Using PHP strings as element ID in HTML

I’m trying to dynamically create  elementss with different ids to "fill" them with JavaScript’s innerHTML property.

Firstly I try to create 50 times:

echo "<p id=space1></p>";

But the number up going like space2, space3 etc.

I made this loop to generate the incrementing numbers and put it together with the string, and echo it:

for ($x = 1; $x <= 50; $x++) {

   $xarray = array($x => "space" . $x);

   echo "<p id=$xarrray[$x]></p>";

}

But when I run it and check the HTML code on the page, all I see is:

<p id=""></p>

But when I check the arrays by directly echoing them, the names get displayed correctly.

So how can I use my php strings as element ID in HTML?

edited Jan 2 at 1:19

Funk Forty Niner

asked Jan 1 at 23:55

Snoeny

add a comment |

I’m trying to dynamically create  elementss with different ids to "fill" them with JavaScript’s innerHTML property.

Firstly I try to create 50 times:

echo "<p id=space1></p>";

But the number up going like space2, space3 etc.

I made this loop to generate the incrementing numbers and put it together with the string, and echo it:

for ($x = 1; $x <= 50; $x++) {

   $xarray = array($x => "space" . $x);

   echo "<p id=$xarrray[$x]></p>";

}

But when I run it and check the HTML code on the page, all I see is:

<p id=""></p>

But when I check the arrays by directly echoing them, the names get displayed correctly.

So how can I use my php strings as element ID in HTML?

edited Jan 2 at 1:19

Funk Forty Niner

asked Jan 1 at 23:55

Snoeny

add a comment |

I’m trying to dynamically create  elementss with different ids to "fill" them with JavaScript’s innerHTML property.

Firstly I try to create 50 times:

echo "<p id=space1></p>";

But the number up going like space2, space3 etc.

I made this loop to generate the incrementing numbers and put it together with the string, and echo it:

for ($x = 1; $x <= 50; $x++) {

   $xarray = array($x => "space" . $x);

   echo "<p id=$xarrray[$x]></p>";

}

But when I run it and check the HTML code on the page, all I see is:

<p id=""></p>

But when I check the arrays by directly echoing them, the names get displayed correctly.

So how can I use my php strings as element ID in HTML?

edited Jan 2 at 1:19

Funk Forty Niner

asked Jan 1 at 23:55

Snoeny

I’m trying to dynamically create  elementss with different ids to "fill" them with JavaScript’s innerHTML property.

Firstly I try to create 50 times:

echo "<p id=space1></p>";

But the number up going like space2, space3 etc.

I made this loop to generate the incrementing numbers and put it together with the string, and echo it:

for ($x = 1; $x <= 50; $x++) {

   $xarray = array($x => "space" . $x);

   echo "<p id=$xarrray[$x]></p>";

}

But when I run it and check the HTML code on the page, all I see is:

<p id=""></p>

But when I check the arrays by directly echoing them, the names get displayed correctly.

So how can I use my php strings as element ID in HTML?

php html arrays

edited Jan 2 at 1:19

Funk Forty Niner

asked Jan 1 at 23:55

Snoeny

edited Jan 2 at 1:19

Funk Forty Niner

asked Jan 1 at 23:55

Snoeny

edited Jan 2 at 1:19

Funk Forty Niner

edited Jan 2 at 1:19

Funk Forty Niner

edited Jan 2 at 1:19

Funk Forty Niner

asked Jan 1 at 23:55

Snoeny

asked Jan 1 at 23:55

Snoeny

asked Jan 1 at 23:55

Snoeny

add a comment |

3 Answers
3

active

oldest

votes

-1

You have a typo in retrieving your array.

$xarray = array($x => "space" . $x);

echo "<p id=$xarrray[$x]></p>";

Your first line it's called $xarray and your second line it's called $xarrray (extra r)

$xarray=array();

for ($x = 1; $x <= 50; $x++) {

    $xarray[$x] = "space" . $x;

    echo '<p id="{$xarray[$x]}"></p>';

}

answered Jan 2 at 0:30

Second2None

1,1281416

Thanks, I totally oversaw that :/ After removing this typo it works exactly the way i tried it in the question.

