----- Original Message -----
From: "Ray Cantwell" <[EMAIL PROTECTED]>
To: <php-general@lists.php.net>
Sent: Tuesday, February 21, 2006 1:27 PM
Subject: [PHP] Parse Error
> Hi all,
> I am a noob and super confused right now. I have some really simple code
> and i am getting an error that reads:
> *Parse error*: syntax error, unexpected T_VARIABLE in
> */var/www/mysql_up.php* on line
>
> here is the code:
>
> <html>
> <head><title>Test MySQL</title></head>
> <body>
> <!-- mysql_up.php -->
> <?php
> $host="localhost"
> $user="ray"
> $password="*****"
>
> mysql_connect($host,$user,$password) ;
> $sql="show status";
> $result = mysql_query($sql);
> if ($result == 0)
> echo "<b>Error " . mysql_errno() . ": "
> . mysql_error() . "</b>";
> else
> {
> ?>
> <!-- Table That Displays the results -->
> <table border="1">
> <tr><td><b>Variable_name</b></td><td><b>Value</b>
> </td></tr>
> <?php
> for ($i = 0; $i < mysql_num_rows($result); $i++) {
> echo "<TR>";
> $row_array = mysql_fetch_row($result);
> for ($j = 0; $j < mysql_num_fields(result); $j++)
> {/
> echo "<TD>" . $row_array[$j] . "</td>";
> }
> echo "</tr>";
> }
> ?>
> </table>
> <?php } ?>
> </body></html>
>
> I am really confused because i cannot see any obvious errors.
> any help would be appreciated.
>
> Ray.
>
>
>
>
you need to add a ; to every variable you assign...
you had:
$host="localhost"
$user="ray"
$password="*****"
it should be:
$host="localhost";
$user="ray";
$password="*****";
also i belive a better way to get the result of your query would be
mysql_connect($host, $user, $pass) or die(mysql_error());
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