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。

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);

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。

ES2015

let someArray = [
               {name:"Kristian", lines:"2,5,10"},
               {name:"John", lines:"1,19,26,96"},
               {name:"Kristian", lines:"2,58,160"},
               {name:"Felix", lines:"1,19,26,96"}
            ];

someArray = someArray.filter(person => person.name != 'John');

它会除掉约翰!

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>

干净的解决方案是使用Array.filter:

var filtered = someArray.filter(function(el) { return el.Name != "Kristian"; }); 

问题是它不能在IE < 9上工作。然而，你可以包括来自Javascript库(例如:underscore.js)的代码，为任何浏览器实现这一点。