Can't get SHOW TABLES to work !

Hi,

I'm using PHP, MYSQL

My code - - -

$donor_db = "donor_database";
$new_db = "new_database";
$db_conn = mysql_connect($host, $user, $pass);

//get a list of all tables in the donor database
mysql_select_db($donor_db, $db_conn);
$query2 = "Show tables from ".$donor_db;
$result2 = mysql_query($query2);
$array2 = mysql_fetch_array($result2);

//there are 40 tables in the donor database

My Problem 1 ---
foreach ($array2 as $table_name)
{
echo $table_name."<br>";
}

//in the browser this returns the first table listed twice - -
//
//accounts
//accounts


My problem 2 - - -

while ($array2 = mysql_fetch_row($result2))
{
echo $array2[0]."<br>";
}

//in the browser this returns 39 table names excluding the first one
(accounts)

TIA - Any help much appreciated

== Quote from Robbo (ian@fds7.com)'s article
> Hi,
> I'm using PHP, MYSQL
> My code - - -
> $donor_db = "donor_database";
> $new_db = "new_database";
> $db_conn = mysql_connect($host, $user, $pass);
> //get a list of all tables in the donor database
> mysql_select_db($donor_db, $db_conn);
> $query2 = "Show tables from ".$donor_db;
> $result2 = mysql_query($query2);
> $array2 = mysql_fetch_array($result2);
> //there are 40 tables in the donor database
> My Problem 1 ---
> foreach ($array2 as $table_name)
> {
> echo $table_name."<br>";
> }
> //in the browser this returns the first table listed twice - -
> //
> //accounts
> //accounts
> My problem 2 - - -
> while ($array2 = mysql_fetch_row($result2))
> {
> echo $array2[0]."<br>";
> }
> //in the browser this returns 39 table names excluding the first one
> (accounts)
> TIA - Any help much appreciated

this is a PHP problem!
--
POST BY: lark with PHP News Reader

On Jul 16, 4:52 pm, lark <ham...@sbcglobal.net> wrote:
> == Quote from Robbo (i...@fds7.com)'s article
>
>
>
>
>
> > Hi,
> > I'm using PHP, MYSQL
> > My code - - -
> > $donor_db = "donor_database";
> > $new_db = "new_database";
> > $db_conn = mysql_connect($host, $user, $pass);
> > //get a list of all tables in the donor database
> > mysql_select_db($donor_db, $db_conn);
> > $query2 = "Show tables from ".$donor_db;
> > $result2 = mysql_query($query2);
> > $array2 = mysql_fetch_array($result2);
> > //there are 40 tables in the donor database
> > My Problem 1 ---
> > foreach ($array2 as $table_name)
> > {
> > echo $table_name."<br>";
> > }
> > //in the browser this returns the first table listed twice - -
> > //
> > //accounts
> > //accounts
> > My problem 2 - - -
> > while ($array2 = mysql_fetch_row($result2))
> > {
> > echo $array2[0]."<br>";
> > }
> > //in the browser this returns 39 table names excluding the first one
> > (accounts)
> > TIA - Any help much appreciated
>
> this is a PHP problem!
> --
> POST BY: lark with PHP News Reader- Hide quoted text -
>
> - Show quoted text -



Great, they - the PHP group tell me it's a Mysql problem - which is
why I posted here !

Thanks anyway

On Mon, 16 Jul 2007 21:24:33 +0200, Robbo <ian@fds7.com> wrote:
>> this is a PHP problem!
>>
>
> Great, they - the PHP group tell me it's a Mysql problem - which is
> why I posted here !

Indeed just a PHP-issue. Answered in comp.lang.php
--
Rik Wasmus

Am 16.07.2007, 12:08 Uhr, schrieb Robbo <ian@fds7.com>:

> Hi,
>
> I'm using PHP, MYSQL
>
> My code - - -
>
> $donor_db = "donor_database";
> $new_db = "new_database";
> $db_conn = mysql_connect($host, $user, $pass);
>
> //get a list of all tables in the donor database
> mysql_select_db($donor_db, $db_conn);
> $query2 = "Show tables from ".$donor_db;
> $result2 = mysql_query($query2);
> $array2 = mysql_fetch_array($result2);
>
> //there are 40 tables in the donor database
>
> My Problem 1 ---
> foreach ($array2 as $table_name)
> {
> echo $table_name."<br>";
> }
>
> //in the browser this returns the first table listed twice - -
> //
> //accounts
> //accounts
>
>
> My problem 2 - - -
>
> while ($array2 = mysql_fetch_row($result2))
> {
> echo $array2[0]."<br>";
> }
>
> //in the browser this returns 39 table names excluding the first one
> (accounts)
>
> TIA - Any help much appreciated
>

Concerning problem 2: If you execute both problematic parts in one script,
the first mysql_fetch_row has already fetched the first row from the
result resource, which is "accounts". It cannot turn up in the second loop
unless you use my_sql_data_seek to reposition the index for array2 before.
I am still a beginner, so I can't tell about problem 1.

--

Am 16.07.2007, 12:08 Uhr, schrieb Robbo <ian@fds7.com>:

> Hi,
>
> I'm using PHP, MYSQL
>
> My code - - -
>
> $donor_db = "donor_database";
> $new_db = "new_database";
> $db_conn = mysql_connect($host, $user, $pass);
>
> //get a list of all tables in the donor database
> mysql_select_db($donor_db, $db_conn);
> $query2 = "Show tables from ".$donor_db;
> $result2 = mysql_query($query2);
> $array2 = mysql_fetch_array($result2);
>
> //there are 40 tables in the donor database
>
> My Problem 1 ---
> foreach ($array2 as $table_name)
> {
> echo $table_name."<br>";
> }
>
> //in the browser this returns the first table listed twice - -
> //
> //accounts
> //accounts
>
>
> My problem 2 - - -
>
> while ($array2 = mysql_fetch_row($result2))
> {
> echo $array2[0]."<br>";
> }
>
> //in the browser this returns 39 table names excluding the first one
> (accounts)
>
> TIA - Any help much appreciated
>

Concerning problem 2: If you execute both problematic parts in one script,
the first mysql_fetch_row has already fetched the first row from the
result resource, which is "accounts". It cannot turn up in the second loop
unless you use my_sql_data_seek to reposition the index for array2 before.
I am still a beginner, so I can't tell about problem 1.

--
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