虽然这可能不适合这种情况，我发现前几天，如果你不需要改变数组的大小，你也可以使用delete关键字从数组中删除一个项目。

var myArray = [1,2,3];

delete myArray[1];

console.log(myArray[1]); //undefined

console.log(myArray.length); //3 - doesn't actually shrink the array down

这个怎么样?

$.each(someArray, function(i){
    if(someArray[i].name === 'Kristian') {
        someArray.splice(i,1);
        return false;
    }
});

最简单的解决方案是创建一个映射，按名称存储每个对象的索引，如下所示:

//adding to array
var newPerson = {name:"Kristian", lines:"2,5,10"}
someMap[ newPerson.name ] = someArray.length;
someArray.push( newPerson );

//deleting from the array
var index = someMap[ 'Kristian' ];
someArray.splice( index, 1 );

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

我们得到对象的索引它的name属性值为"Kristian"

const index = someArray.findIndex(key => key.name === "Kristian");
console.log(index); // 0

通过使用拼接函数我们删除了name属性值为“Kristian”的对象

someArray.splice(index,1);
console.log(someArray); // [{name:"John", lines:"1,19,26,96"}]

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"));

在数组上使用splice函数。指定开始元素的位置和要删除的子序列的长度。

someArray.splice(pos, 1);