MySQL returns null values for certain rows

Question

I am having a problem with MySQL and PHP. I'm trying to create something that gets values from a database, encode it in JSON and then print it. I have that down, but there's something that is keeping from one certain row in the database from displaying. It always returns NULL except for the id value I set. Here's my code, am I doing something wrong?

$srv = mysql_query("SELECT * FROM `players` WHERE `name` LIKE '%" . mysql_real_escape_string($_GET['q']) . "%'");
while ($record = mysql_fetch_array($srv)) {
    $playerInfo = array('id' => $playerarray['id'], 'name' => $playerarray['name'], 'server' => $playerarray['server']);
    echo(json_encode($playerInfo));
}

If you want to take a look at it, it's hosted here. The funny thing is, this page uses the exact same code, but doesn't return null. Any ideas?

Edit:

Here's what is in $playerInfo (when I use geekygamer14)

array (size=3)
  'id' => null
  'name' => null
  'server' => null

It seems that whatever rows that have verified set to 1 (integer), it gives NULL.

What you have in $playerInfo? var_dump it and post it in the question — Awlad Liton
– Awlad Liton, Commented Dec 31, 2013 at 7:15
what is $playerarray?, it should be $playerInfo = array('id' => $record['id'], 'name' => $record['name'], 'server' => $record['server']); — Vikas Arora
– Vikas Arora, Commented Dec 31, 2013 at 7:18
Even the id is null in the results, I think your query doesn't matched any row in the table players — Sawsan
– Sawsan, Commented Dec 31, 2013 at 7:18
So where does $playerarray come from and what has the while-loop to do with it? — Menno
– Menno, Commented Dec 31, 2013 at 7:24
Found the problem... $playerarray was equal to mysql_fetch_array($srv). Whoops. — tjhorner
– tjhorner, Commented Dec 31, 2013 at 7:26

Ahmed Siouani · Accepted Answer · 2013-12-31 13:31:47Z

1

You're using variable $record in your loop, though this doesn't show in your generated array. Instead you're using a variable named $playerarray. I assume the code should be:

while ($record = mysql_fetch_array($srv)) {
    $playerInfo = array('id' => $record['id'], 'name' => $record['name'], 'server' => $record['server']);
    echo(json_encode($playerInfo));
}

Note: Consider using an alternative to mysql-functions, for example mysqli. Mysql-functions are deprecated.

edited Dec 31, 2013 at 13:31

Ahmed Siouani

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answered Dec 31, 2013 at 7:20

Menno

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1 Comment

tjhorner Over a year ago

Oh wow. I really am that stupid. I guess I should get some sleep before I completely break everything.

Ahmed Siouani · Accepted Answer · 2013-12-31 13:32:40Z

0

Change this

$srv = mysql_query("SELECT * FROM `players` WHERE `name` LIKE '%" . mysql_real_escape_string($_GET['q']) . "%'");
while ($record = mysql_fetch_array($srv)) {
    $playerInfo = array('id' => $playerarray['id'], 'name' => $playerarray['name'], 'server' => $playerarray['server']);
    echo(json_encode($playerInfo));
}

to this

$srv = mysql_query("SELECT * FROM `players` WHERE `name` LIKE '%" . mysql_real_escape_string($_GET['q']) . "%'");
while ($record = mysql_fetch_array($srv)) {
    $playerInfo = array('id' => $record['id'], 'name' => $record['name'], 'server' => $record['server']);
    echo(json_encode($playerInfo));
}

Your variables name $playerarray['id'] is wrong this is write $record['id']. Your data is in $record because of loop so you should use $record instead of $playerarray.

edited Dec 31, 2013 at 13:32

Ahmed Siouani

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answered Dec 31, 2013 at 7:31

user2727841

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