ajax success not getting php records

Question

I am trying to get my db rows from php. Currently the js success function isn't being reached. Is there anything obvious in the php that may be causing an issue?

The php

$id=$_GET['id'];

$stmt = $db->prepare("SELECT * FROM brand_members WHERE Id = :id");
$stmt->bindValue(':id',$id,PDO::PARAM_INT);
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);

echo json_encode($result);

The js

$.ajax({                                      
      url: 'dead.php',                  //the script to call to get data          
      data: "id=8",                      
      dataType: 'json',                //data format      
      success: function(data)          //on recieve of reply
      {
        var id = data[0];              //get id
        var vname = data[1];           //get name
        alert("a");         //get name
      }
});

Try `console.log(data)` inside your `success` function to see what exactly is the structure of your `data` variable. — Dekel, Dec 26 '16 at 01:59
did you successfully connect to your database? before running that query — Beginner, Dec 26 '16 at 02:00
can you view your network response like this https://gyazo.com/cad3c8bddc33b05c1d856d796532a32d — Beginner, Dec 26 '16 at 02:07
you got `FETCH_ASSOC` activated, so your data will be something like `data[0]['id']` and `data[0]['name']` and not only `data[0]`! What does `console.log(data);` really show? — Jeff, Dec 26 '16 at 02:08
@Fred-ii- why not? `$result` is an array according to the [docs](http://php.net/manual/de/pdostatement.fetchall.php). — Jeff, Dec 26 '16 at 02:11
[How to pass mysql result as jSON via ajax](http://stackoverflow.com/questions/30072794/how-to-pass-mysql-result-as-json-via-ajax) and http://stackoverflow.com/questions/25593541/pdo-doesnt-return-data-from-mysql-to-jquery-ajax — Funk Forty Niner, Dec 26 '16 at 02:13
`data: "id=8"` might by the/one mistake: shouldn't it be `data: { "id": 8 }`? — Jeff, Dec 26 '16 at 02:13
@Fred-ii- ah, ok, sorry. I've read it the other way round... (as "the data from php back to client") — Jeff, Dec 26 '16 at 02:16
ok console log is returning [{"Id":"8","email":"test@gmail.com"}] — Buy Naural, Dec 26 '16 at 02:17
the console log returns my entire html page before [{"Id":"8","email":"test@gmail.com"}] which is why there is an issue. How can I just get the array — Buy Naural, Dec 26 '16 at 02:23
as we don't see the rest of the php-script it's hard to tell! But asccording to your description it seems that you might want to seperate data-getting from html. — Jeff, Dec 26 '16 at 02:25
@BuyNaural you returns an array so in your callback you must specified index 0 `data[0]['column'];` also add `limit 1` into your query to optimize it. — Beginner, Dec 26 '16 at 02:30
can you `console.log(data);` see whats the result or also `console.log(typeof data);` see if it's result converted as json object — Beginner, Dec 26 '16 at 02:36
console log returns all of my html followed by [{"Id":"8","email":"test@gmail.com"}] and type of returns a string — Buy Naural, Dec 26 '16 at 02:40
try changing it to `json_encode( $result, JSON_FORCE_OBJECT );` — Beginner, Dec 26 '16 at 02:48

Beginner · Answer 1 · 2016-12-26T02:37:18.970

0

The problem is you didn't passed properly your $_GET parameter

To solve that

Change your

url: 'dead.php'

to

url: 'dead.php?id=8',

also data: your_data is no longer needed

Or this method change

data: "id=8"

to

data: {id: 8}

Also in your callback since your ajax response returns an array specified the index 0

success: function(data)          //on recieve of reply
  {
    var id = data[0]['Id'];              //get id
    var vname = data[0]['name'];           //get name
    alert("a"); 
  }

edited Dec 26 '16 at 02:37

answered Dec 26 '16 at 02:18

Beginner

4,118
3
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26

maybe add `var id = data[0]['id'];` to your answer – Jeff Dec 26 '16 at 02:23

ajax success not getting php records

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