从jquery get方法中传递数据 [英] pass data out from jquery get method

问题描述

  var transactionID; 
 $ .get（'http://127.0.0.1/getId'，null，function（data）{
 transactionID = data; 
 alert（transactionID）; 
}） ; 
 alert（transactionID）; 
  

get方法中的警报正确返回值。但是，get方法之外的第二个alert中的transactionID仍然为空时？如何从get方法中正确地传递返回数据？

解决方案 $。get

code>是异步执行的，即它将请求发送到服务器并执行后面的行。
因此，外部警报将在服务器返回答复之前执行，并且当它执行时，它将执行 get 方法中的函数。

为了使用返回的值，将所有需要的代码放在 get 调用中，例如调用另一个函数，将transactionID作为参数传递给一个参数：

  var transactionID; 
 $ .get（'http://127.0.0.1/getId'，null，function（data）{
 transactionID = data; 
 alert（transactionID）; 
 doSomethingWithID（ transactionID）; 
}）; 
  

或者只是调用一个函数来使用它：

  $ .get（'http://127.0.0.1/getId'，null，function（data）{
 transactionID = data; 
 alert（transactionID ）; 
 doSomethingWithID（）; 
}）; 
函数doSomethingWithID（）
 {
 //使用transationID的代码
} 
  

var transactionID;
$.get('http://127.0.0.1/getId', null,function(data) {
    transactionID = data;
    alert(transactionID);
});
alert(transactionID);

The alert inside the get method returns value correctly. However, when the transactionID in the 2nd alert outside of the get method is still null? How to correctly pass out the return data from the get method?

解决方案

The $.get is executed asynchronously, i.e. it sends the request to the server and executes the lines after it. So the alert outside will be executed before the server returns the reply, and when it does it will execute the function in the get method.

In order to use the returned value, place any needed code inside the get call, for example call another function passing it the transactionID as an argument:

var transactionID;
$.get('http://127.0.0.1/getId', null,function(data) {
    transactionID = data;
    alert(transactionID);
    doSomethingWithID(transactionID);
});

Or just call a function the uses it:

$.get('http://127.0.0.1/getId', null,function(data) {
    transactionID = data;
    alert(transactionID);
    doSomethingWithID();
});
function doSomethingWithID()
{
  // code that uses transationID
}

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