How do I insert PHP variable inside jQuery/JavaScript properly?

问题

This code works when running the application, but Dreamweaver is giving a syntax error. It doesn't like the question mark there. I like DW to be syntax error free. Is there a different way to write this? I have DW cs5.5 I can't upgrade Dreamweaver version.

    if ( $('#postage6').val() == "Your Permit Standard" ) {
        $('#postage6rate').val('<?php echo $your_permit_standard; ?>');
    }

Putting a backslash before the question mark just makes it print like this, which is not right.

    if ( $('#postage6').val() == "Your Permit Standard" ) {
        $('#postage6rate').val('<\?php echo $your_permit_standard; ?>');
    }

when it renders, there is supposed to be a value like this:

    if ( $('#postage6').val() == "Your Permit Standard" ) {
        $('#postage6rate').val('0.333');
    }

Also this doesn't work:

    if ( $('#postage6').val() == "Your Permit Standard" ) {
        var somevar = "<?php echo $your_permit_standard; ?>";
        $('#postage6rate').val(somevar);
    }

The syntax error just transfers from the line where the PHP variable was to the new line where the PHP variable is.

回答1:

You could define the value in a separate php block:

<script type="text/javascript">
   var value = '<?=$your_permit_standard?>';
</script>

And then use it in your JS:

if ( $('#postage6').val() == "Your Permit Standard" ) {
    $('#postage6rate').val(value);
}

But then you would be introducing JS dependency in PHP, which I wouldn't recommend, but since you're mixing both anyway...

回答2:

Looks like you’re setting an input value with JavaScript, after having set the .val() method argument with PHP. Why not set the value of the input with PHP directly?

<input type="text" name="postage6rate" value="<?php echo $your_permit_standard; ?>">

If you need to run this script at some time other than page load, you could bind the data to an element with the data attribute.

<input type="text" name="postage6rate" data-permit="<?php echo $your_permit_standard; ?>">

And then when you need to run your script…

window.addEventListener('onSomeEvent', function addTheData() {
  var $input = $('input[name="postage6rate"]');
  $input.val($input.data('permit'));
});

回答3:

I assume that your javascript isn't in-line (in the same PHP file) which prevents PHP from executing

Try:

<?php 
$your_permit_standard = "0.335";
?>
<html>
  <body>
    <input id="postage6" name="postage6" value="Your Permit Standard"/>
    <input id="postage6rate" name="postage6rate"/>
    <script src="http://code.jquery.com/jquery-2.0.3.min.js"></script>
    <script>
    if ( $('#postage6').val() == "Your Permit Standard" ) {
        $('#postage6rate').val('<?php echo $your_permit_standard; ?>');
    }
    </script>
  </body>
<html>

And here's the working example

来源：https://stackoverflow.com/questions/19912467/how-do-i-insert-php-variable-inside-jquery-javascript-properly

标签

javascript

php

jquery

dreamweaver