unnecessary confusion.
$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>$(document).ready(function() {
$('.DaysLeft').each(function() {
var $sla = $(this).closest('tr').find('.slaClass');
var daysLeft = $(this).text();
if (daysLeft != "" && daysLeft != undefined) {
if (parseInt(daysLeft) <= 30) {
console.log('Red');
$sla.text('Red');
} else if (parseInt(daysLeft) > 30 && parseInt(daysLeft) < 90) {
console.log('Warning');
$sla.text('Warning');
} else if (parseInt(daysLeft) >= 90) {
console.log('Green');
$sla.text('Green');
}
} else(alert("Error"));
})
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<table>
<tbody>
<tr>
<th>Applicants</th>
<th>Interviews</th>
<th>Offers</th>
<th>Days Left to Fill </th>
<th>SLA</th>
</tr>
<tr>
<td>530</td>
<td>50</td>
<td>1</td>
<td class="DaysLeft">125</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>56</td>
<td>0</td>
<td>0</td>
<td class="DaysLeft">25</td>
<td class="slaClass"></td>
</tr>
<tr>
<td>82</td>
<td>6</td>
<td>0</td>
<td class="DaysLeft">62</td>
<td class="slaClass"></td>
</tr>
</tbody>
</table>answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
edited Nov 21 '18 at 16:05
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
answered Nov 21 '18 at 16:02
Rory McCrossanRory McCrossan
245k29212248
245k29212248
May I be allowed to add my opinion to your answer? I suggest to use.next()
– Foo
Nov 21 '18 at 16:05
1
@TânNguyễn that's an alternative, but I prefer the use ofclosest()andfind()as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Usingnext()would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.
– Rory McCrossan
Nov 21 '18 at 16:07
add a comment |
May I be allowed to add my opinion to your answer? I suggest to use.next()
– Foo
Nov 21 '18 at 16:05
1
@TânNguyễn that's an alternative, but I prefer the use ofclosest()andfind()as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Usingnext()would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.
– Rory McCrossan
Nov 21 '18 at 16:07
May I be allowed to add my opinion to your answer? I suggest to use
.next()– Foo
Nov 21 '18 at 16:05
May I be allowed to add my opinion to your answer? I suggest to use
.next()– Foo
Nov 21 '18 at 16:05
1
1
@TânNguyễn that's an alternative, but I prefer the use of
closest() and find() as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Using next() would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.– Rory McCrossan
Nov 21 '18 at 16:07
@TânNguyễn that's an alternative, but I prefer the use of
closest() and find() as it's more robust. If a designer/HTML editor changes the order of the fields, or adds a new one in between, this logic will still work. Using next() would break. Perhaps not important when working alone or in a small team, but in enterprise level apps with a large distributed team, it can be a big problem.– Rory McCrossan
Nov 21 '18 at 16:07
add a comment |
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Does your demo code not work in a StackOverflow snippet?
– connexo
Nov 21 '18 at 15:57