如何获得角函数外的变量值 [英] How to get variable value outside angular function

问题描述

我已经花了数小时的时间去了解为什么我不能在Angular.js函数之外获得变量的值。首先我从Firebase数据库中获取值，这里是我的参考：

$ p $ var $ refspaedtSe​​rv = new Firebase（FBURLSPA + $ routeParams。 ID）;
$ scope.spaedServ = $ firebaseObject（refspaedtSe​​rv）;

然后我有函数：

  /////图片参考//// 
 var lafoto =''; 
 refspaedtSe​​rv.on（value，function（rootSnapshot）{
 lafoto = rootSnapshot.val（）。foto; 
 console.log（Inside image，lafoto）
 }）; 
  

正如您所看到的，我将变量'lafoto'定义为全局
使用控制台.log里面的图像，我可以看到值是正确的

但是当我试图获得lafoto变量的值，在功能外，我得到

  console.log（Outside Image，lafoto）
  

这看起来很愚蠢，但是我正为此疯狂。
有人可以给我一个提示吗？

问候，
Victor

解决方案

非常简单，让我们看看你的代码，假设你的函数在'value'事件中需要1500ms完成：

  var lafoto =''; 
 console.log（'start'）; 
 
 //监听事件
 refspaedtSe​​rv.on（value，function（rootSnapshot）{
 // after〜1500ms ... 
 lafoto = rootSnapshot .val（）。foto; 
 console.log（Inside，lafoto）
}）; 
 
 console.log（'outside'，lafoto）; //立即执行
  

在控制台中，您将收到结果：

 '开始'
'在'$ b $'之外'
  

这是因为你在等待异步事件：因此，外的.on函数你执行代码之前

所以，您的变量lafoto总是会导致undefined，因为它还没有被分配。

编辑1

您可以使用回调函数在.on事件之后执行代码：

  var lafoto =''; 
 console.log（'start'）; 
 
 //监听事件
 refspaedtSe​​rv.on（value，function（rootSnapshot）{
 // after〜1500ms ... 
 lafoto = rootSnapshot .val（）。foto; 
 console.log（Inside，lafoto）
 myAfterFunction（）; 
}）; 
 
函数myAfterFunction（）{
 console.log（'myAfterFunction'，lafoto）; 
} 
 
 console.log（'outside'，lafoto）; //立即执行
  

在控制台中，您将收到结果：

 '开始'
'外部'$ b $'内'
'myAfterFunction'
  

I have been hours trying to understand why I don't get the value of a variable, outside an Angular.js function. First I get the value from a Firebase database, here is my reference :

var refspaedtServ = new Firebase(FBURLSPA + $routeParams.id);
$scope.spaedServ = $firebaseObject(refspaedtServ);

Then I have the function:

        /////  IMAGE REFERENCE  ////
        var lafoto = ''; 
        refspaedtServ.on("value", function(rootSnapshot) {
           lafoto = rootSnapshot.val().foto;
           console.log("Inside image ", lafoto)
       });

As you can see, I define my variable 'lafoto' as global With the console.log Inside image, I can see the value is correct

But when I try to get the value of "lafoto" variable, outside the function, I'm getting "undefined", I mean no value.

console.log("Outside Image ", lafoto)

It seems silly, but I'm reaching madness for that. Can anybody give me a hint please?

Regards, Victor

解决方案

It's pretty simple, let's see it by your code, assuming your function in the 'value' event needs 1500ms to complete:

var lafoto = '';
console.log('start');

//listening for the event
refspaedtServ.on("value", function(rootSnapshot) {
    //after ~1500ms...
    lafoto = rootSnapshot.val().foto;
    console.log("Inside", lafoto)
});

console.log('outside', lafoto); //executed immediately

In console, you will receive as result:

'Start'
'outside'
'inside'

that's because you are waiting asynchronously for the event: hence, "outside" the .on function you are executing the code before the code inside.

So, your variable lafoto will always result undefined, because it has not been already assigned.

.

Edit 1

You can use a callback function to perform code after the .on event:

 var lafoto = '';
console.log('start');

//listening for the event
refspaedtServ.on("value", function(rootSnapshot) {
    //after ~1500ms...
    lafoto = rootSnapshot.val().foto;
    console.log("Inside", lafoto)
    myAfterFunction();
});

function myAfterFunction(){
    console.log('myAfterFunction', lafoto);
}

console.log('outside', lafoto); //executed immediately

In console, you will receive as result:

'Start'
'outside'
'inside'
'myAfterFunction'

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