如何使用Javascript逐个加载URL

Question

如何通过javascript ajax或jquery从数组中逐个触摸url？因为如果你一触摸触摸大进程的php就会超时，那么如何逐个处理？

实施例

select group_concat(concat('date:',DATE,':',RESOURCE_NAME,':',MARKS)
 separator ' , ' )from mytable group by DATE order by DATE desc;

可能是如果grape.php完成，然后在苹果旁边，如果苹果完成，然后是橙色。

然后如果所有过程都完成，则显示警报成功。

Answer 1

你是说这个？

var myurls = [
"http://example.com/grape.php",
"http://example.com/apple.php",
"http://example.com/orange.php",
"http://example.com/banana.php"],cnt=0;
function process(data) {
  console.log(data);
}
function loadUrl() {
  if (cnt>=myurls.length) return;
  $.get(myurls[cnt++],function(data) {
    process(data);
    loadUrl();
  });
}
$(function() {
  loadUrl();
})

Answer 2

根据您的问题以及我们在您的评论中进行的讨论，我认为您希望根据您拥有的一系列网址执行AJAX调用顺序。

解决方案1：使用重复的自引用$.ajax()请求

这无疑是更容易解决的问题。我们在这里得到的是我们跟踪我们所在数组的位置，并在迭代数组时停止发出AJAX请求。

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$(function() {

  // Array of URLs
  var myurls = [
    "https://jsonplaceholder.typicode.com/posts/1",
    "https://jsonplaceholder.typicode.com/posts/2",
    "https://jsonplaceholder.typicode.com/posts/3",
    "https://jsonplaceholder.typicode.com/posts/4"
  ];

  // Iterate through array, and keep a cursor on which item we are at
  var urlCount = 0,
    ajaxCall = function() {
      if (urlCount < myurls.length) {
        console.log('Making AJAX call to url: '+myurls[urlCount]);
        $.ajax({
            url: myurls[urlCount]
          })
          .done(function(returnedData) {
            console.log(returnedData);
            urlCount++;
            ajaxCall();
          });
      }
    };

  ajaxCall();

});

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

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解决方案2：使用for循环

使用.then()链接设置

对于您的用例，我有adapted the code provided in this answer，因为对于那些不熟悉延迟对象和返回承诺的人来说，这个例子很难理解。

所以诀窍如下：

1. 设置主延期对象
2. 设置一个将为您调用AJAX的常规函数​​
3. 循环遍历数组，使用.then()
链接它们
4. 在主延迟对象上启动第一个AJAX调用
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$(function() {

  // Array of URLs
  var myurls = [
    "https://jsonplaceholder.typicode.com/posts/1",
    "https://jsonplaceholder.typicode.com/posts/2",
    "https://jsonplaceholder.typicode.com/posts/3",
    "https://jsonplaceholder.typicode.com/posts/4"
  ];

  // Set up chain of AJAX requests
  var d = $.Deferred(),
    _d = d,
    ajaxRequest = function(ajaxUrl) {
      // Log in browser console that AJAX call is being made
      console.log('Making AJAX call to: ' + ajaxUrl);

      // Return deferred object for .then() chaining
      return $.ajax({
        url: ajaxUrl
      });
    };

  // We chain each AJAX call to the next one
  for (var i in myurls) {
    // Use IIFE so that reference to `i` is fixed
    (function(j) {
      // Update _d for chaining
      _d = _d.then(function() {
        // The _request is a defered object returned
        // So we can chain deferred methods such as .done()
        var _request = ajaxRequest(myurls[j]).done(function(ajaxData) {
          // Just to show that data is being returned after each call
          console.log(ajaxData);
        });
        
        // Return the deferred object for chaining
        return _request;
      });
    })(i);
  }

  // Kick start sequential ajax call
  d.resolve();
});
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<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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