Question

我从这样的表中提取了一个json格式数据：

[ { id:"3",
"Date Extracted":"12/3",
Experiment:"normal"},
{ id:"2",
"Date Extracted":"12/3",
Experiment:"powder",
Notes:"" },
{ id:"1",
"Date Extracted":"12/3",
Experiment:"fine",
Notes:"" }
]

我使用下面的功能只获取每个ID的属性，例如＆＃34; 12/3＆＃34;，＆＃34;普通＆＃34;，不包括名称，如＆＃34; Date Extracted＆＃34;，实验。不知怎的，这不起作用。但我不知道具体细节如何。

var j = JSON.parse(data);

function(){
for(name in j.property.name)
console.log(j.property.value);
}

Answer 1

提取的json数据与属性名称有一些验证问题。您可以通过jsonlint.com验证您的jsondata。

所以json数据应该是：

var jsonData ='[{ "id":"3", "Date Extracted":"12/3","Experiment":"normal"}, { "id":"2","Date Extracted":"12/3","Experiment":"powder","Notes":"" },{ "id":"1","Date Extracted":"12/3","Experiment":"fine","Notes":"" }]';

功能：

var json = $.parseJSON(jsonData);
$(json).each(function(i,val){
    $.each(val,function(name,value){
        console.log(name + " : " + value);       
    });
});

Answer 2

你不能像那样使用“for”。

var j = JSON.parse('[{  "id": "3",  "Date Extracted": "12/3",   "Experiment": "normal"}, {  "id": "2",  "Date Extracted": "12/3",   "Experiment": "powder", "Notes": ""}, { "id": "1",  "Date Extracted": "12/3",   "Experiment": "fine",   "Notes": ""}]');for(var i=0;i<j.length;i++){
var obj = j[i];
for(var key in obj){
    var attrName = key;
    var attrValue = obj[key];
    console.log(attrValue);
}}

Answer 3

＆＃13;

var data = [{
  id: "3",
  "Date Extracted": "12/3",
  Experiment: "normal"
}, {
  id: "2",
  "Date Extracted": "12/3",
  Experiment: "powder",
  Notes: ""
}, {
  id: "1",
  "Date Extracted": "12/3",
  Experiment: "fine",
  Notes: ""
}];





for (var i = 0; i < data.length; i++) {
  console.log("Date Extracted " + data[i]['Date Extracted'])//for index Date Extracted
  console.log("Experiment " + data[i]['Experiment'])//for index Experiment
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

你可以像这样使用for循环

如何获取json格式字符串（从表中提取）的值

3 个答案: