[英]Get the name of variable function was assigned to
I'm trying to return the name of the variable that a function is assigned to. 我正在尝试返回分配函数的变量的名称。
I have included an example below. 我在下面列举了一个例子。 The end result is I would like modelPerson.title()
to return the variable name title
. 最终结果是我希望modelPerson.title()
返回变量名称title
。
For example, I have the following code: 例如,我有以下代码:
Defining some base model types 定义一些基本模型类型
var types = {
string: function() {
return function() {
return "I want this to return 'title'";
}
}
};
Using the model types 使用模型类型
var modelPerson = {
title: types.string(),
firstName: types.string(),
surname: types.string(),
position: types.string()
};
Trying to return the title 试图返回标题
console.log(modelPerson.title());
Sorry if this is a little unclear. 对不起,如果这有点不清楚。 I have included a JSFiddle if it helps: http://jsfiddle.net/4f6VE/ 我有一个JSFiddle,如果它有帮助: http : //jsfiddle.net/4f6VE/
Thanks for any help you can give 谢谢你提供的所有帮助
That's actually possible, but involves some v8 specific stuff: 这实际上是可能的,但涉及一些v8特定的东西:
var types = {
string: function() {
return function() {
var obj = {};
var prepare = Error.prepareStackTrace;
Error.prepareStackTrace = function (_, stack) {
return stack
}
Error.captureStackTrace(obj)
var method = obj.stack[0].getMethodName();
Error.prepareStackTrace = prepare;
return method;
}
}
};
var modelPerson = {
title: types.string(),
firstName: types.string(),
surname: types.string(),
position: types.string()
};
console.log(modelPerson.title());
console.log(modelPerson.firstName());
but you probably should use something less insane 但你可能应该使用一些不那么疯狂的东西
I don't really know what is this for, but 我真的不知道这是为了什么,但是
var modelPerson = {
title : function title(){ return arguments.callee.name; },
firstName : function firstName(){ return arguments.callee.name; },
surname : function surname(){ return arguments.callee.name; },
position : function position(){ return arguments.callee.name; },
}
should do what you say. 应该做你说的。
EDIT 编辑
Banzaaai~ ! Banzaaai~!
var types = {
string: function(){
eval('var x = function '+arguments.callee.caller.name+'(){var x = function(){return arguments.callee.caller.name;}; return x();}');
return x();
}
};
var modelPerson = {
title: function title(){ return types.string(); },
firstName: function firstName(){ return types.string(); },
surname: function surname(){ return types.string(); },
position: function position(){ return types.string(); }
};
SRSLY THOUGH 很幸运的
var types = {
string: function(x){
return function(){ return x; }
}
};
var modelPerson = {
title: types.string('title'),
firstName: types.string('firstName'),
surname: types.string('surname'),
position: types.string('position')
};
I'm trying to return the name of the variable that a function is assigned to 我正在尝试返回分配函数的变量的名称
You can't, not reliably. 你不能,不可靠。 Several variables or properties can reference the same object, and some objects exists without ever being assigned to a variable (such as function expressions without a name that are called immediately). 多个变量或属性可以引用同一个对象,并且某些对象不会被分配给变量(例如没有立即调用的名称的函数表达式)。
The end result is I would like modelPerson.title() to return the variable name title. 最终结果是我希望modelPerson.title()返回变量名称标题。
Then use something like this: 然后使用这样的东西:
function define(obj, name) {
obj[name] = function() {
return name;
};
}
var modelPerson = {};
define(modelPerson, "title");
define(modelPerson, "firstName");
define(modelPerson, "surname");
define(modelPerson, "position");
// … - a loop maybe?
> console.log(modelPerson.title());
"title"
here is a method that can work in strict mode (without deprecated arguments.callee or proprietary arguments.callee.caller properties), using your code with minimal re-factoring and no hard-coded names: 这是一个可以在严格模式下工作的方法(没有弃用的arguments.callee或专有的arguments.callee.caller属性),使用您的代码进行最少的重新分解并且没有硬编码的名称:
var types={
string: function types(){
return function me() {
for(var it in this){
if(me==this[it]) return it;
}
};
}
};
var modelPerson = {
title: types.string(),
firstName: types.string(),
surname: types.string(),
position: types.string()
};
alert( modelPerson.title() ); // shows: "title"
alert( modelPerson.surname() ); // shows: "surname"
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