[PHP] Little Problem with my Guestbook
Hi. I'm new to PHP. I need some help from you, because, I'm creating a guestbook. But I have a little problem. I have a table, where the different My-SQL-Entries are read out. But now I get only one Entry, but not more. How can do it, that the table read out 4 times, when 4 entries are in the database? And the cursor from the database are jumping to the next entry. Sorry for my bad english. :) MFG Florian Paucke -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Little Problem with my Guestbook
Florian Paucke wrote: I need some help from you, because, I'm creating a guestbook. But I have a little problem. I have a table, where the different My-SQL-Entries are read out. But now I get only one Entry, but not more. How can do it, that the table read out 4 times, when 4 entries are in the database? And the cursor from the database are jumping to the next entry. Sorry for my bad english. :) Please provide the code that is causing problems so that we can understand your problem and help you fix it. -- Jasper Bryant-Greene Freelance web developer http://jasper.bryant-greene.name/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Little Problem with my Guestbook
?php
include(sql.inc.php);
include(config.inc.php);
$connection = mysql_connect($sql['host'],$sql['uid'],$sql['pwd']);
$select_db = mysql_select_db($sql['db']);
$select = mysql_query('SELECT * FROM comments');
$data = mysql_fetch_array($select);
$result = mysql_query('SELECT * FROM comments');
$rows = mysql_num_rows($result);
?
?php
echo Gauml;stebucheintrauml;ge gesamt: $rows;
?
table border=1 width=95%
tr
td
ID:
/td
td
?php echo $data[id]; ?
/td
/tr
tr
td
Name:
/td
td
?php echo $data[autor]; ?
/td
/tr
tr
td
Titel:
/td
td
?php echo $data[titel]; ?
/td
/tr
tr
td
Kommentar:
/td
td
pre?php echo $data[comment]; ?/pre
/td
/tr
/table
Thats the code i have written.
And I would like to post the table/table how often
I have SQL-entries in my database.
I hope you understand what I would like to say.
And in each cycle it ought to put out the next entry of the database.
Florian P.
Jasper Bryant-Greene [EMAIL PROTECTED] schrieb im Newsbeitrag
news:[EMAIL PROTECTED]
Florian Paucke wrote:
I need some help from you, because, I'm creating a
guestbook. But I have a little problem.
I have a table, where the different My-SQL-Entries
are read out.
But now I get only one Entry, but not more.
How can do it, that the table read out 4 times, when
4 entries are in the database?
And the cursor from the database are jumping to
the next entry.
Sorry for my bad english. :)
Please provide the code that is causing problems so that we can
understand your problem and help you fix it.
--
Jasper Bryant-Greene
Freelance web developer
http://jasper.bryant-greene.name/
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Little Problem with my Guestbook
Florian P. wrote:
?php
include(sql.inc.php);
include(config.inc.php);
$connection = mysql_connect($sql['host'],$sql['uid'],$sql['pwd']);
$select_db = mysql_select_db($sql['db']);
$select = mysql_query('SELECT * FROM comments');
$data = mysql_fetch_array($select);
$result = mysql_query('SELECT * FROM comments');
$rows = mysql_num_rows($result);
snip
No reason to run this twice
$data = array();
if ( $result = mysql_query ( 'SELECT * FROM comments' ) ) {
while ( $temp = mysql_fetch_array ( $result, MYSQL_ASSOC ) ) {
$data[] = $temp;
}
$rows = mysql_num_rows ( $result );
mysql_free_result ( $result );
} else {
echo ( mysql_error() );
exit;
}
The returned rows will be in the $data array.
--
John C. Nichel
ÜberGeek
KegWorks.com
716.856.9675
[EMAIL PROTECTED]
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