Mixing
creating dynamic table using jquery i want to append model values inside a model
0
contentType: "application/json",
dataType: "json",
data: JSON.stringify(reportCriteria),
success: function (response) {
console.log(response);
if (response.reportResult != null) {
for (var i = 0 ; i < response.reportResult.length; i++) {
var data = "<tr>" +
"<td class='reportTbl'>" + moment(new Date(parseInt(response.reportResult[i].InvoiceDate.substr(6))).toLocaleDateString()).format('YYYY-MMMM-DD') + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].InvoiceNumber + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].TotalValueWithVAT + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].**PartialPayments.ChequeNumber** + "</td>" +
"</tr>";
$('#completedPaymentReportTbl tbody').after(data);
}
}
}
"PartialPayments" is the inner model how can i append the data
jquery asp.net model-view-controller
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
add a comment |
0
contentType: "application/json",
dataType: "json",
data: JSON.stringify(reportCriteria),
success: function (response) {
console.log(response);
if (response.reportResult != null) {
for (var i = 0 ; i < response.reportResult.length; i++) {
var data = "<tr>" +
"<td class='reportTbl'>" + moment(new Date(parseInt(response.reportResult[i].InvoiceDate.substr(6))).toLocaleDateString()).format('YYYY-MMMM-DD') + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].InvoiceNumber + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].TotalValueWithVAT + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].**PartialPayments.ChequeNumber** + "</td>" +
"</tr>";
$('#completedPaymentReportTbl tbody').after(data);
}
}
}
"PartialPayments" is the inner model how can i append the data
jquery asp.net model-view-controller
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
add a comment |
0
0
0
contentType: "application/json",
dataType: "json",
data: JSON.stringify(reportCriteria),
success: function (response) {
console.log(response);
if (response.reportResult != null) {
for (var i = 0 ; i < response.reportResult.length; i++) {
var data = "<tr>" +
"<td class='reportTbl'>" + moment(new Date(parseInt(response.reportResult[i].InvoiceDate.substr(6))).toLocaleDateString()).format('YYYY-MMMM-DD') + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].InvoiceNumber + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].TotalValueWithVAT + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].**PartialPayments.ChequeNumber** + "</td>" +
"</tr>";
$('#completedPaymentReportTbl tbody').after(data);
}
}
}
"PartialPayments" is the inner model how can i append the data
jquery asp.net model-view-controller
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
contentType: "application/json",
dataType: "json",
data: JSON.stringify(reportCriteria),
success: function (response) {
console.log(response);
if (response.reportResult != null) {
for (var i = 0 ; i < response.reportResult.length; i++) {
var data = "<tr>" +
"<td class='reportTbl'>" + moment(new Date(parseInt(response.reportResult[i].InvoiceDate.substr(6))).toLocaleDateString()).format('YYYY-MMMM-DD') + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].InvoiceNumber + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].TotalValueWithVAT + "</td>" +
"<td class='reportTbl'>" + response.reportResult[i].**PartialPayments.ChequeNumber** + "</td>" +
"</tr>";
$('#completedPaymentReportTbl tbody').after(data);
}
}
}
"PartialPayments" is the inner model how can i append the data
jquery asp.net model-view-controller
jquery asp.net model-view-controller
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
edited Jan 1 at 8:17
Rajesh Pandya
1,3413921
1,3413921
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
asked Jan 1 at 8:15
Romeo_SLRomeo_SL
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
After messing with a JQuery sandbox I found that $('#completedPaymentReportTbl').append(data); should work for you. As reported in Appending rows to table, Jquery 1.4+ will automatically work out that you have a tbody there and know to append the rows inside the tbody.
I have two suggestions with things that may help prevent this bug and/or future bugs:
- Try and introduce c# variables in your model or static variables for things like table names. Instead of typing #completedPaymentReportTbl, you'd type something like '@completedPaymentReportTableId' (example is for Razor templates). This will keep your table HTML and your JQuery "in sync".
- Instead of sending back the information of your model and constructing the row in your javascript, render the row on the server side and send back the HTML of the row. This ensures that when you change the row structure, all the rows will be consistent.
answered Jan 1 at 10:13
Dev243Dev243
336
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
After messing with a JQuery sandbox I found that $('#completedPaymentReportTbl').append(data); should work for you. As reported in Appending rows to table, Jquery 1.4+ will automatically work out that you have a tbody there and know to append the rows inside the tbody.
I have two suggestions with things that may help prevent this bug and/or future bugs:
- Try and introduce c# variables in your model or static variables for things like table names. Instead of typing #completedPaymentReportTbl, you'd type something like '@completedPaymentReportTableId' (example is for Razor templates). This will keep your table HTML and your JQuery "in sync".
- Instead of sending back the information of your model and constructing the row in your javascript, render the row on the server side and send back the HTML of the row. This ensures that when you change the row structure, all the rows will be consistent.
answered Jan 1 at 10:13
Dev243Dev243
336
add a comment |
1
After messing with a JQuery sandbox I found that $('#completedPaymentReportTbl').append(data); should work for you. As reported in Appending rows to table, Jquery 1.4+ will automatically work out that you have a tbody there and know to append the rows inside the tbody.
I have two suggestions with things that may help prevent this bug and/or future bugs:
- Try and introduce c# variables in your model or static variables for things like table names. Instead of typing #completedPaymentReportTbl, you'd type something like '@completedPaymentReportTableId' (example is for Razor templates). This will keep your table HTML and your JQuery "in sync".
- Instead of sending back the information of your model and constructing the row in your javascript, render the row on the server side and send back the HTML of the row. This ensures that when you change the row structure, all the rows will be consistent.
answered Jan 1 at 10:13
Dev243Dev243
336
add a comment |
1
1
1
After messing with a JQuery sandbox I found that $('#completedPaymentReportTbl').append(data); should work for you. As reported in Appending rows to table, Jquery 1.4+ will automatically work out that you have a tbody there and know to append the rows inside the tbody.
I have two suggestions with things that may help prevent this bug and/or future bugs:
- Try and introduce c# variables in your model or static variables for things like table names. Instead of typing #completedPaymentReportTbl, you'd type something like '@completedPaymentReportTableId' (example is for Razor templates). This will keep your table HTML and your JQuery "in sync".
- Instead of sending back the information of your model and constructing the row in your javascript, render the row on the server side and send back the HTML of the row. This ensures that when you change the row structure, all the rows will be consistent.
answered Jan 1 at 10:13
Dev243Dev243
336
After messing with a JQuery sandbox I found that $('#completedPaymentReportTbl').append(data); should work for you. As reported in Appending rows to table, Jquery 1.4+ will automatically work out that you have a tbody there and know to append the rows inside the tbody.
I have two suggestions with things that may help prevent this bug and/or future bugs:
- Try and introduce c# variables in your model or static variables for things like table names. Instead of typing #completedPaymentReportTbl, you'd type something like '@completedPaymentReportTableId' (example is for Razor templates). This will keep your table HTML and your JQuery "in sync".
- Instead of sending back the information of your model and constructing the row in your javascript, render the row on the server side and send back the HTML of the row. This ensures that when you change the row structure, all the rows will be consistent.
answered Jan 1 at 10:13
Dev243Dev243
336
answered Jan 1 at 10:13
Dev243Dev243
336
answered Jan 1 at 10:13
Dev243Dev243
336
answered Jan 1 at 10:13
Dev243Dev243
336
336
add a comment |
add a comment |
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