Question

我有一个数组，其中包含一个名称列表。如果这些名称与另一个数组中的键/值对匹配，我想删除包含匹配项的对象，并返回没有该对象的数组。

这就是我迄今为止的例子：

var supervisorsToRemove = ["Supervisor1", "Supervisor2"];
var data = [{id: "Name1", supervisor: "Supervisor1"},{id: "Name2", supervisor: "Supervisor1"},{id: "Name3", supervisor: "Supervisor2"},{id: "Name4", supervisor: "Supervisor3"}]

我希望得到的数据数组为：

data = [{id: "Name4", supervisor: "Supervisor3"}]

我可以使用以下内容删除一个主管的对象：

var supervisorsToRemove = "Supervisor1";
data = $.grep(data, function(e){ 
     return e.supervisor != supervisorsToRemove; 
});

但是当我尝试删除数组中的所有内容时，它不起作用：

var supervisorsToRemove = ["Supervisor1","Supervisor2"];
data = $.grep(data, function(e){ 
     return e.supervisor != supervisorsToRemove; 
});

Answer 1

使用Array#indexOf方法（或Array#includes方法）检查数组中存在的元素

data = $.grep(data, function(e){ 
  return supervisorsToRemove.indexOf(e.supervisor) == -1; 
  // or return supervisorsToRemove.includes(e.supervisor); 
});

＆＃13;

var supervisorsToRemove = ["Supervisor1", "Supervisor2"];
var data = [{
  id: "Name1",
  supervisor: "Supervisor1"
}, {
  id: "Name2",
  supervisor: "Supervisor1"
}, {
  id: "Name3",
  supervisor: "Supervisor2"
}, {
  id: "Name4",
  supervisor: "Supervisor3"
}]
console.log($.grep(data, function(e) {
  return supervisorsToRemove.indexOf(e.supervisor) == -1;
  // or return supervisorsToRemove.includes(e.supervisor); 
}));

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

更好的方法是使用一个将超级用户值保存为属性的对象，这有助于使其更快，因为Array#indexOf方法较慢。

var supervisorsToRemove = {"Supervisor1" : true, "Supervisor2" : true };

data = $.grep(data, function(e){ 
  return !supervisorsToRemove[e.supervisor]; 
});

＆＃13;

var data = [{
  id: "Name1",
  supervisor: "Supervisor1"
}, {
  id: "Name2",
  supervisor: "Supervisor1"
}, {
  id: "Name3",
  supervisor: "Supervisor2"
}, {
  id: "Name4",
  supervisor: "Supervisor3"
}]
var supervisorsToRemove = {
  "Supervisor1": true,
  "Supervisor2": true
};

console.log(
  $.grep(data, function(e) {
    return !supervisorsToRemove[e.supervisor];
  })
);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

如果您无法控制数组，请使用Array#reduce方法生成对象。

var obj = supervisorsToRemove.reduce(function(obj, v){ obj[v] = true; return obj; }, {});

data = $.grep(data, function(e){ 
  return !obj[e.supervisor]; 
});

＆＃13;

var supervisorsToRemove = ["Supervisor1", "Supervisor2"];
var data = [{
  id: "Name1",
  supervisor: "Supervisor1"
}, {
  id: "Name2",
  supervisor: "Supervisor1"
}, {
  id: "Name3",
  supervisor: "Supervisor2"
}, {
  id: "Name4",
  supervisor: "Supervisor3"
}]
var supervisorsToRemove = {
  "Supervisor1": true,
  "Supervisor2": true
};


var obj = supervisorsToRemove.reduce(function(obj, v) {
  obj[v] = true;
  return obj;
}, {});

console.log(
  $.grep(data, function(e) {
    return !obj[e.supervisor];
  })
);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

您甚至可以使用原生Javascript Array#filter方法。

＆＃13;

var supervisorsToRemove = ["Supervisor1", "Supervisor2"];
var data = [{
  id: "Name1",
  supervisor: "Supervisor1"
}, {
  id: "Name2",
  supervisor: "Supervisor1"
}, {
  id: "Name3",
  supervisor: "Supervisor2"
}, {
  id: "Name4",
  supervisor: "Supervisor3"
}]
var supervisorsToRemove = {
  "Supervisor1": true,
  "Supervisor2": true
};

console.log(
  data.filter(function(e) {
    return !supervisorsToRemove[e.supervisor];
  })
);

＆＃13;

Answer 2

在任何合理的现代浏览器中，都不需要jQuery。 filter()将完成这项工作。

var supervisorsToRemove = ["Supervisor1", "Supervisor2"];
var data = [{
  id: "Name1",
  supervisor: "Supervisor1"
}, {
  id: "Name2",
  supervisor: "Supervisor1"
}, {
  id: "Name3",
  supervisor: "Supervisor2"
}, {
  id: "Name4",
  supervisor: "Supervisor3"
}]


var results = data.filter(function(o) {
  return (supervisorsToRemove.indexOf(o.supervisor) < 0);
});

console.log(results);

jQuery基于另一个数组从数组中删除多个对象

2 个答案: