CODEWITHSUNDEEP
jquery
DJ180
DJ180

Reputation: 19854

JQuery: find and replace of attribute values

I am having difficulty performing a find and replace of attribute values.

Consider this html:

<tr id="rules[0]">
    <td>
        <input id="rules0.isActive1" name="rules[0].isActive" type="checkbox" value="true"/><input type="hidden" name="_rules[0].isActive" value="on"/>
    </td>
    <td>
        <select id="rules0.leftCondition.name" name="rules[0].leftCondition.name">
            ...
        </select>
    </td>

I am looking to update each 's name attribute with the appropriate count e.g. rules[0].isActive is updated to rules[10].isActive

I am using this JQuery snippet:

$(newRow).find('[id*="rules"]').each(function(){
    // Update the 'rules[0]' part of the name attribute to contain the latest count 
    this.name.replace('rules\[0\]','rules\[$count\]');
});

where newRow is a variable holding the entire block and count holds the latest counter value. However, when I debug this code it seems that "this.name" is undefined.

Where am I going wrong here?

Thanks

Upvotes: 24

Views: 81880

Answers (4)

Muru Bakthavachalam
Muru Bakthavachalam

Reputation: 1350

try this: Assume row attribute name is 'filterCol' then you can set the new value to that attribute

$("#trId").attr("filterCol", newValue);

Upvotes: 11

kaz
kaz

Reputation: 1943

To update attribute of element you should use the .attr function. Also, when using this in jQuery context you must use $(this) syntax. If I understand your intentions so it should be something like that:

$(newRow).find('[id*="rules"]').each(function(){
    // Update the 'rules[0]' part of the name attribute to contain the latest count 
    $(this).attr('name',$(this).attr('name').replace('rules\[0\]','rules\[$count\]'));
});

Upvotes: 33

Erik Escobedo
Erik Escobedo

Reputation: 2803

You should use $(this).attr('name') instead of this.name.

Upvotes: 6

techfoobar
techfoobar

Reputation: 66663

Try this:

$(this).attr('name', $(this).attr('name').replace('rules\[0\]','rules\[$count\]'));

Upvotes: 3

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