问题
I don't think I am passing the variable the right way between my separate PHP and AJAX files.
I am debugging this by triggering the second condition $status = 'info';
in my PHP file.
Currently, status
is coming up as "undefined" for alert(data.status);
signup_process.php
if (condition){
$status = 'success';
else {
$status = 'info';
}
AJAX
function send() {
var data = $('#signup_form').serialize();
$.ajax({
type: "POST",
url: "signup_process.php",
data: data,
success: function (data) {
alert(data.status);
if (data.status == 'success') {
// everything went alright, submit
$('#signup_form').submit();
} else if (data.status == 'info')
{
console.log(data.status);
$("label#email_error").show();
return false;
}
}
});
return false;
};
I know that the 2nd condition is being triggered because I put a header redirect there just for testing and it worked fine.
回答1:
Good to use json while return back data from php to ajax.
$return_data = array();
if (condition){
$return_data['status'] = 'success';
} else {
$return_data['status'] = 'info';
}
echo json_encode($return_data);
exit();
Now, if you are return back json data to ajax, then you need to specify return data type into ajax call as below
function send() {
var data = $('#signup_form').serialize();
$.ajax({
type: "POST",
url: "signup_process.php",
data: data,
dataType: 'json',
success: function (data) {
alert(data.status);
if (data.status == 'success') {
// everything went alright, submit
$('#signup_form').submit();
} else if (data.status == 'info')
{
console.log(data.status);
$("label#email_error").show();
return false;
}
}
});
return false;
};
回答2:
You should send a JSON object back from php:
$data = array();
if (condition){
$data['status'] = 'success';
else {
$data['status'] = 'info';
}
header('Content-type: application/json');
echo json_encode($data);
The json_encode() method converts the array to a JSON object so you can access each array key by name on the js side.
来源:https://stackoverflow.com/questions/12762609/how-can-i-get-a-php-variable-to-ajax