尝试使用ajax表单而不是表单操作(解决)

Question

我想改善我的网站的用户体验。因此，我试图改变表单动作ajax，我已经尝试了一些教程，但我仍然被困住了。

我正在使用一个php论坛程序/源代码调用!Discuz，它来自中国。下面是我的编码。

在html中。

<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>

在PHP中，文件名为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.
  }
}
?>

以上脚本在使用form action时运行良好，并重定向到我的输出页面。如果我想要更改为ajax，如何调整下面的源代码？

<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>

它必须调整plugin.php源代码吗？

更新，下面是我的工作，谢谢fayis003。html更改<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

不能像处理同步请求那样，在ajax请求上使用服务器端代码启动直接浏览器重定向。相反，您必须返回要重定向到的URL，然后在结果回调中执行类似location.href = result.link的操作。

对于ajax请求，最简单的选项是如下所示

$.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');
});

现在，在服务器端代码中，不要创建来自PHP的重定向，而是返回以下内容

return json_encode(['link' => 'somlink']);

就像往常一样返回成功信息。