PHP link
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- RobertGonzalez
- Site Administrator
- Posts: 14293
- Joined: Tue Sep 09, 2003 6:04 pm
- Location: Fremont, CA, USA
Right. I have successfully generate the variable I need with this code:
But I need to include the $row['id'] variable in a link.
The link should be in the following format:
http://forum.example.com/index.php?showuser=$row['id']
But I don't know how to write that in php.
I would also like 'View your forum CP' to be shown as the 'link'
I hope someone understands me, because even I'm not sure!
Code: Select all
$query = sprintf("SELECT id FROM ibf_members WHERE name='%s'",
mysql_real_escape_string($username));
// Perform Query
$result = mysql_query($query);
// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
// Use result
// Attempting to print $result won't allow access to information in the resource
// One of the mysql result functions must be used
// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc.
while ($row = mysql_fetch_assoc($result)) {
print $row['id'];
}
// Free the resources associated with the result set
// This is done automatically at the end of the script
mysql_free_result($result);The link should be in the following format:
http://forum.example.com/index.php?showuser=$row['id']
But I don't know how to write that in php.
I would also like 'View your forum CP' to be shown as the 'link'
I hope someone understands me, because even I'm not sure!
- RobertGonzalez
- Site Administrator
- Posts: 14293
- Joined: Tue Sep 09, 2003 6:04 pm
- Location: Fremont, CA, USA
Code: Select all
<?php
$query = "SELECT id FROM ibf_members WHERE name='" . mysql_real_escape_string($username) . "'";
// Perform Query & check result
if (!$result = mysql_query($query))
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
// Use result
$user = mysql_fetch_array($result);
// Make the link
$link = 'http://forum.example.com/index.php?showuser=' . $user['id'];
// Make the markup
$href = '<a href="' . $link . '">View your forum CP</a>';
// Free the resources associated with the result set
// This is done automatically at the end of the script
mysql_free_result($result);
// Now show the markup
echo $href;
?>Had to change it a bit:
But it works fine now
Thanks!
Code: Select all
<?php
$query = sprintf("SELECT id FROM ibf_members WHERE name='%s'",
mysql_real_escape_string($username));
// Perform Query
$result = mysql_query($query);
// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
// Use result
// Attempting to print $result won't allow access to information in the resource
// One of the mysql result functions must be used
// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc.
while ($row = mysql_fetch_assoc($result))
$link = "http://forum.askscholey.com/index.php?showuser=".$row['id'];
$href = '<a href="' . $link . '">View your forum CP</a>';
echo $href;
// Free the resources associated with the result set
// This is done automatically at the end of the script
mysql_free_result($result);
?>Thanks!
- RobertGonzalez
- Site Administrator
- Posts: 14293
- Joined: Tue Sep 09, 2003 6:04 pm
- Location: Fremont, CA, USA