Re: PHP5 is coming out!!!
| From: | jdwright at mmm dot com | Date: | Tue, 01 May 2001 12:29:18 +0000 |
| Subject: | Re: PHP5 is coming out!!! | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-9020@lists.php.net to get a copy of this message | ||
Hiya,
> what's wrong with the following script -
>
> <?php
>
> if (empty($hidden)) {
> displayform();
> exit;
> }
>
> $site= 'db.alphabot.f2s.com';
> $usernames = 'alphabot';
> $passwords = 'money';
> $dbName = 'alphabot';
> $connect = mysql_connect($site,$usernames,$passwords);
> mysql_select_db($dbName,$connect);
> $result = mysql_query("SELECT ID, rank, description FROM sites WHERE
description LIKE '%$query%' ",$connect);
> echo $result; <<<<<<<<< This is the problem.
$result is just a pointer to the MySQL results form the query. You must do
something like:
-------start---------
?>
<table width=100%>
<tr>
<td>ID</td>
<td>rank</td>
<td>description</td>
</tr>
while ($row = mysql_fetch_row($result) {
echo " <tr>\n <td>$row[0]</td>\n <td>$row[1]</td>\n
<td>$row[2]</td>\n
</tr>\n";
}
echo "</table>";
--------end----------
That's will print out the values for your query.
Note: I haven't checked the HTML or code. This should give you a starting point
though.
> function displayform() {
> global $PHP_SELF;
> ?>
> <form method="GET" action="<?php echo $php_self;?>">
> <input type="text" size="20" name="query">
> <input type="hidden" name="hidden" value="true">
> <input type="hidden" name="shown" value="0">
> <input type="submit" name="B1" value="Submit"><br>
> <table cellpadding="0" cellspacing="0" border="0">
> <td><input type=radio name=all checked value=no></td>
> <td><font face="Arial" size="1">Any Of the
> words</font></td>
> <td><input type=radio name=all value=yes></td>
> <td><font face="Arial" size="1">All Of The
> Words</font></td>
> </form>
> <?php
> }
> ?>
Good Luck!
Jonathan Wright..