我做了一个动态函数，以对象数组，键和值，并返回相同的数组后删除所需的对象:

function removeFunction (myObjects,prop,valu)
        {
             return myObjects.filter(function (val) {
              return val[prop] !== valu;
          });

        }

完整示例:DEMO

var obj = {
            "results": [
              {
                  "id": "460",
                  "name": "Widget 1",
                  "loc": "Shed"
              }, {
                  "id": "461",
                  "name": "Widget 2",
                  "loc": "Kitchen"
              }, {
                  "id": "462",
                  "name": "Widget 3",
                  "loc": "bath"
              }
            ]
            };


        function removeFunction (myObjects,prop,valu)
        {
             return myObjects.filter(function (val) {
              return val[prop] !== valu;
          });

        }


console.log(removeFunction(obj.results,"id","460"));

你可以使用以下几种方法从数组中删除项:

//1
someArray.shift(); // first element removed
//2
someArray = someArray.slice(1); // first element removed
//3
someArray.splice(0, 1); // first element removed
//4
someArray.pop(); // last element removed
//5
someArray = someArray.slice(0, someArray.length - 1); // last element removed
//6
someArray.length = someArray.length - 1; // last element removed

如果你想移除x位置的元素，使用:

someArray.splice(x, 1);

Or

someArray = someArray.slice(0, x).concat(someArray.slice(-x));

回复@chill182的评论:您可以使用array从数组中删除一个或多个元素。过滤器或数组。拼接结合数组。findIndex(参见MDN)。

请看这个Stackblitz项目或下面的代码片段:

// non destructive filter > noJohn = John removed, but someArray will not change let someArray = getArray(); let noJohn = someArray.filter( el => el.name !== "John" ); log(`let noJohn = someArray.filter( el => el.name !== "John")`, `non destructive filter [noJohn] =`, format(noJohn)); log(`**someArray.length ${someArray.length}`); // destructive filter/reassign John removed > someArray2 = let someArray2 = getArray(); someArray2 = someArray2.filter( el => el.name !== "John" ); log("", `someArray2 = someArray2.filter( el => el.name !== "John" )`, `destructive filter/reassign John removed [someArray2] =`, format(someArray2)); log(`**someArray2.length after filter ${someArray2.length}`); // destructive splice /w findIndex Brian remains > someArray3 = let someArray3 = getArray(); someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1); someArray3.splice(someArray3.findIndex(v => v.name === "John"), 1); log("", `someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1),`, `destructive splice /w findIndex Brian remains [someArray3] =`, format(someArray3)); log(`**someArray3.length after splice ${someArray3.length}`); // if you're not sure about the contents of your array, // you should check the results of findIndex first let someArray4 = getArray(); const indx = someArray4.findIndex(v => v.name === "Michael"); someArray4.splice(indx, indx >= 0 ? 1 : 0); log("", `someArray4.splice(indx, indx >= 0 ? 1 : 0)`, `check findIndex result first [someArray4] = (nothing is removed)`, format(someArray4)); log(`**someArray4.length (should still be 3) ${someArray4.length}`); // -- helpers -- function format(obj) { return JSON.stringify(obj, null, " "); } function log(...txt) { document.querySelector("pre").textContent += `${txt.join("\n")}\n` } function getArray() { return [ {name: "Kristian", lines: "2,5,10"}, {name: "John", lines: "1,19,26,96"}, {name: "Brian", lines: "3,9,62,36"} ]; } <pre> **Results** </pre>

你也可以使用地图功能。

someArray = [{name:"Kristian", lines:"2,5,10"},{name:"John",lines:"1,19,26,96"}];
newArray=[];
someArray.map(function(obj, index){
    if(obj.name !== "Kristian"){
       newArray.push(obj);
    }
});
someArray = newArray;
console.log(someArray);

如果你想删除给定对象的所有出现(基于某些条件)，那么使用for循环中的javascript splice方法。

因为删除对象会影响数组长度，所以请确保减少计数器一步，以保持长度检查不变。

var objArr=[{Name:"Alex", Age:62},
  {Name:"Robert", Age:18},
  {Name:"Prince", Age:28},
  {Name:"Cesar", Age:38},
  {Name:"Sam", Age:42},
  {Name:"David", Age:52}
];

for(var i = 0;i < objArr.length; i ++)
{
  if(objArr[i].Age > 20)
  {
    objArr.splice(i, 1);
    i--;  //re-adjust the counter.
  }
}

上面的代码片段删除了年龄大于20的所有对象。

你所显示的“数组”是无效的JavaScript语法。花括号{}表示具有属性名/值对的对象，而方括号[]表示数组，就像这样:

someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];

在这种情况下，您可以使用.splice()方法删除一个项。要删除第一项(索引0)，如下所示:

someArray.splice(0,1);

// someArray = [{name:"John", lines:"1,19,26,96"}];

如果你不知道索引，但想通过数组搜索找到名称为“Kristian”的项来删除，你可以这样做:

for (var i =0; i < someArray.length; i++)
   if (someArray[i].name === "Kristian") {
      someArray.splice(i,1);
      break;
   }

编辑:我刚刚注意到你的问题被标记为“jQuery”，所以你可以尝试$.grep()方法:

someArray = $.grep(someArray,
                   function(o,i) { return o.name === "Kristian"; },
                   true);

你也可以用一些:

someArray = [{name:"Kristian", lines:"2,5,10"},
             {name:"John", lines:"1,19,26,96"}];

someArray.some(item => { 
    if(item.name === "Kristian") // Case sensitive, will only remove first instance
        someArray.splice(someArray.indexOf(item),1) 
})