Проверка за изпълнена заявка

jaffy

Registered
Става на въпрос да проверява след заявката дали я е изпълнило.. демек нещо подобно на:


Код:
$zaiawka = mysql_query("INSERT INTO `info` (`title`,`text`) VALUES ('".$title."', '".$text."');
$nomer = mysql_num_rows($zaiawka);
if ($nomer>0){
echo "its ok";
}
else
{
echo "dude... learn php..";
}

Обаче този скрипт вади следното:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in - on line 58

+1 .. :)
 
Return Values

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.

The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.

Use mysql_num_rows() to find out how many rows were returned for a SELECT statement or mysql_affected_rows() to find out how many rows were affected by a DELETE, INSERT, REPLACE, or UPDATE statement.

mysql_query() will also fail and return FALSE if the user does not have permission to access the table(s) referenced by the query.

Код:
$zaiawka = mysql_query("INSERT INTO `info` (`title`,`text`) VALUES ('".$title."', '".$text."');
if ($zaiawka){
echo "its ok";
}
else
{
echo "dude... learn php..";
}
 
Сега го разбрах.. при вкарване на резултати не връща number rows ,а връща "да" или "не".

Благодаря ти!

+1
 

Горе