异步递归函数结束后的回调

时间:2011-02-19 04:34:43

标签: javascript asynchronous recursion callback

以下功能以递归方式在文件夹中打印Chrome书签。在处理完最终递归循环后,如何更改下面的函数来调用另一个函数? chrome.bookmarks.getChildren()是异步的,这使得很难知道函数何时完成处理所有内容。

感谢。

    for (var i = 0; i < foldersArray.length; i++) {
        // The loop makes several calls with different folder IDs.
        printBookmarks(foldersArray[i]);  
    }

    // I'd like any code here to be run only after the above has 
    //finished processing    

    function printBookmarks(id) {
        chrome.bookmarks.getChildren(id, function(children) {
           children.forEach(function(bookmark) { 
               console.debug(bookmark.title);
               printBookmarks(bookmark.id);
           });
        });
    }
编辑:对不起,我不认为我在初始代码示例中已经清楚了。我已经更新了代码,通过多次调用函数来显示我对异步函数的问题。我想在printBookmarks函数调用之后的任何代码等待所有printBookmarks函数完成处理。

5 个答案:

答案 0 :(得分:26)

你的异步方法实例可能都在同时执行,你不知道预先有多少个。因此,您必须保持计数,然后在最后一个异步方法完成时使用回调。

for (var i = 0; i < foldersArray.length; i++) {
    // The loop makes several calls with different folder IDs.
    printBookmarks(foldersArray[i], thingsToDoAfter);
}

function thingsToDoAfter() {
    // I'd like any code here to be run only after the above has 
    // finished processing.
}

var count = 0;

function printBookmarks(id, callback) {
    count++;
    chrome.bookmarks.getChildren(id, function(children) {
        children.forEach(function(bookmark) { 
            console.debug(bookmark.title);
            printBookmarks(bookmark.id, callback);
        });
        count--;
        if (count === 0 && callback)
            callback();
    });
}

答案 1 :(得分:2)

我最近不得不解决这个问题。解决方案类似于Eric的,但我发现我需要count变量是函数的本地变量。以下是我将如何解决这个问题:

for(var i=0;i<foldersArray.length; i++){
  // The loop make's several call's with different folder ID's.
  var printed_children = 0;
  printBookmarks(foldersArray[i],function() {
    printed_children++;
    if(printed_children == foldersArray.length){
      // You know you are done!
    }
  });  
}
// I'd like any code here to be run only after the above has 
//finished processing.


function printBookmarks(id,finished_callback) {
  // the printed_children count is local so that it can keep track of 
  // whether or not this level of recursion is complete and should call 
  // back to the previous level
  var printed_children = 0;
  chrome.bookmarks.getChildren(id, function(children) {
    children.forEach(function(bookmark) { 
      console.debug(bookmark.title);
      // added a callback function to the printBookmarks so that it can
      // check to see if all upstream recursions have completed.
      printBookmarks(bookmark.id,function() {
        printed_children++;
        if(printed_children == children.length){
          finished_callback();
        }
      });
    });
    if(children.length == 0){
      finished_callback();
    }
  });
}

这有点难看,但它应该有用。

答案 2 :(得分:0)

您可以将所有已完成的调用保存到变量中,并根据要处理的书签数量进行测试。到达结尾时(完成次数等于要处理的书签数量),然后执行最终功能。

这里有类似问题的答案,您可以使用代码作为指南:

How do I load a variable number of scripts with jQuery.getScript() before running javascript code?

答案 3 :(得分:0)

你可以做这样的事情JQFAQ.com。我正在更新以备将来使用。

答案 4 :(得分:-1)

可能是一个更好的方法,但你可以添加一个深度参数,如

printBookmarks('0', 0);

function printBookmarks(id, depth) {
    chrome.bookmarks.getChildren(id, function(children) {
        children.forEach(function(bookmark) { 
            console.debug(bookmark.title);
            printBookmarks(bookmark.id, depth + 1);
        });
        if(depth == 0) yourFunction();
    });     
}

编辑以回应评论

对于略有不同的方法,这是另一个答案的变体。

runCount = 0;
for (var i = 0; i < foldersArray.length; i++) {
    // The loop makes several calls with different folder IDs.
    printBookmarks(foldersArray[i]);  
    runCount++;
}

while(runCount > 0) { // sleep for 10ms or whatnot}
// I'd like any code here to be run only after the above has 
// finished processing.    

function printBookmarks(id) {
    chrome.bookmarks.getChildren(id, function(children) {
        children.forEach(function(bookmark) { 
            console.debug(bookmark.title);
            printBookmarks(bookmark.id);
            runCount--;
        });
    });
}