– Snoeny
Jan 2 at 0:36

It happens :) If you plan on using the array you should initiate it outside the loop to stop anything unexpected from happening. Also don't forget to wrap your p id in either single or double quotes

– Second2None
Jan 2 at 0:38

1

@Snoeny you may want to look at using an IDE - many will warn you about undefined variables, etc. even though PHP is very loose with declaration vs use much less data type etc

– ivanivan
Jan 2 at 0:42

add a comment |

-2

<p id="<?php echo $php_variable; ?>"></p>

This is because the server runs and compile before the browser renders the view/template. On rendering the above P element in your DOM, $php_variable would have been resolved and echoed into the quotes.

answered Jan 1 at 23:59

Joshua

1025

add a comment |

-2

Try this:

   $xarray=array();

   for ($x = 1; $x <= 50; $x++) 

   {

        $xarray[$x] = "space" . $x;

        echo "<p id='" . $xarray[$x] . "'></p>";

   }

edited Jan 2 at 0:34

answered Jan 2 at 0:01

Ass3mbler

3,11611617

tried it and still just get this as html code: 

– Snoeny
Jan 2 at 0:29

@Snoeny there was a typo, sorry, try again now :)

– Ass3mbler
Jan 2 at 0:35

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

-1

You have a typo in retrieving your array.

$xarray = array($x => "space" . $x);

echo "<p id=$xarrray[$x]></p>";

Your first line it's called $xarray and your second line it's called $xarrray (extra r)

$xarray=array();

for ($x = 1; $x <= 50; $x++) {

    $xarray[$x] = "space" . $x;

    echo '<p id="{$xarray[$x]}"></p>';

}

answered Jan 2 at 0:30

Second2None

1,1281416

Thanks, I totally oversaw that :/ After removing this typo it works exactly the way i tried it in the question.

– Snoeny
Jan 2 at 0:36

It happens :) If you plan on using the array you should initiate it outside the loop to stop anything unexpected from happening. Also don't forget to wrap your p id in either single or double quotes

– Second2None
Jan 2 at 0:38

1

@Snoeny you may want to look at using an IDE - many will warn you about undefined variables, etc. even though PHP is very loose with declaration vs use much less data type etc

– ivanivan
Jan 2 at 0:42

add a comment |

-1

You have a typo in retrieving your array.

$xarray = array($x => "space" . $x);

echo "<p id=$xarrray[$x]></p>";

Your first line it's called $xarray and your second line it's called $xarrray (extra r)

$xarray=array();

for ($x = 1; $x <= 50; $x++) {

    $xarray[$x] = "space" . $x;

    echo '<p id="{$xarray[$x]}"></p>';

}

answered Jan 2 at 0:30

Second2None

1,1281416

Thanks, I totally oversaw that :/ After removing this typo it works exactly the way i tried it in the question.

– Snoeny
Jan 2 at 0:36

It happens :) If you plan on using the array you should initiate it outside the loop to stop anything unexpected from happening. Also don't forget to wrap your p id in either single or double quotes

– Second2None
Jan 2 at 0:38

1

@Snoeny you may want to look at using an IDE - many will warn you about undefined variables, etc. even though PHP is very loose with declaration vs use much less data type etc

– ivanivan
Jan 2 at 0:42

add a comment |

-1

You have a typo in retrieving your array.

$xarray = array($x => "space" . $x);

echo "<p id=$xarrray[$x]></p>";

Your first line it's called $xarray and your second line it's called $xarrray (extra r)

$xarray=array();

for ($x = 1; $x <= 50; $x++) {

    $xarray[$x] = "space" . $x;

    echo '<p id="{$xarray[$x]}"></p>';

}

answered Jan 2 at 0:30

Second2None

1,1281416

You have a typo in retrieving your array.

$xarray = array($x => "space" . $x);

echo "<p id=$xarrray[$x]></p>";

Your first line it's called $xarray and your second line it's called $xarrray (extra r)

$xarray=array();

for ($x = 1; $x <= 50; $x++) {

    $xarray[$x] = "space" . $x;

    echo '<p id="{$xarray[$x]}"></p>';

}

answered Jan 2 at 0:30

Second2None

1,1281416

answered Jan 2 at 0:30

Second2None

1,1281416

answered Jan 2 at 0:30

Second2None

1,1281416

answered Jan 2 at 0:30

Second2None

1,1281416

Thanks, I totally oversaw that :/ After removing this typo it works exactly the way i tried it in the question.

– Snoeny
Jan 2 at 0:36

It happens :) If you plan on using the array you should initiate it outside the loop to stop anything unexpected from happening. Also don't forget to wrap your p id in either single or double quotes

– Second2None
Jan 2 at 0:38

1

@Snoeny you may want to look at using an IDE - many will warn you about undefined variables, etc. even though PHP is very loose with declaration vs use much less data type etc

– ivanivan
Jan 2 at 0:42

add a comment |

Thanks, I totally oversaw that :/ After removing this typo it works exactly the way i tried it in the question.

– Snoeny
Jan 2 at 0:36

It happens :) If you plan on using the array you should initiate it outside the loop to stop anything unexpected from happening. Also don't forget to wrap your p id in either single or double quotes

– Second2None
Jan 2 at 0:38

1

@Snoeny you may want to look at using an IDE - many will warn you about undefined variables, etc. even though PHP is very loose with declaration vs use much less data type etc

– ivanivan
Jan 2 at 0:42

Thanks, I totally oversaw that :/ After removing this typo it works exactly the way i tried it in the question.

– Snoeny
Jan 2 at 0:36

It happens :) If you plan on using the array you should initiate it outside the loop to stop anything unexpected from happening. Also don't forget to wrap your p id in either single or double quotes

– Second2None
Jan 2 at 0:38

@Snoeny you may want to look at using an IDE - many will warn you about undefined variables, etc. even though PHP is very loose with declaration vs use much less data type etc

– ivanivan
Jan 2 at 0:42

add a comment |

-2

<p id="<?php echo $php_variable; ?>"></p>

answered Jan 1 at 23:59

Joshua

1025

add a comment |

-2

<p id="<?php echo $php_variable; ?>"></p>

answered Jan 1 at 23:59

Joshua

1025

add a comment |

-2

<p id="<?php echo $php_variable; ?>"></p>

answered Jan 1 at 23:59

Joshua

1025

<p id="<?php echo $php_variable; ?>"></p>

answered Jan 1 at 23:59

Joshua

1025

answered Jan 1 at 23:59

Joshua

1025

answered Jan 1 at 23:59

Joshua

1025

answered Jan 1 at 23:59

Joshua

1025

add a comment |

-2

Try this:

   $xarray=array();

   for ($x = 1; $x <= 50; $x++) 

   {

        $xarray[$x] = "space" . $x;

        echo "<p id='" . $xarray[$x] . "'></p>";

   }

edited Jan 2 at 0:34

answered Jan 2 at 0:01

Ass3mbler

3,11611617

tried it and still just get this as html code: 

– Snoeny
Jan 2 at 0:29

@Snoeny there was a typo, sorry, try again now :)

– Ass3mbler
Jan 2 at 0:35

add a comment |

-2

Try this:

   $xarray=array();

   for ($x = 1; $x <= 50; $x++) 

   {

        $xarray[$x] = "space" . $x;

        echo "<p id='" . $xarray[$x] . "'></p>";

   }

edited Jan 2 at 0:34

answered Jan 2 at 0:01

Ass3mbler

3,11611617

tried it and still just get this as html code: 

– Snoeny
Jan 2 at 0:29

@Snoeny there was a typo, sorry, try again now :)

– Ass3mbler
Jan 2 at 0:35

add a comment |

-2

Try this:

   $xarray=array();

   for ($x = 1; $x <= 50; $x++) 

   {

        $xarray[$x] = "space" . $x;

        echo "<p id='" . $xarray[$x] . "'></p>";

   }

edited Jan 2 at 0:34

answered Jan 2 at 0:01

Ass3mbler

3,11611617

Try this:

   $xarray=array();

   for ($x = 1; $x <= 50; $x++) 

   {

        $xarray[$x] = "space" . $x;

        echo "<p id='" . $xarray[$x] . "'></p>";

   }

edited Jan 2 at 0:34

answered Jan 2 at 0:01

Ass3mbler

3,11611617

edited Jan 2 at 0:34

answered Jan 2 at 0:01

Ass3mbler

3,11611617

answered Jan 2 at 0:01

Ass3mbler

3,11611617

answered Jan 2 at 0:01

Ass3mbler

3,11611617

tried it and still just get this as html code: 

– Snoeny
Jan 2 at 0:29

@Snoeny there was a typo, sorry, try again now :)

– Ass3mbler
Jan 2 at 0:35

add a comment |

tried it and still just get this as html code: 

– Snoeny
Jan 2 at 0:29

@Snoeny there was a typo, sorry, try again now :)

– Ass3mbler
Jan 2 at 0:35

tried it and still just get this as html code: 

– Snoeny
Jan 2 at 0:29

@Snoeny there was a typo, sorry, try again now :)

– Ass3mbler
Jan 2 at 0:35

add a comment |

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