Reputation: 1114
I have an JQuery function which submits data from a php form to a different php file(submitForm.php
). The submission works fine and there isn't any problem whatsoever, below is the handler:
submitHandler : function(){
$.ajax({
type: "POST",
cache:false,
url: "submitForm.php",
data: $('#form').serialize(),
success: function(data) {
}
});
}
Now, I want to be able get data from the submit form (submitForm.php
), and load it into a different page.
This is my submitForm.php
<?php
$name="Amanda";
$age="23";
$data = array("name"=>"$name","age"=>"$age");
header("Content-Type: application/json");
echo json_encode($data);
?>
This is how my new submitHandler
looks like
submitHandler : function(){
$.ajax({
type: "POST",
cache:false,
url: "submitForm.php",
data: $('#form').serialize(),
success: function(data) {
var name= html(name);
var age=html(age);
$("#message").load("newpage.php",{name:name,age:age});
}
});
}
I think I am doing it wrong, I would be grateful if somebody could correct me or give an idea as to how to do this. Thanks
Upvotes: 1
Views: 629
Reputation: 2226
It should be like this. If you want to take your returner data you should use formal parameter of success function.
submitHandler : function(){
$.ajax({
type: "POST",
cache:false,
url: "submitForm.php",
data: $('#form').serialize(),
dataType : 'json',
success: function(data) {
var name= data.name;
var age=data.age;
$("#message").load("newpage.php",{name:name,age:age});
}
});
}
Edit: Also you need dataType: 'json' line.
Upvotes: 3