Bug #15899 Updated: RC from INSERT INTO if incorrect
| From: | php-bugs at lists dot php dot net | Date: | Sun, 02 Jun 2002 04:00:07 +0000 |
| Subject: | Bug #15899 Updated: RC from INSERT INTO if incorrect | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-9131@lists.php.net to get a copy of this message | ||
ID: 15899
Updated by: php-bugs@lists.php.net
Reported By: rene.wunderlich@milaro.net
-Status: Feedback
+Status: No Feedback
Bug Type: MySQL related
Operating System: linux
PHP Version: 4.1.2
New Comment:
No feedback was provided for this bug for over a month, so it is
being suspended automatically. If you are able to provide the
information that was originally requested, please do so and change
the status of the bug back to "Open".
Previous Comments:
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[2002-03-06 10:55:49] hholzgra@php.net
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
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[2002-03-06 10:35:43] andreas@milaro.net
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.
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[2002-03-06 09:54:20] mfischer@php.net
I failed to see what problem you try to describe?
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[2002-03-06 09:53:49] mfischer@php.net
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
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[2002-03-06 09:39:37] rene.wunderlich@milaro.net
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
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Edit this bug report at http://bugs.php.net/?id=15899&edit=1