2016-11-24 81 views
0

我试图使用JSON.parse数据:[{“id”:“ABC123”,“provider_type”:“Center”,“name” :“Test1”,“phone”:“0322229999”,“mobile”:“1111123123”,“email”:“[email protected]”,“address”:[“3999,Victoria”,“Est Danvale”, “113 HD streed”, “块d,K屋”]}]AngularJS:JSON.parse(data).success is not defined

和代码:

var callBackProviderSuccess = function(data){ 
     var providerObj = JSON.parse(data); 
     if(providerObj.success) { 
      $scope.providerInfo = providerObj.provider; 
      $localStorage.providerJPayInfos = $scope.providerInfo;    IBMService.getStaffsOfProvider(providerID).then(callBackStaffsSuccess,callBackStaffsError); 

     } else { 
      $scope.infoLoadingText = "Provider not found on our database"; 
      //try again 
      IBMService.getProviderById(providerID).then(callBackProviderSuccess,callBackProviderError); 
     } 
    }; 

调试时,如果(providerObj.success),我有一个消息:providerObj.success undefined。请帮我提出你的想法。 谢谢

回答

2

是的,JSON内没有success属性。

更改它,

if(providerObj) { 
+0

哇,太感谢你了 –

+0

@PhạmQuốcBảo标记为答案,如果它帮助 – Sajeetharan

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