Try to using ajax form instead of form action (solve) - php

Question

I would like to improve user experience at my website. So I try to change the form action ajax, and I has been try some tutorial but I still getting stuck.
I am using a php forum program/source code call !Discuz and it was from China. Below is my coding now.
In html.
<form method="post" id="jnfarm_pop" action="plugin.php?id=cc&do=shop">
<input type="hidden" name="shopsubmit" value="yes">
<!--first item-->
<input type="checkbox" name="jsid[1]" value="1">
<input type="number" style="width:3em;" name="qty[1]">
<!--second item-->
<input type="checkbox" name="jsid[2]" value="1">
<input type="number" style="width:3em;" name="qty[2]">
...continue 50 item
<button type="submit" class="layui-btn layui-btn-fluid" name="submitbutn">submit</button>
</form>
in PHP, file name plugin.php
<?php
if($_GET['id'] == 'cc'){
if(submitcheck('shopsubmit')){ //core function in !Discuz
for($x=1;$x<=50;$x++){
if($_GET['jsid'][$x] == '1'){
$qty[$x] = intval($_GET['qty'][$x]);
//process....
}
}
showmessage('message here','redirectlink');//this is !Discuz program function and it is fine.
}
}
?>
The above script is working fine while using form action, and redirect to my output page. If I would like to change to ajax, how do I adjust the below source code?
<script type="text/javascript">
function login() {
$.ajax({
type: "POST",
dataType: "json",//? is it can use json? since my form data can get as array
url: "plugin.php?id=cc&do=shop" ,//url
data: $('#jnfarm_pop').serialize(),
success: function (result) {
console.log(result);
if (result.resultCode == 200) {
alert("SUCCESS");
}
;
},
error : function() {
alert("ERROR");
}
});
}
</script>
<form method="post" id="jnfarm_pop" action="plugin.php?id=cc&do=shop">
<input type="hidden" name="shopsubmit" value="yes">
<!--first item-->
<input type="checkbox" name="jsid[1]" value="1">
<input type="number" style="width:3em;" name="qty[1]">
<!--second item-->
<input type="checkbox" name="jsid[2]" value="1">
<input type="number" style="width:3em;" name="qty[2]">
...continue 50 item
<button type="submit" class="layui-btn layui-btn-fluid" name="submitbutn" onclick="login()">submit</button>
</form>
And is it have to adjust the plugin.php source code?
Updated, below is work for me, thanks fayis003.
html change the <script></script>
$.get('plugin.php?id=cc&do=shop', $('#jnfarm_pop').serialize(), result => {
//alert('success');
console.log(result); // check the result in console and if you can see it as a JS object you don't need to parse
result = JSON.parse(result); // Parse is required if you return the result as plain text otherwise you can omit this step in case you are returning the result as content type json
alert(result.final);//alert message here
location.href = result.link;// if you need to get redirected
}).fail(result => {
alert('fail');
});
PHP
<?php
if($_GET['id'] == 'cc'){
if(submitcheck('shopsubmit')){ //core function in !Discuz
for($x=1;$x<=50;$x++){
if($_GET['jsid'][$x] == '1'){
$qty[$x] = intval($_GET['qty'][$x]);
//process....
}
}
$final = 'message here';
echo json_encode(['final' => $final]);
}
}
?>

Answer 1

You can not initiate a direct browser redirect using server-side code on ajax request like you do with synchronous requests. instead, you have to return a URL to which you want to get redirected to and then do something like location.href = result.link in the result callback.
for ajax request, the simplest option is using as follows
$.get('plugin.php?id=cc&do=shop', $('#jnfarm_pop').serialize(), result => {
//alert('success');
console.log(result); // check the result in console and if you can see it as a JS object you don't need to parse
result = JSON.parse(result); // Parse is required if you return the result as plain text otherwise you can omit this step in case you are returning the result as content type json
let final = result.final;
location.href = result.link;// if you need to get redirected
}).fail(result => {
alert('fail');
});
now in the server-side code instead of creating a redirect from PHP return something like
return json_encode(['link' => 'somlink']);
of just return success message as usual.