Question

如果我有两个从ajax调用的函数我想传递第一个函数返回到另一个变量的东西，并且只使用它一次只有一次

使用returnResult（）函数$ .each（）的代码bellow循环并且一些json数据丢失，就像“labe”中没有文字一样：{}

 $.ajax({
          type: "POST",
          url: "WebService.asmx/MyEventData",
          data: "{}",
          dataType: "json",
          contentType: "application/json; charset=utf-8",
          success: function(data) {
          returnResult(data.d);  
         }

      });

       $.ajax({
          type: "POST",
          url: "WebService.asmx/Getlines",
          data: "{}",
          dataType: "json",
          contentType: "application/json; charset=utf-8",
          success: function(data) {
              lineReturn(data.d);
          }

      });



      });

      function lineReturn(result) {
      var jsonObj = [] ; 
      $.each(result, function(index, value) {
         jsonObj.push({ "value": value.dateTime, "color": '#D2691E', "width": 1,    "label": { "text": value.LineEvent} });
      });
      return JSON.stringify(jsonObj) ; << this is the return that should go to myline variable and be used on the highchart >>>> 

  }

      function returnResult(result) {
      console.log(result) ;
      var myLine = lineReturn(result) ;
      var datetemp = new Array ;
      var dateflow = new Array ;
      var datepressure = new Array;
      var datecond = new Array ;

Answer 1

尝试将第二个ajax调用放在第一个回调中。

$.ajax({
      type: "POST",
      url: "WebService.asmx/Getlines",
      data: "{}",
      dataType: "json",
      contentType: "application/json; charset=utf-8",
      success: function(dataA) {
           $.ajax({
               type: "POST",
               url: "WebService.asmx/MyEventData",
               data: "{}",
               dataType: "json",
               contentType: "application/json; charset=utf-8",
               success: function(dataB) {
                   // From here you can access both dataA.d and dataB.d
                   var myLineA = lineReturn(dataA.d);
                   var myLineB = lineReturn(dataB.d);
              }
           });
      }
});

Answer 2

您可以在第一个函数的.success上使用第二个函数。

在函数之间传递变量

2 个答案: