Jquery 如何将文本输入val()添加到JSON数组中

Jquery 如何将文本输入val()添加到JSON数组中,jquery,json,Jquery,Json,您能看一下并告诉我如何将数据添加/插入JSON数组用户吗 "users":[ {"user":"John", "age":36, "grade":"C", "mark":88 }, {"user":"Rose", "age":28, "grade":"B", "mark":90 }, .... ] $(document).ready(function () { $("#addtoJSON").click(function () { var u

您能看一下并告诉我如何将
数据添加/插入JSON数组用户吗

"users":[
    {"user":"John", "age":36, "grade":"C", "mark":88 }, 
    {"user":"Rose", "age":28, "grade":"B", "mark":90 }, 
    ....
]

$(document).ready(function () {
    $("#addtoJSON").click(function () {
        var user = $("#user").val();
        var age = $("#age").val();
        var grade = $("#grade").val();
        var mark = $("#mark").val();
        // alert(user);
        var data = 'name=' + user + '&age=' + age + '&grade=' + grade + '&mark=' + mark;
        alert(data);
    });
});
谢谢使用此代码:

var dataArr = [];
var data = JSON.stringify({  // use JSON.stringify to convert object to JSON
    name: user,
    age: age,
    grade: grade,
    mark: +mark
});
dataArr.push(data);
alert(data); // well formed JSON

您只需将新的
对象
添加到“用户”
对象
的数组中即可:

这将有助于-:

$(document).ready(function () {
  var jsonArg1 = new Object();
  var jsonArg2 = new Object();
   $("#addtoJSON").click(function () {
     jsonArg1.user = $("#user").val();
     jsonArg1.age = $("#age").val();
     jsonArg1.grade = $("#grade").val();
     jsonArg1.mark = $("#mark").val();

     jsonArg12.users = jsonArg1;

     var pluginArrayArg = new Array();
     pluginArrayArg.push(jsonArg12);
     pluginArray = (JSON.stringify(pluginArrayArg))
     console.log(pluginArray);
  });
});

嗨,阿米特,谢谢你的回复,但是我如何才能将每个新的输入序列添加到数据中呢?我的意思是我需要创建JSON对象的数组
$(document).ready(function () {
  var jsonArg1 = new Object();
  var jsonArg2 = new Object();
   $("#addtoJSON").click(function () {
     jsonArg1.user = $("#user").val();
     jsonArg1.age = $("#age").val();
     jsonArg1.grade = $("#grade").val();
     jsonArg1.mark = $("#mark").val();

     jsonArg12.users = jsonArg1;

     var pluginArrayArg = new Array();
     pluginArrayArg.push(jsonArg12);
     pluginArray = (JSON.stringify(pluginArrayArg))
     console.log(pluginArray);
  });
});