如何获取json格式字符串（从表中提取）的值

Question

I have extract a json format data from a table like this:

[ { 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:"" }
]

And I have used function like below to get only the attributes of each id like "12/3","normal", not include the name like "Date Extracted", Experiments. Somehow this does not work. but I have no idea how do it exactly in details.

var j = JSON.parse(data);

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

Answer 1

The extracted json data had some validation issues with the property names. You could verify your jsondata via jsonlint.com .

So the json data should likely be:

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":"" }]';

Function:

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

Answer 2

You can't use "for" like that.

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

 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 } 

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

You can do using for loop like this