Question

I'm developing an Android application that makes some request to a server in which I have programmed a database. In the server I work with PHP to make query. Now I wrote this code to make a query and to push the result into an array that I can use later in "Android part" by using java language. This is my PHP code:

$query="SELECT * FROM $tbl_name WHERE user='$myusername'";
$res=mysql_query($query);   
$posts = array();
if(mysql_num_rows($res)) {
  while($post[] = mysql_fetch_assoc($res)) {
     array_push($posts, $post);
  }
} 
echo json_encode(array('posts'=>$posts));

The problem is that when I try to get the result in java I can see only the first row of the result. I'm sure that I have more than 1 row because I tried to print mysql_num_rows($res) and the result was bigger than 1. How can I fix this problem??

Edit. Just to clarify. I make a call to my server in which I use php from Android AsynTask. In onPostExecute method I make something like this:

String res = result.toString();

where result is the JSONObject that I obtain from doInBackground method. Maybe I'm doing error Here because all your solution give me always the first row of the query's result only.

Answer 1

You don't need to specify array here.

Change: while($post[] = mysql_fetch_assoc($res)) {

To: while($post = mysql_fetch_assoc($res)) {

You can always var_dump($posts) after while instruction and check if you get what you want.

php json_encode return only first row of an array

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