Question

当前返回所有结果，但是当未找到_miner时，它将在控制台中返回undefined并继续显示时间和块。没有返回_miner时如何不输出？

输出应该是这样的，其中时间“ 3”没有_miner匹配，因此没有任何输出。

$.ajax({
  url: "https://xxxxxxxxxxxx/api/payments",
  dataType: 'json',
  success: function(data) {

    var _miner = "t1XHpNtYY2N3EMDRoX9RM2hq4DWWPZSmawJ";

    var addr = _miner;

    for (var i = 0; i < data.length; i++) {

      //Name
      console.log(data[i].name);

      var payments = data[i].payments;
      for (var j = 0; j < payments.length; j++) {

        //Time
        console.log(payments[j].time);

        //blocks
        var blocks = payments[j].blocks;
        for (var b = 0; b < blocks.length; b++) {
          console.log(blocks[b])
        }

        console.log(payments[j].amounts[_miner]);


      }
    }

  },
  error: function() {
    //alert("Was unable to get info!");
  }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

Answer 1

只需检查是否存在：

<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react-dom.min.js"></script>
<div id="root"></div>

或更妙的是：

if (payments[j].amounts[_miner] === undefined)
    console.log(payments[j].amounts[_miner]);

甚至：

if (payments[j].amounts[_miner] === undefined) 
   console.log(payments[j].amounts[_miner]) 
else
   console.log("no payments")

排除“未定义”结果导致JSON上的for循环

1 个答案: