无法在 TypeScript 中执行连接

Question

我有一个对象数组，如下所示

var tempObj =  [{
"isAvailable": true,
"receipent": [{
    "id": "a6aedf0c34",
    "receipentName": "ABC"
}, {
    "id": "a6aedbc34",
    "receipentName": "XYZ"
}]  }]

而且我想要以逗号分隔的字符串形式的名称值。

我使用下面的代码来实现：

 var b = Array.prototype.map.call(tempObj.receipent, function (item) {
          return item.receipentName;
        }).join(",");

但我收到以下错误：

未捕获（承诺）TypeError：Array.prototype.map 调用 null 或 undefined

我也试过这个：

  var to = tempObj.receipent;
   var b = to.map(e => e.receipentName).join(",");

为此，我收到以下错误：

无法读取未定义的属性'map'

Answer 1

您的 tempObj 本身就是一个数组。如果你只想处理它的第一个元素，那么你必须使用 tempObj[0] 然后按如下方式做

var tempObj =  [{
"isAvailable": true,
"receipent": [{
    "id": "a6aedf0c34",
    "receipentName": "ABC"
}, {
    "id": "a6aedbc34",
    "receipentName": "XYZ"
}]  }];

//tempObj = JSON.parse(tempObj);
var b = tempObj[0].receipent.map(o => o.receipentName).join(",");

console.log(b);

// In case you want to do it for every tempObj then do as follows
tempObj.forEach(obj => {
  const _b = obj.receipent.map(o => o.receipentName).join(",");
  console.log(_b);
});

无法在 TypeScript 中执行连接

Not able to perform join in TypeScript

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