Question

81

Calling a PHP variable in jQuery

rated 0 times [ 83] [ 2] / answers: 1 / hits: 37621 / 11 Years ago, wed, august 14, 2013, 12:00:00

I'am developing a zoom tool in my shopping cart and I am stuck on how to call a PHP variable in a jQuery function.

Here's my code :

jQuery(document).ready(function($){

$('#image1').addimagezoom({ // single image zoom

    zoomrange: [3, 10],

    magnifiersize: [800,300], 

    magnifierpos: 'right',

    cursorshade: true,

    largeimage: php variable //we add the directory of the image.

});

});

I need to put

$src =images/products/.mysql_result($execute_select_product_query,0,'image1').

in my function where I put PHP variable.

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zahrafrancisr

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answered 11 Years ago adrianobeds · Accepted Answer

You have two or three options: if the Javascript is in the php file, you can

var phpVar = <?php echo $var; ?>;

Otherwise if the Javascript is anywhere at all, you can do:

<input type=hidden id=phpVar value=<?php echo $var; ?>>

and then access it as

$('#phpVar').val();

Example 1:

jQuery(document).ready(function($){

$('#image1').addimagezoom({ // single image zoom

    zoomrange: [3, 10],

    magnifiersize: [800,300], 

    magnifierpos: 'right',

    cursorshade: true,

    largeimage: <?php echo $var; ?> //we add the directory of the image.

});

});

Example 2:

Html:

<input type=hidden id=phpVar value=<?php echo $var; ?>>

Javascript

jQuery(document).ready(function($){

$('#image1').addimagezoom({ // single image zoom

    zoomrange: [3, 10],

    magnifiersize: [800,300], 

    magnifierpos: 'right',

    cursorshade: true,

    largeimage: $('#phpVar').val(); //we add the directory of the image.

});

});