CODEWITHSUNDEEP
phpjquery
luke
luke

Reputation: 53

How can i use jquery code in php file?

I need to add a property to the input[name=$key] after a control inside a PHP file

Code: 

if ( !isset( $field_obj[ $key ] ) ) 
{
?>
    <script> $('input[name=$key]').prop("data-toggle", "popover");</script>
<?php

but $key is declared in PHP section and i can't use it inside the script section. So how can I do it?

Upvotes: 0

Views: 178

Answers (5)

jiboulex
jiboulex

Reputation: 3021

Try to echo $key:

<?php
if ( !isset( $field_obj[ $key ] ) ) {
    ?>
    <script>
        $('input[name=<?= $key ?>').prop("data-toggle", "popover");
    </script>
    <?php
}
?>

Upvotes: -1

Digpal Singh
Digpal Singh

Reputation: 166

<?php    
if ( !isset( $field_obj[ $key ] ) ) {
?>

<script>
    $('input[name="<?php echo $key ?>" ]').prop("data-toggle", "popover"); 
</script>

<?php
}
?>

Upvotes: -1

Simba
Simba

Reputation: 5032

The bad practice that everyone keeps going on about is the practice of generating your Javascript code.

Best practice would be to have all of your JS code as static as possible, loaded in a separate .js file.

In order to get this bit of code to trigger when the PHP variable is set, you would do something like this:

In your main JS code file, you would have a block like this which runs when the page is loaded:

for(var key in window.pageConfig.keys) {
    $('input[name='+key+']').prop("data-toggle", "popover");
}

This code would be loaded exactly the same every time the page loads. However, your PHP code would generate a small chunk of JS code that populates the window.pageConfig object with the relevant data.

Your PHP code might look something like this:

$jsonKeys = json_encode(['keys'=>array_keys($field_obj)]);
print "<script>var pageConfig = {$jsonKeys};</script>";

This would produce the pageConfig object in Javascript which the rest of your code could then use to work out which keys to process.

Note, I've made some assumptions in this PHP code; eg I'm assuming you want to run the jQuery code for every field_obj. If that assumption is wrong, then feel free to change the code to suit, but hopefully the example I've given you will take you along the right path.

They key point here is to minimise the amount of JS code that your PHP code generates; certainly avoid generating anything other than JS data objects in PHP; the rest of your JS code should be static.

Upvotes: 0

Naumov
Naumov

Reputation: 1167

<script> $('input[name=<?php echo $key ?>]').prop("data-toggle", "popover"); </script>

but this is a bad practic..

Upvotes: 3

Nishan Senevirathna
Nishan Senevirathna

Reputation: 719

Not recommended but try this,

<?php    
if ( !isset( $field_obj[ $key ] ) ) {
?>
<script> $('input[name=<?php echo $key ?> ]').prop("data-toggle", "popover"); </script>
<?php
}
?>

Upvotes: 1

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