Thanks! Tell me though, what does the following mean or should I ask how
does it work?
$row->user_id;
Is it somehow relate to the following?
$row["user_id"];
I tried to look for this in the manual but without any luck... perhaps I'm
going blind ;)
SED
-----Original Message-----
From: Benjamin Munoz [mailto:[EMAIL PROTECTED]]
Sent: 17. janúar 2001 01:46
To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
Subject: RE: [PHP] MySQL - How to transfer a query resault into a
variable?
http://www.php.net/manual/en/function.mysql-fetch-object.php
Modified example from the manual above...
<?php
mysql_connect ($host, $user, $password);
$result = mysql_db_query ("database", "select * from table");
while ($row = mysql_fetch_object ($result)) {
// save database fields into variables here
$user_id = $row->user_id;
$name = $row->name;
// print them out if you want, or store them in an array to return
if this is a function
print ("<a href=user_details.php?user_id=$user_id>$name</a><br>\n");
}
mysql_free_result ($result);
?>
-Ben
-----Original Message-----
From: SED [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, January 16, 2001 5:21 PM
To: [EMAIL PROTECTED]
Subject: [PHP] MySQL - How to transfer a query resault into a variable?
Hi,
Lets say I have columns e.g.
| name | address | phone |
I want to make a list of all the names and put the result into a HTML page
where each name is a link.
When the name is clicked (the link), it will open another HTML page with
full information (name, address and phone).
I know how to send the query to list the names (and into a formatted
HTML-file), but I don't know how to put each name into a variable or a link
for the next step.
Can you point me to functions I should focus at or even a solution?
Regards,
Sumarlidi Einar Dadason
SED - Graphic Design
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