What is "Array to string error"?

Question

$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE user_id = '".$id."' ");
while($rows = mysql_fetch_array($data))
{ 
 $pkmn_id = $rows['pkmn_id'];
 $path = mysql_query(" SELECT path FROM pokemons WHERE pk_id = '".$pkmn_id."' ");
 $poke = mysql_result($path, 0, "path");
 echo $poke; 
 echo "<br />";
 $level = $rows['level']; 
 echo $level;
 echo "<br />";
 $exp = $rows['exp']; 
 echo $exp;

This is my PHP code, its showing an error: Array to string conversion in C:\wamp\www\slots.php on line 18

Line 18 is this:while($rows = mysql_fetch_array($data))

I haven't used any array?? but this used to work! but suddenly this error started coming??

Walk through the code in a debugger, inspect the variable values on each line... — user1233508
– user1233508, Commented Apr 19, 2014 at 3:43
How are you setting $id that is in you query? Is it possible that it is an array and not a string/integer. — Sean
– Sean, Commented Apr 19, 2014 at 3:43
Stop using mysql_ functions and use PDO. Also don't interpolate variables into SQL queries. — siride
– siride, Commented Apr 19, 2014 at 3:46
$id is an array! I want to convert it into a variable!?? how can i do that? — user3545779
– user3545779, Commented Apr 19, 2014 at 3:46

souslov · Accepted Answer · 2014-04-19 03:57:18Z

3

Check if $id is an array, it seems that it causes this problem.

$id = "'" . implode("', '", $id) . "'";
$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE user_id IN ({$id}) ");

edited Apr 19, 2014 at 3:57

answered Apr 19, 2014 at 3:42

souslov

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