json encode multidimensional array get value

Question

 Array
(
    [0] => Array
        (
            [sysid] => 1
            [code] => 140101000
            [name] => China
            [parentid] => 1
        )

    [1] => Array
        (
            [sysid] => 2
            [code] => 140102000
            [name] => Japan
            [parentid] => 1
        )

    [2] => Array
        (
            [sysid] => 3
            [code] => 140103000
            [name] => Hongkong
            [parentid] => 1
        )
)

This is my array that i get from my print_r request it is from php it is a multidimensional array that i used json_encode now this is the data i get from success. I want to get all the value of sysid and name from this json to be put in select option how is this possible

I used this code below but i get undefined and i get 4,000 result

PAIR 0: undefined

PAIR 0: undefined

PAIR 1: undefined

PAIR 1: undefined

for (var i = 0; i < data.length; i++) {
    console.log("PAIR " + i + ": " + data[i].sysid);
    console.log("PAIR " + i + ": " + data[i].name);
}

UPDATE

my bad the first one is the print_r this is the json

[{"sysid":"1","code":"140101000","name":"China","parentid":"1"},{"sysid":"2","code":"140102000","name":"Japan","parentid":"1"},
{"sysid":"3","code":"140103000","name":"Hongkong","parentid":"1"}]

ajax is

$.ajax({
    type: 'POST',
    url: '../include/country.php',
    data: {
        id: id
    },
    success: function(data) {
        // the next thing you want to do 
        var obj = data;
        console.log(obj);
        //for (var i = 0; i < data.length; i++) {
        //console.log("PAIR " + i + ": " + data[i].sysid);
        //console.log("PAIR " + i + ": " + data[i].name);
        //}


    }
});

in my console.log i get Pair 0: undefined how is that ok sir @MrBearAndBeer but in regular console log if the console log is only console.log(data); i get result what i want to show in console log is only the sysid and name — Brownman Revival
– Brownman Revival, Commented Jan 17, 2015 at 4:10
Am I right that you want to put the results into a select input? That's what I feel like you're asking — thedouglenz
– thedouglenz, Commented Jan 17, 2015 at 5:46

score 1 · Accepted Answer · 2015-01-17 06:04:48Z

1

$.ajax({
    type: 'GET',
    url: '../include/country.php',
    dataType : "json",
    data: {
        id: id
    },
    success: function(data) {
       for(var i = 0; i < data.length; i++) {
           console.log("PAIR " + i + ": " + data[i].sysid);
           console.log("PAIR " + i + ": " + data[i].name);
       }
    }
});

edited Jan 17, 2015 at 6:04

answered Jan 17, 2015 at 3:58

user3998237

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10 Comments

adeneo Over a year ago

I'm guessing the OP is showing us the PHP code, not the returned JSON.

user3998237 Over a year ago

why? just delete the var obj = data, or update your loop to use obj instead data.

Brownman Revival Over a year ago

i did try to just use the json alone i mean i copied it it worked but when i use data it is undefined what seems to be the problem?

Brownman Revival Over a year ago

also i noticed that the first item it has no json equivalent it has a result for print_r but for json none it is blank..

user3998237 Over a year ago

why u are using the POST method, if you want GET the data, and not pass?

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