javascript-在每次地图迭代的开始处添加对象

Question

所以，让我们假设我有这个：

var officers = [
  { id: 20, name: 'Captain', lastName: 'Piett' },
  { id: 24, name: 'General', lastName: 'Veers'  },
  { id: 56, name: 'Admiral', lastName: 'Ozzel'  },
  { id: 88, name: 'Commander', lastName: 'Jerjerrod'  }
];

如果我这样做：var officersIds = officers.map(x => [x.name, x.lastName]);officersIds的结果将是这样：

[ "Captain", "Piett" ], [ "General", "Veers" ], [ "Admiral", "Ozzel" ], [ "Commander", "Jerjerrod" ]

对吗？

所以，我需要做的是在每次地图迭代中放置一个对象，因此现在的输出是这样，例如：

[["x", "y"], [ "Captain", "Piett" ]], 
[["x", "y"], [ "General", "Veers" ]],  
[["x", "y"], [ "Admiral", "Ozzel" ]], 
[["x", "y"], [ "Commander", "Jerjerrod" ]]

我为什么需要这个？最好不要问;）但这是一个复杂的问题，如果您能帮助我解决这个简单的问题，我可以将您的解决方案转移到我的复杂问题上。

重要提示：有没有办法在一行中做到这一点？

Answer 1

在特定索引处使用拼接。

var officers = [
  { id: 20, name: 'Captain', lastName: 'Piett' },
  { id: 24, name: 'General', lastName: 'Veers'  },
  { id: 56, name: 'Admiral', lastName: 'Ozzel'  },
  { id: 88, name: 'Commander', lastName: 'Jerjerrod'  }
];

var officersIds = officers.map(x => [x.name, x.lastName]);
officersIds.splice(0, 0, ['x','y']);
// or officersIds.unshift(['x','y'])
console.log(officersIds);

Answer 2

您可以使用spread operator将其设置为单线：

var officersIds = [['x', 'y'], ...officers.map(x => [x.name, x.lastName])];

请参见下面的概念证明：

var officers = [
  { id: 20, name: 'Captain', lastName: 'Piett' },
  { id: 24, name: 'General', lastName: 'Veers'  },
  { id: 56, name: 'Admiral', lastName: 'Ozzel'  },
  { id: 88, name: 'Commander', lastName: 'Jerjerrod'  }
];

var officersIds = [['x', 'y'], ...officers.map(x => [x.name, x.lastName])];

console.log(officersIds);

Answer 3

var officers = [
  { id: 20, name: 'Captain', lastName: 'Piett' },
  { id: 24, name: 'General', lastName: 'Veers'  },
  { id: 56, name: 'Admiral', lastName: 'Ozzel'  },
  { id: 88, name: 'Commander', lastName: 'Jerjerrod' },
];

officers.unshift({ name: x, lastName: y });

var officersIds = officers.map(x => [x.name, x.lastName]);

或

var officers = [
      { id: 20, name: 'Captain', lastName: 'Piett' },
      { id: 24, name: 'General', lastName: 'Veers'  },
      { id: 56, name: 'Admiral', lastName: 'Ozzel'  },
      { id: 88, name: 'Commander', lastName: 'Jerjerrod' },
    ];


    var officersIds = officers.map(x => [x.name, x.lastName]);

    officersIds.unshift(["x", "y" ]);