function main()
{
   Hello();
}

function Hello()
{
  // How do you find out the caller function is 'main'?
}

有办法找到调用堆栈吗?


当前回答

使用*arguments.callee更安全。调用自参数。不赞成调用…

其他回答

你可以使用函数。调用方获取调用函数。旧的方法使用参数。Caller被认为是过时的。

下面的代码说明了它的用法:

function Hello() { return Hello.caller;}

Hello2 = function NamedFunc() { return NamedFunc.caller; };

function main()
{
   Hello();  //both return main()
   Hello2();
}

关于废参数的注释。打电话者:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments/caller

功能。呼叫方为非标准:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/caller

在这里,除了函数名之外的所有内容都被RegExp从caller.toString()中剥离。

<!DOCTYPE html>
<meta charset="UTF-8">
<title>Show the callers name</title><!-- This validates as html5! -->
<script>
main();
function main() { Hello(); }
function Hello(){
  var name = Hello.caller.toString().replace(/\s\([^#]+$|^[^\s]+\s/g,'');
  name = name.replace(/\s/g,'');
  if ( typeof window[name] !== 'function' )
    alert ("sorry, the type of "+name+" is "+ typeof window[name]);
  else
    alert ("The name of the "+typeof window[name]+" that called is "+name);
}
</script>

看起来这是一个相当解决的问题,但我最近发现,callee是不允许在“严格模式”,所以为了我自己的使用,我写了一个类,将从它被调用的路径。它是一个小型助手库的一部分,如果你想单独使用代码,请更改用于返回调用者堆栈跟踪的偏移量(使用1而不是2)。

function ScriptPath() {
  var scriptPath = '';
  try {
    //Throw an error to generate a stack trace
    throw new Error();
  }
  catch(e) {
    //Split the stack trace into each line
    var stackLines = e.stack.split('\n');
    var callerIndex = 0;
    //Now walk though each line until we find a path reference
    for(var i in stackLines){
      if(!stackLines[i].match(/http[s]?:\/\//)) continue;
      //We skipped all the lines with out an http so we now have a script reference
      //This one is the class constructor, the next is the getScriptPath() call
      //The one after that is the user code requesting the path info (so offset by 2)
      callerIndex = Number(i) + 2;
      break;
    }
    //Now parse the string for each section we want to return
    pathParts = stackLines[callerIndex].match(/((http[s]?:\/\/.+\/)([^\/]+\.js)):/);
  }

  this.fullPath = function() {
    return pathParts[1];
  };

  this.path = function() {
    return pathParts[2];
  };

  this.file = function() {
    return pathParts[3];
  };

  this.fileNoExt = function() {
    var parts = this.file().split('.');
    parts.length = parts.length != 1 ? parts.length - 1 : 1;
    return parts.join('.');
  };
}

我认为下面的代码段可能会有帮助:

window.fnPureLog = function(sStatement, anyVariable) {
    if (arguments.length < 1) { 
        throw new Error('Arguments sStatement and anyVariable are expected'); 
    }
    if (typeof sStatement !== 'string') { 
        throw new Error('The type of sStatement is not match, please use string');
    }
    var oCallStackTrack = new Error();
    console.log(oCallStackTrack.stack.replace('Error', 'Call Stack:'), '\n' + sStatement + ':', anyVariable);
}

执行以下代码:

window.fnPureLog = function(sStatement, anyVariable) {
    if (arguments.length < 1) { 
        throw new Error('Arguments sStatement and anyVariable are expected'); 
    }
    if (typeof sStatement !== 'string') { 
        throw new Error('The type of sStatement is not match, please use string');
    }
    var oCallStackTrack = new Error();
    console.log(oCallStackTrack.stack.replace('Error', 'Call Stack:'), '\n' + sStatement + ':', anyVariable);
}

function fnBsnCallStack1() {
    fnPureLog('Stock Count', 100)
}

function fnBsnCallStack2() {
    fnBsnCallStack1()
}

fnBsnCallStack2();

日志是这样的:

Call Stack:
    at window.fnPureLog (<anonymous>:8:27)
    at fnBsnCallStack1 (<anonymous>:13:5)
    at fnBsnCallStack2 (<anonymous>:17:5)
    at <anonymous>:20:1 
Stock Count: 100

试着访问这个:

arguments.callee.caller.name