尝试使用forEach(item,index) helper

var projects = [
    {
        value: "jquery",
        label: "jQuery",
        desc: "the write less, do more, JavaScript library",
        icon: "jquery_32x32.png"
    },
    {
        value: "jquery-ui",
        label: "jQuery UI",
        desc: "the official user interface library for jQuery",
        icon: "jqueryui_32x32.png"
    },
    {
        value: "sizzlejs",
        label: "Sizzle JS",
        desc: "a pure-JavaScript CSS selector engine",
        icon: "sizzlejs_32x32.png"
    }
];

let search_to_change = 'jquery'

projects.forEach((item,index)=>{
   if(item.value == search_to_change )
      projects[index].desc = 'your description ' 
})

我认为这样比较好

const index = projects.findIndex(project => project.value==='jquery-ui');
projects[index].desc = "updated desc";

你必须像这样在数组中搜索:

function changeDesc( value, desc ) {
   for (var i in projects) {
     if (projects[i].value == value) {
        projects[i].desc = desc;
        break; //Stop this loop, we found it!
     }
   }
}

然后像这样使用它

var projects = [ ... ];
changeDesc ( 'jquery-ui', 'new description' );

更新:

为了更快:

var projects = {
   jqueryUi : {
      value:  'lol1',
      desc:   'lol2'
   }
};

projects.jqueryUi.desc = 'new string';

(根据Frédéric的评论，你不应该在对象键中使用连字符，或者你应该使用“jquery-ui”和项目["jquery-ui"]符号。)

你可以使用$.each()遍历数组并定位你感兴趣的对象:

$.each(projects, function() {
    if (this.value == "jquery-ui") {
        this.desc = "Your new description";
    }
});

试试这段代码。它使用jQuery的grep函数

array = $.grep(array, function (a) {
    if (a.Id == id) {
        a.Value= newValue;
    }
    return a;
});

让你想更新数组[2]= "data"的值

    for(i=0;i<array.length;i++){
      if(i == 2){
         array[i] = "data";
        }
    }