javascript - Ajax 表单提交给两个脚本

Question

基本上我有一个表单，我想将变量传递给两个单独的 php 脚本以便在提交时进行处理。我所做的工作对于我的目的来说似乎工作正常，我只是想知道是否有一种方法可以在一个 jquery 脚本中执行此操作，而不必像我所做的那样重复脚本和更改 URL。还是 iv 的方式很好？在谷歌上找不到很多这种用法的例子，但也许我没有在搜索正确的东西。此外，如果我正在做一些不好的做法，请告诉我，我是 Ajax 的新手，我遵循的每个教程似乎都以不同的方式做事。

<script>
$(document).ready(function(){
    $('#sform').submit(function(){

        // show that something is loading
        $('#response').html("<b>Loading response...</b>");

        /*
         * 'post_receiver.php' - where you will pass the form data
         * $(this).serialize() - to easily read form data
         * function(data){... - data contains the response from post_receiver.php
         */
        $.ajax({
            type: 'POST',
            url: 'move_to_lease.php', 
            data: $(this).serialize()
        })



        .done(function(data){

            // show the response
            $('#response').html(data);

        })



        .fail(function() {

            // just in case posting your form failed
            alert( "Posting failed." );

        });

        // to prevent refreshing the whole page page
        return false;

    });
});
</script>

<script>
$(document).ready(function(){
    $('#sform').submit(function(){

        // show that something is loading
        $('#txtHint').html("<b>Loading response...</b>");

        /*
         * 'post_receiver.php' - where you will pass the form data
         * $(this).serialize() - to easily read form data
         * function(data){... - data contains the response from post_receiver.php
         */
        $.ajax({
            type: 'POST',
            url: 'show_lease_sub.php', 
            data: $(this).serialize()
        })
        .done(function(data){

            // show the response
            $('#txtHint').html(data);

        })
        .fail(function() {

            // just in case posting your form failed
            alert( "Posting failed." );

        });

        // to prevent refreshing the whole page page
        return false;

    });
});

</script>

Answer 1

最佳做法是使用一个后端脚本处理这两个调用。您可以使用数组返回 2 个单独的响应。 Protip - 写一个类。

ajax.php

class leaseStuff{
    public static function moveToLease($data){
        // add code from move_to_lease.php, change $_POST to $data
    }
    public static function showLeaseSub($data){
        // add code from show_lease_sub.php, change $_POST to $data
    }
}

echo json_encode(array(
    "moveToLease"=>leaseStuff::moveToLease($_POST),
    "showLeaseSub"=>leaseStuff::showLeaseSub($_POST)
));

然后您可以将您的 Ajax 合并到一个调用中:

$.ajax({
    type: 'POST',
    url: 'ajax.php', 
    data: $(this).serialize()
}).done(function(data){
    data = JSON.parse(data);
    $('#response').html(data.moveToLease);
    $('#txtHint').html(data.showLeaseSub);
});