JSON encoded array, how to use in jQuery

Question

Ok I've got this code in a file test.php:

<?php
$con = mysqli_connect("localhost", "user", "password", "DB");
if (mysqli_connect_errno()){
    echo "failed to connect:" . mysqli_connect_error();
}

$grab = mysqli_query($con, "SELECT * FROM DB");
$cars = array();
while($row = mysqli_fetch_assoc($grab)){
    array_push($cars, array("id" => $row["Id"], "name" => $row["Name"], "color" => $row["Color"];
}
echo json.encode($cars);

And I've got the jQuery code on my HTML page:

$.get("test.php", function($cars){
    $("itemNameLink1").html($cars.name);
    console.log($cars.name);
});

My next question is how do I access the data in my json_encoded array and use it in my jQuery on the HTML page. Right now I only get back undefined in console log. It was only my first try so wasn't disheartened by it. But any advise would be appreciated.

json.encode? that a typo?, any way, you need to loop the response, since this is most likely to have many rows — Kevin
– Kevin, Commented Sep 16, 2014 at 14:02
apparently cars is an array ? you are accessing name, which is probably a property of an individual car, not of the array. — njzk2
– njzk2, Commented Sep 16, 2014 at 14:04

Roman Bekkiev · Accepted Answer · 2014-09-16 14:17:41Z

2

That's because you try to access name property of $cars array instead of name property of each car. You should to iterate through your array first:

for(var car in $cars){
    console.log(car.name);
}

or

$cars.forEach(function(e){
    console.log(e);
});

if you maintain only modern browsers. Or use $.each method as @StartCoding mentioned.

AND, again as @StartCoding mentioned use $.getJSON instead of $.get or $.parseJSON($cars) if you've got a string in you response (it possible if your server returned wrong MIME-type [text/plain for example instead application/json]).

edited Sep 16, 2014 at 14:17

answered Sep 16, 2014 at 14:07

Roman Bekkiev

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Comments

StartCoding · Accepted Answer · 2014-09-16 14:09:56Z

1

$.get("test.php", function($cars){
 $.each($.parseJSON($cars), function(i, val){
$("itemNameLink1").html(val.name);
console.log(val.name);
});
});

answered Sep 16, 2014 at 14:09

StartCoding

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1 Comment

Tom Over a year ago

doesn't work, console log just gives a list of names. no change to itemNameLink1, it's ok, it'll take a bit to get my head round this.

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