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up vote 0 down vote favorite

Asked this before, but I've narrowed down the issue to this bit of code. Here's my code, when I run it, it just says "null"..

$getmsg = "SELECT * FROM user WHERE account_id = $id";      
$showmsg = @mysqli_query ($dbc, $getmsg);
        while ($row = mysqli_fetch_array($showmsg, MYSQLI_ASSOC)) {

$arrResults = array($row['user_username']);


} // END WHILE


// Print them out, one per line
echo json_encode($arrResults);
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I think I found your previously abandoned question here where we narrowed it down: stackoverflow.com/questions/5902397/… That's shady. –  Beez May 5 '11 at 19:16
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2 Answers 2

up vote 4 down vote

First of all you have put the echo outside the loop which just echoes the last item instead of everyone and you don't check if there is a error with your query.

Instead this would be sufficient: 

$getmsg = "SELECT * FROM user WHERE account_id = $id";      
$result = @mysqli_query($dbc, $getmsg) or die("Error: " . mysql_error());
$result = mysql_fetch_assoc($result);
echo json_encode($result);

It puts the result in one assoc array and then converts the whole array to json and prints it.

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Not related to author's problem, but using @ is a bad style, as well as outputting mysql_error() and mixing MySQLi (mysqli_query()) with MySQL (mysql_error()). –  binaryLV May 5 '11 at 19:15
    
I just copied his code for proof of concept. –  rzetterberg May 5 '11 at 19:18
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up vote 1 down vote

The problem you are likely having is in your assignment statement:

$arrResults = array($row['user_username']);

You should change it to the following:

$arrResults[] = $row['user_username'];

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