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I have the following code

dt.ajax.url( 'test.php?status=' ).load();

I defined a variable

var status= 55 

I wand to add status_2 in the link like this

dt.ajax.url( 'test.php?status=&status' ).load();

How can I do it ?

marvillous
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1 Answers1

1

Try this. You need to concatenate the variable with url

dt.ajax.url('test.php?status='+status).load();
Mairaj Ahmad
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