jQuery Ajax 等待每个函数

Question

$xy('#simpan').click(function() {

  $xy('input[id="cekbok[]"]:checked').each(function() {

    var data = (this.value);
    var div = (this.value);
    var str = window.location.href;
    var res = str.replace("wp-admin/options-general.php?page=katalogu-options", "/wp-content/plugins/katalog/includes/img/loading.gif");
    var loading = ('<img src="'+res+'">') ;


    $xy.ajax({
        type : 'POST',
        url : '../wp-content/plugins/katalogunique/proses2.php',           
        data: {
            id  : (this.value)
        },

        success:function (data) {
            $xy('#result'+div).empty();
            $xy('#result'+div).append(data);
            $xy('#tz'+div).remove();
        }          

    });  

  });
});

我的函数在循环中将复选框值发送到 proses2.php，但是当我运行这个脚本时，它将运行所有 ajax POST 立即调用.我想要运行 ajax 请求 1 到 1 或等到完成，我该如何解决这个问题？

Answer 1

您可以使用这样的递归函数一个接一个地执行所有调用：

$xy('#simpan').click(function() {
    var str = window.location.href;
    var i=0; things = $xy('input[id="cekbok[]"]:checked');
    (function doOneTask(){
        if (i++>=things.length) return;
        var thing = things[i];
        var data = thing.value;
        var div = thing.value;
        var res = str.replace("wp-admin/options-general.php?page=katalogu-options", "/wp-content/plugins/katalog/includes/img/loading.gif");
        var loading = ('<img src="'+res+'">') ;
        $xy.ajax({
            type : 'POST',
            url : '../wp-content/plugins/katalogunique/proses2.php',           
            data: {
                id  : thing.value
            },
            success:function (data) {
                $xy('#result'+div).empty();
                $xy('#result'+div).append(data);
                $xy('#tz'+div).remove();
            }          
        }).always(doOneTask);
    })();
});

注意：如果您想在第一次失败时停止，而不是像我一样进行下一次调用，请将 always 替换为 done。

Answer 2

一种方法是使用递归函数，每次都将列表（减去第一个条目）传回。

例如调用代码简化为：

$xy('#simpan').click(function() {
    LoadAjax($xy('input[id="cekbok[]"]:checked'));

});

并且 LoadAjax 是这样递归的：

function LoadAjax(elements) {
    // Exit when done
    if (!elements.length) return;
    var $this = elements.first();
    var data = ($this.value);
    var div = ($this.value);
    var str = window.location.href;
    var res = str.replace("wp-admin/options-general.php?page=katalogu-options", "/wp-content/plugins/katalog/includes/img/loading.gif");
    var loading = ('<img src="' + res + '">');

    $xy.ajax({
        type: 'POST',
        url: '../wp-content/plugins/katalogunique/proses2.php',
        data: {
            id: ($this.value)
        },

        success: function (data) {
            $xy('#result' + div).empty();
            $xy('#result' + div).append(data);
            $xy('#tz' + div).remove();
            // Go recursive with the rest of the list
            LoadAjax(elements.slice(1));
        }

    });
}

Answer 3

这里有一种没有递归并使用简单循环的方法：

$xy('#simpan').click(function() {

    var url = '../wp-content/plugins/katalogunique/proses2.php';
    var d = $.Deferred().resolve(); // empty promise
    $xy('input[id="cekbok[]"]:checked').each(function() {
        var div = this.value;
        d = d.then(function(data){
            $xy('#result'+div).empty().append(data);
            $xy('#tz'+div).remove();
            return $xy.post(url, {id: div}); // this will make the next request wait
        });
    });
    // can d.then here for knowing when all of the requests are done.
});

请注意，我可以 "clever this up" 使用 .reduce 将行数从 6 缩短到 4，但老实说，我宁愿保持循环结构让 OP 感到舒服。这是有效的，因为承诺链接 - 基本上当你 return 来自 then 的动作时，它会在执行下一个 then 之前等待它，你将它链接到。

让我们举个例子：

function log(msg){ // simple function to log stuff
    document.body.innerHTML += msg + "<br />";
}

var delay = function(ms){ // this is an async request, simulating your AJAX
   var d = $.Deferred();
   setTimeout(d.resolve, ms); // resolving a deferred makes its then handlers execute
   return d;
};

// $.Deferred.resolve() starts a new chain so handlers execute
var p = $.Deferred().resolve().then(function(){ 
    log("1");
    return delay(1000); // like in your code, we're waiting for it when we chain
}).then(function(){ // in the above code, this is the d = d.then part,
    log("2"); // this only runs when the above delay completes
    return delay(1000);
});

// and more like in the above example, we can chain it with a loop:
[3,4,5,6,7,8,9,10].forEach(function(i){
     p = p.then(function(){
         log(i);
         return delay(1000);
     });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

请注意，执行 $xy('#result'+div) 可能不是一个好主意，因为您正在查询视图层以查找放置在那里的内容 - 考虑将相关的 div 保留为数组并保留引用:)

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