ID: 15899 Updated by: [EMAIL PROTECTED] Reported By: [EMAIL PROTECTED] -Status: Bogus +Status: Feedback Bug Type: MySQL related Operating System: linux PHP Version: 4.1.2 New Comment:
1) from the manual http://php.net/mysql_query: "Only for SELECT,SHOW,DESCRIBE or EXPLAIN statements, mysql_query() returns a new result identifier that you can pass to mysql_fetch_array() and other functions dealing with result tables." so the error checkeing for mysql_query() doesn't make sense for INSERT queries 2) if mysql_insert_id() still returns 0 after removing the error checking then you might have a mysql client lib problem when using php with mysql as an apache module, and other apache modules use mysql, too, you have to make sure php is compiled against the system-wide mysql client lib and not the one that is bundled with php by specifying '--with-mysql=/path/to/mysql/inst' when configuring php, usually path is /usr or /usr/local Previous Comments: ------------------------------------------------------------------------ [2002-03-06 10:35:43] [EMAIL PROTECTED] Hi I try to describe the problem. The MySQL warning is no problem. The result is 0 but it should be have a value not zero!!! All of these three functions should be a correct result and not 0. They gaves me false values. Please try it with an older version of PHP. ------------------------------------------------------------------------ [2002-03-06 09:54:20] [EMAIL PROTECTED] I failed to see what problem you try to describe? ------------------------------------------------------------------------ [2002-03-06 09:53:49] [EMAIL PROTECTED] The bug system is not the appropriate forum for asking support questions. For a list of a range of more appropriate places to ask for help using PHP, please visit http://www.php.net/support.php ------------------------------------------------------------------------ [2002-03-06 09:39:37] [EMAIL PROTECTED] hi all $my_sql = "INSERT INTO $table ( my_time ) values( NOW())"; if(!$my_res = mysql_query($my_sql,$my_db)) {echo "ERROR query<BR>".mysql_error($my_db);} $test1 = mysql_insert_id(); $test2 = mysql_affected_rows($my_db); echo "<p>my result $test1 $test2 <br>"; the correct result from $test1 if 1..9 and from $test2 = 1 the entry in the db is correct plz test this link http://62.72.17.147/bug.php4 http://62.72.17.147/bug.phps and the mysql db for local test's http://62.72.17.147/bug.txt sorry for my bad englich ;) Rene ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=15899&edit=1
