Question

I’m using jquery form plugin. and this is my sample form

<form id='formId' action='<?php echo $_SERVER['PHP_SELF']l?>' method='POST'>
<input type='text' name='username'/>
</form>

And this my Javascript

$('#formId').ajaxForm({
    success:function(data){
       console.log(data)
    }
})

In this case i want to pass another data value dynamically without implement it in HTML. For example i want to pass the password field. Is there any method to implement this option?

Answer 1

From This Page, the beforeSubmit function allows you to hook into the form data submitted and inject / modify data.

It allows for 3 arguments, the form presented as an array(this means we can push into it), the form presented as a jquery Object (for serialization), and options object that you use with ajaxForm/ajaxSubmit. Working example ->

$('#formId').ajaxForm({
  beforeSubmit: function(formData, formObject, formOptions){
    //formData is the form array. We can push the password into the array before we submit the form.
    formData.push({
      password: //this is where you put the password you want
    });
  },
  success:function(data){
   console.log(data)
  }
});