MySQL & PHP Parameter 1 as Resource

Question

Alright, PHP is throwing this error at me (in the log) when I run the code mentioned below:

Error

mysql_num_rows() expects parameter 1 to be resource, string given in (place) on line 10

Line 9-11

$queryFP = ("SELECT * FROM db");
$countFP = mysql_num_rows($queryFP);
$aID = rand(1, $countFP);

I think it has something to do with the $queryFP's syntax, but I'm not completely sure how to fix it since $queryFP's syntax is the simplest query I've ever seen.

+1 for the perfectly formed question. Both problem code and error message. Everyone to follow this example. — Your Common Sense, May 30 '10 at 12:48
possible duplicate of [Warning: mysql_fetch_* expects parameter 1 to be resource, boolean given error](http://stackoverflow.com/questions/11674312/warning-mysql-fetch-expects-parameter-1-to-be-resource-boolean-given-error) — John Conde, Jul 28 '12 at 16:47
possible duplicate of [mysql\_fetch\_array() expects parameter 1 to be resource, boolean given in select](http://stackoverflow.com/questions/2973202/mysql-fetch-array-expects-parameter-1-to-be-resource-boolean-given-in-select) — Glavić, Jan 07 '14 at 22:26

score 3 · Accepted Answer · answered May 30 '10 at 12:37

3

You need to query the database first.

$queryFP = ("SELECT * FROM db");

Should be:

$queryFP = mysql_query("SELECT * FROM db");

answered May 30 '10 at 12:37

svens

11,438
6
36
55

Wow, I should have known that. Thanks. – Nik May 30 '10 at 12:41

score 0 · Answer 2 · answered May 30 '10 at 12:38

0

You are missing the mysql_query function, it should be like this:

$queryFP = "SELECT * FROM table_name_here";
$queryFP = mysql_query($queryFP) or die(mysql_error());
$countFP = mysql_num_rows($queryFP);
$aID = rand(1, $countFP);

answered May 30 '10 at 12:38

Sarfraz

377,238
77
533
578

That `or die(..)` stuff is so terrible. – whiskeysierra Nov 14 '10 at 00:11
True, this is just an example and should be avoided at *production* level, good idea to see if there is any error in the query while in *development* :) – Sarfraz Nov 14 '10 at 07:45

score 0 · Answer 3 · answered May 30 '10 at 12:46

0

As it been said, you're missing mysql_query function.
Though whole approach is wrong. You shouldn't select whole load of ata if you need only number of rows.
So, it must be

$sql = "SELECT count(*) FROM db";
$res = mysql_query($sql) or trigger_error(mysql_error().$sql);
$row = mysql_fetch_row($res);
$countFP = $row[0];
$aID = rand(1, $countFP);

And I hope you won't use $aID for any database related action

answered May 30 '10 at 12:46

Your Common Sense

156,878
40
214
345

$aID is used to select the row of which needs displayed. – Nik May 30 '10 at 12:48
It actually needs the entire row, not just the number of them, which is why $aID is required. – Nik May 30 '10 at 12:49
@Nik there can be gaps with no ids. `SELECT * FROM db, limit $aID,1` query could do the trick. – Your Common Sense May 30 '10 at 12:50
In this very example you don't get any rows. Only number. – Your Common Sense May 30 '10 at 12:51
I'm aware, I only posted the necessary code to fix the issue, the next block of code pulls the array of which is required then displays it. – Nik May 30 '10 at 12:53
What should I do then, because this is the only logical way to achieve what the rest of the block does. – Nik May 30 '10 at 13:23
@Nik if you need only ONE random row off the whole table, you need 2 steps: 1. Get the NUMBER of rows. 2. Get ONE random row. First step I posted in my answer and query for the second in the comment. – Your Common Sense May 30 '10 at 13:53
Oh, I see, why did you say that I shouldn't use $aID for any database related action? – Nik May 30 '10 at 14:26
@Nik because name of the variable makes me think if `id` database field while you cannot use this random value as id. – Your Common Sense May 30 '10 at 14:34
And my main intention was to tell you that you shouldn't request whole table data if you need only number of rows. – Your Common Sense May 30 '10 at 14:41

MySQL & PHP Parameter 1 as Resource

3 Answers3

Linked

Related