views:

49

answers:

2

I have a form which contains the method "POST" and action ="abc.php" and button type of <input type ="button"> I have a handler when i cick that button i want to send a request to abc.php but nothing is happening no action is being prformed.I dont want to change the <input type ="button"> to <input type="submit>.How do i submit the Form .Here is the code

<form name= "form1" id ="form1" action ="abc.php" method="post">
<input type ="button" id="mybutton" value ="Add"> 
......
//All form Elements.
</form>

$(document).ready(function() {
    //Load all elements
});
$("#mybutton").click(function(e){
    alert(true);
    //$("#frmpohdr").submit();
});

The Above Statement is giving error and i know we need to have button type of submit for this method.How do i submit the Form to the abc.php when i click the button .I have tried all $.ajax methods

+2  A: 

Have you tried putting all the code inside ready?

Also, if the form's id is form 1 you should do this:

$("#form1").submit();

And to avoid the buttons default's behaviour you should also add this link inside click's function:

e.preventDefault();

I also recommend you having a look at jQuery Form Plugin: http://jquery.malsup.com/form/

I hope i helped :)

ozke
"if the form's **id** is form1" - the # selector in jQuery is for ids, not names.
GalacticCowboy
You're right, silly mistake. Fixed :)
ozke
A: 

You will either submit the form through the HTML form (i.e. change the input type = submit) Or you could use the $.post/$.ajax

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