这个怎么样?

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

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>

Splice (i, 1)，其中i是数组的增量索引，将删除对象。 但是请记住，splice也会重置数组长度，所以要注意'undefined'。用你的例子，如果你删除了“Kristian”，那么在循环中的下一次执行中，i将是2，但someArray将是1的长度，因此，如果你试图删除“John”，你会得到一个“undefined”错误。解决这个问题的一个方法是使用单独的计数器来跟踪要删除的元素的索引。

具有es6箭头功能

let someArray = [
                 {name:"Kristian", lines:"2,5,10"},
                 {name:"John", lines:"1,19,26,96"}
                ];
let arrayToRemove={name:"Kristian", lines:"2,5,10"};
someArray=someArray.filter((e)=>e.name !=arrayToRemove.name && e.lines!= arrayToRemove.lines)

只从数组中返回属性名不是“Kristian”的对象

var noKristianArray = $.grep(someArray, function (el) { return el.name!= "Kristian"; });

演示:

var someArray = [ {名称:“克里斯蒂安”,行:“2、5、10”}, {名称:“约翰”,行:“1,19日,26日96”}, {名称:“克里斯蒂安”,行:" 58160 "}, {名称:“费利克斯”,行:“96”1,19日,26日} ]; var noKristianArray = $。grep(someArray, function (el){返回el.name!= "克里斯蒂安”;})； console.log (noKristianArray); < script src = " https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js " > < /脚本>

在你的数组语法中似乎有一个错误，所以假设你的意思是一个数组，而不是一个对象，数组。Splice是你的朋友:

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