PHP: Warning: sort() expects parameter 1 to be array, resource given

Question

I wanted to arrange the array of table list with sort() function but i am getting same kind of warning my code as follows :

 <?PHP 
     require_once("lib/connection.php"); 

     $result = mysql_query("SHOW TABLES FROM `st_db_1`");

     sort($result);
     foreach ($result as $result){
     echo $result ;
     } 
 ?>

and the warning i am getting are :

Warning: sort() expects parameter 1 to be array, resource given in C:\wamp\www\Copy (4)       of st_db_1\test_2.php on line 9
Warning: Invalid argument supplied for foreach() in C:\wamp\www\Copy (4) of st_db_1\test_2.php on line 10

You need to [fetch](http://php.net/manual/en/function.msql-fetch-array.php) the result first. E.g. using `$row = mysql_fetch_array($result):` — Jürgen Thelen, May 29 '11 at 17:16
you need to lookup on what `mysql_query` returns, as well as `mysql_fetch_assoc`. By the way you'll need to use `print_r` instead of `echo` to see the structure of. — onteria_, May 29 '11 at 17:17
As well in your foreach loop, you are overwriting the `$result` variable which is probably not what you want at all. — judda, May 29 '11 at 17:26

score 2 · Answer 1 · answered May 29 '11 at 17:18

The warning is pretty clear: mysql_query does not return an array with results from the query, but a resource. You need a function like mysql_fetch_array() to return the data you need (and on which you can perform a sort operation).

See the manual for the use of mysql_query() http://nl3.php.net/mysql_query

And maybe unrelated, but you can sort your results in MySQL right away by adding ORDER BY <fieldname> to your query.

Trott · Answer 2 · 2011-05-29T18:02:23.137

2

I'm not providing the most efficient code imaginable, but this should make it clear what's going on and solve your problem:

 $result = mysql_query("SHOW TABLES FROM `st_db_1`");

 $my_array_of_table_names = array();
 while ( $row = mysql_fetch_array($result, MYSQL_NUM)) {
     $my_array_of_table_names[] = $row[0];
 }
 sort($my_array_of_table_names);

 foreach ($my_array_of_table_names as $table_name){
     echo "$table_name\n";
 }

edited May 29 '11 at 18:02

answered May 29 '11 at 17:22

Trott

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Yup, moved the sort to the right place. That's what I get for not testing before posting. Uh, will go test now to make sure I didn't just make it worse. – Trott May 29 '11 at 17:32
No problem, the main problem is that the person asking this question was not fetching the rows. And you showed this. Now we have almost the same answer. I gave you a point and might delete my answer later because they are too similar. – Raffael Luthiger May 29 '11 at 18:02

score 2 · Answer 3 · answered May 29 '11 at 17:24

The variable $result is only a resource of the type result. You need to fetch then the data from the result set with e.g. mysql_fetch_assoc().

$result = mysql_query("SHOW TABLES FROM `st_db_1`");
$array = array();
while ($row = mysql_fetch_assoc($result)) {
    $array[] = $row["Tables_in_st_db_1"];
}
sort($array);
foreach ($array as $item) {
   echo $item;
}

score 1 · Answer 4 · answered May 29 '11 at 17:22

Your problem is that you aren't actually getting the data from the query.

mysql_query() doesn't give you a recordset.

What it does is query the database and returns a database resource which you can then use to get the data.

What you need is after calling mysql_query(), you then need to also call mysql_fetch_array() or similar. (there are a range of functions available, but that's probably the best one to use in this case). Then sort() the data from that, not $result.

score 0 · Answer 5 · edited May 30 '11 at 08:44

0

It clearly says: it expects an array and you pass something else.

If you had checked the type of $result you would have seen that it is not an array, intead a resource.

edited May 30 '11 at 08:44

kapa

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answered May 29 '11 at 17:16

vbence

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i checked ,,its a array if table names but i am successfully able to fetch array of table using while loop – ravi May 29 '11 at 17:21

PHP: Warning: sort() expects parameter 1 to be array, resource given

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