I have this javascript in one of my forms. This one is the login form I get the URL from the form itself I'm using API style, ("not sure if I got the concept right") but I will attach the code down
var formup = $('#loginfrom');
formup.submit(function () {
$.ajax({
type: formup.attr('method'),
url: formup.attr('action'),
data: formup.serialize(),
success: function (data) {
alert('you have successfuly logged in');
window.location = "profile.html";
}
});
return false;
});
here is my login function called from api file
if(!mysql_fetch_array($result)){
return $res;
}
else
return $resc;
where $res returns "error", and $resc returns "correct"
and my api.php
if(!isset($function))
$resp['err'] = "Function: ".$action." does not exist";
else
$resp = $function($request);
when I enter the correct user name and password, it works fine and I'm directed to the user profile.
However, I don't know how to handle if I got the user name or password incorrect I don't want to be directed to profile
I do think the return from php should do something to tell the javascript that the login did not work then do something else
please provide me some help regarding this problem, I will appreciate it.
The request is always a success so, you have to check the value of the data
variable to redirect to an error page
...
success: function (data) {
if ( data == 'correct') {
alert('you have successfuly logged in');
window.location = "profile.html";
} else {
// handle error
}
}
...
Change your login code to this
if(!mysql_fetch_array($result)){
echo 'failure'; die;
}
else {
echo 'success'; die;
}
And change your ajax success to this
success: function (data) {
if (data == 'success') {
alert('you have successfuly logged in');
window.location = "profile.html";
} else {
alert('Login Failed');
}
}
Use datatype as json to get the return value from php. Check my suggestions in the below code.
$.ajax({
type: formup.attr('method'),
url: formup.attr('action'),
data: formup.serialize(),
dataType: 'json',
success: function (data) {
if(data.error == 0) {
alert('you have successfuly logged in');
window.location = "profile.html";
} else {
alert("error");
}
}
});
In your php file, convert $res and $resc into an array and then later use json_encode. Like this,
if(!mysql_fetch_array($result)){
echo array('value'=>$res,'error'=>0);
}
else
echo array('value'=>$res,'error'=>1);
You can do something with the data or you can have PHP return a 401 status, if you return something with your PHP instead of a 401 status than you can have your JavaScript check the returned text:
var formup = $('#loginfrom');
formup.submit(function () {
$.ajax({
type: formup.attr('method'),
url: formup.attr('action'),
data: formup.serialize(),
success: function (data) {
console.log(data);
// check your console (press F12 in Chrome or Firefox with firebug plugin)
// do something if the data is valid.
window.location = "profile.html";
}
});
return false;
});
When the login fails in PHP and you want to return 401 (causing your javascript success function not to be called but you might need a fail/error function in JS.
header("HTTP/1.0 401 Unauthorized");
And you need to add a fail callback in your javaScript so the user knows the log in failed:
$.ajax({
...
success: function (data) {
...
},
error: function(){
//inform the user
}
});
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