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").subm开发者_运维百科it();
});
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
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 :)
jq("#showDetail").click(function() {
jq('#formName').submit();
});
<input type="button" id="showDetail" class="secondarybutton" value="Done" />
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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