Parse.com如何获取查询的返回值 [英] Parse.com how to get return value of query

问题描述

在Parse.Com中，我通常会在函数中获得查询的值。如何将它放在函数的一边？

  var TableObject = Parse.Object.extend（TableId）; 
 var qyNewParse = new Parse.Query（TableObject）; 
 qyNewParse.doesNotExist（last_sync）; 
 qyNewParse.find（{
成功：函数（结果）{
 
 
}，
错误：函数（错误）{
 
} 
}）; 
 var something = results.length 
  

例如，如果我想得到result.length at功能之外。如何获取 var something 值？

解决方案

你不能有一个函数返回结果查询，因为函数将始终在查询完成之前返回。您应该为函数提供自己的回调参数，而不是让函数返回值，该参数将在查询完成时调用，如：

  function some_cback（callback）{
 var TableObject = Parse.Object.extend（TableId）; 
 var qyNewParse = new Parse.Query（TableObject）; 
 qyNewParse.doesNotExist（last_sync）; 
 qyNewParse.find（{
 success：function（results）{
 callback（results）; // call callback 
}，
 error：function（error） {
} 
}）; 
} 
 //调用函数
 var something = some_cback（function（response）{
 console.log（response）; 
}）; 
  

in Parse.Com i usually get value of query in the function. How to get it out side of the function?

var TableObject = Parse.Object.extend(TableId);
var qyNewParse = new Parse.Query(TableObject);    
qyNewParse.doesNotExist("last_sync");
qyNewParse.find({
     success: function(results) {


     },
     error: function(error){

     }
});
var something = results.length

for example if i want to get result.length at outside of function. How to get var something value?

解决方案

You can't have a function return the result of a query, because the function will always return before the query is completed. Instead of having your function return a value, you should give the function its own callback parameter that will be called when the query completes, like:

function some_cback(callback) {
    var TableObject = Parse.Object.extend(TableId);
    var qyNewParse = new Parse.Query(TableObject);    
    qyNewParse.doesNotExist("last_sync");
    qyNewParse.find({
         success: function(results) {
             callback(results); //call the callback
         },
        error: function(error){
        }
    });
}
//call the function
var something = some_cback(function(response) {
    console.log(response);
});

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