PHP MYSQL : why i am getting this warning?

Question

When I use

mysql_affected_rows($result);

...in php then I get below warning how to remove it?

Warning: mysql_affected_rows(): supplied resource is not a valid MySQL-Link resource in C:\wamp\www\st_db_1\search_db.php on line 60

I'm afraid you will need to show us more code than that to get a reasonable answer. The reason it's happening is because $result is not a valid MySQL link resource (so if it's being set to the return value of mysql_query(), then perhaps the query you are running is incorrect causing mysql_query to return false instead). — Nils Luxton, May 21 '11 at 06:34
I'm assuming, from what I've experienced recently, that either what's inside $result is not what is expected, or that there's actually nothing inside $result. Could you please provide more info? — βӔḺṪẶⱫŌŔ, May 21 '11 at 06:35
$query = "INSERT INTO `$clas` (`adm_no`, `adm_dt`, `name`, `dob`, `f_name`, `f_office`, `f_o_no`, `m_name`, `m_office`, `addr`, `pho_no`,`id`) VALUES ('$_SESSION[adm_no]', '$adm_dt', '$name', '$dob', '$f_name', '$f_office', '$f_o_no', '$m_name', '$m_office', '$addr', '$pho_no', '1');"; $result = mysql_query($query, $connection); — ravi, May 21 '11 at 06:45
Please read this question - http://stackoverflow.com/questions/332365/xkcd-sql-injection-please-explain — El Yobo, May 21 '11 at 07:37
You have an error in your query, if you want some real help, provide the query in your question — Ibu, May 21 '11 at 07:40

score 1 · Accepted Answer · answered May 21 '11 at 06:41

1

I assume $result = mysql_query() ??

Don't pass it that variable, you can pass it the connection link $variable or just use mysql_affected_rows();

answered May 21 '11 at 06:41

0xAli

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score 1 · Answer 2 · edited May 21 '11 at 07:35

1

$result must be the link identifier not the query example

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');

// will return for the most recent connection
echo mysql_affected_rows();

// will return for $link connection defined 2 rows up
echo mysql_affected_rows($link);

edited May 21 '11 at 07:35

nico

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answered May 21 '11 at 06:46

Andrei Stanca

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score 1 · Answer 3 · answered May 21 '11 at 07:29

1

if($result)
{
    mysql_affected_rows($result);

}

answered May 21 '11 at 07:29

Pradeep Singh

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score 0 · Answer 4 · answered May 21 '11 at 06:34

0

Your query did not produce a valid result.
Try echo mysql_error() before calling mysql_affected_rows().

answered May 21 '11 at 06:34

gnur

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PHP MYSQL : why i am getting this warning?

4 Answers4