#49573 [Opn]: Mysqli_info doesn't return results when using INSERT .. values
| From: | uw@php.net | Date: | Wed, 16 Sep 2009 23:16:00 +0000 |
| Subject: | #49573 [Opn]: Mysqli_info doesn't return results when using INSERT .. values | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-142439@lists.php.net to get a copy of this message | ||
ID: 49573
Updated by: uw@php.net
Reported By: terryllong at gmail dot com
Status: Open
Bug Type: MySQLi related
Operating System: Windows XP
PHP Version: 5.3.0
New Comment:
It could be a bogus bug because: "Note that mysql_info() returns a
non-NULL value for INSERT ... VALUES only for the multiple-row form of
the statement (that is, only if multiple value lists are specified)",
http://dev.mysql.com/doc/refman/5.1/en/mysql-info.html
Please check what happens for the statements listed in the MySQL
manual.
Previous Comments:
------------------------------------------------------------------------
[2009-09-16 21:12:30] terryllong at gmail dot com
Description:
------------
Mysqli_info doesn't seem to return results
when using INSERT .. values
Reproduce code:
---------------
<?php
$link = mysqli_connect( 'host,'un', 'pw' ) or die("Connect to db
didn't work..."); $crlf=chr(13) .chr(10);
$test = mysqli_query($link, "CREATE DATABASE IF NOT EXISTS test") or
die("Create db didn't work...");
$test = mysqli_query($link, "USE test") or die("Use TEST db
didn't..");
$test = mysqli_query($link, "CREATE TEMPORARY TABLE IF NOT EXISTS t1
(city varchar(30))") or die("Create Temp Table didn't work...");
$data = mysqli_query($link, "INSERT INTO t1 (city) VALUES ('Tulsa')")
or die("Insert data didn't work...");
$info_from_insert = mysqli_info($link);
print_r("mysqli_info values from 'INSERT' = $info_from_insert
$crlf");
print_r("Var_Dump of mysqli_info = ");
var_dump($info_from_insert);
print_r("---------$crlf");
$data= mysqli_query( $link , "SELECT CITY FROM t1 " ) or die("Select
statement didn't...");
$rows = mysqli_affected_rows($link); echo "Rows 'Affected' = $rows
$crlf"; //- Get number of affected rows in previous MySQL operation
$select_info = mysqli_info($link);
print_r("select info: $select_info $crlf");
print_r("Var_Dump of mysqli_info = ");
var_dump($info_from_insert); echo $crlf;
$mysqli_test = mysqli_fetch_all($data, MYSQLI_ASSOC) ; echo "results
from mysqli_fetch_all..$crlf";
var_dump($mysqli_test);
?>
Expected result:
----------------
Expect text string show the results of the insert.
Actual result:
--------------
Results of Insert query returns a string(0) as shown with a Var_dump
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=49573&edit=1