PHP Forum

**Justin-Credible** · 04-20-2005, 07:12 PM

I'm new to working with MySQL through PHP. I was following this tutorial http://www.freewebmasterhelp.com/tutorials/phpmysql/4

I was able to create a database and enter information into with though PHP scripting, but I'm stuck on outputting it (see link).

This is the error I get when I browse to my out.php page:

Quote:

Warning: mysql_query(): Unable to save result set in D:\websites\wwwroot\temp\out.php on line 9

Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in D:\websites\wwwroot\temp\out.php on line 11

Database Output

If I log into MySQL from a command line and type "SELECT * FROM contacts;" it spits out the entries in the database:

Quote:

mysql> SELECT * FROM contacts;
+----+-------+-------+--------------+--------------+--------------+-------------
------------+--------------------------+
| id | first | last | phone | mobile | fax | email
| web |
+----+-------+-------+--------------+--------------+--------------+-------------
------------+--------------------------+
| 1 | John | Smith | 01234 567890 | 00112 334455 | 01234 567891 | johnsmith@go
wansnet.com | http://www.gowansnet.com |
+----+-------+-------+--------------+--------------+--------------+-------------
------------+--------------------------+
1 row in set (0.00 sec)

Any ideas?

**leavethisplace** · 04-20-2005, 07:29 PM

that looks more like a PHP error to me, which means your coding may have something wrong with it. We can help you a little better if you post your code, im not promising anything though!

**Justin-Credible** · 04-20-2005, 07:38 PM

I copy and pasted everying from out.php from the tutorial page and just changed the info for the database/username/password:

OUT.PHP

PHP Code:


			
<?

$username="Justin";

$password="xxxxxx";

$database="justin";



mysql_connect("localhost",$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT * FROM contacts";

$result=mysql_query($query);



$num=mysql_numrows($result);



mysql_close();



echo "<b><center>Database Output</center></b><br><br>";



$i=0;

while ($i < $num) {



$first=mysql_result($result,$i,"first");

$last=mysql_result($result,$i,"last");

$phone=mysql_result($result,$i,"phone");

$mobile=mysql_result($result,$i,"mobile");

$fax=mysql_result($result,$i,"fax");

$email=mysql_result($result,$i,"email");

$web=mysql_result($result,$i,"web");



echo "<b>$first $last</b><br>Phone: $phone<br>Mobile: $mobile<br>Fax: $fax<br>E-mail: $email<br>Web: $web<br><hr><br>";



$i++;

}



?>

I also noticed the page for inputing values to the database through a form on an HTML page doesn't work. When I click submit I get the standard "this page cannot be displayed" message.

Page.html:

HTML Code:

<html>
<body>
<form action="insert.php" method="post">
First Name: <input type="text" name="first"><br>
Last Name: <input type="text" name="last"><br>
Phone: <input type="text" name="phone"><br>
Mobile: <input type="text" name="mobile"><br>
Fax: <input type="text" name="fax"><br>
E-mail: <input type="text" name="email"><br>
Web: <input type="text" name="web"><br>
<input type="Submit">
</form>
</body>
</html>

Insert.php:

PHP Code:


			
<?

$username="Justin";

$password="xxxxx";

$database="justin";



$first=$_POST['first'];

$last=$_POST['last'];

$phone=$_POST['phone'];

$mobile=$_POST['mobile'];

$fax=$_POST['fax'];

$email=$_POST['email'];

$web=$_POST['web'];



mysql_connect("localhost",$username,$password);

@mysql_select_db($database) or die( "Unable to select database");



$query = "INSERT INTO contacts VALUES ('','$first','$last','$phone','$mobile','$fax','$email','$web')";

mysql_query($query);



mysql_close();

?>

**leavethisplace** · 04-20-2005, 07:47 PM

Hmm, im not sure if that PHP code is completely incorrect for the query, but it certainly seems an odd way to go about it, try using this instead:

PHP Code:


			
$query = "SELECT * FROM contacts"; 

$result = mysql_query($query); 



echo "<b><center>Database Output</center></b><br><br>"; 



/* Line below will go through every row of the result, and put them in the array, using the 

cell names as a key, so that they can be called using $row['cellname']; */



while ( $row = mysql_fetch_array($result) ) {



$first = $row['first'];

$last= $row['second'];

$phone = $row['phone'];

$mobile = $row['mobile'];

$fax = $row['fax'];

$email = $row['email'];

$web = $row['web'];

echo "<b>$first $last</b><br>Phone: $phone<br>Mobile: $mobile<br>Fax: $fax<br>E-mail: $email<br>Web: $web<br><hr><br>"; 



} 



// Close mySQL when you've finished with it

mysql_close();

That should sort it out, but do post back if you have problems. If this does fix it, then slam that tutorial and check out this one that i recommend every single time!

**Justin-Credible** · 04-21-2005, 04:51 PM

I tried using the code you gave me and got this error:

Code:

Warning: mysql_query(): Unable to save result set in D:\websites\wwwroot\temp\out2.php on line 12

Database Output



Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\websites\wwwroot\temp\out2.php on line 19

I'm assuming this means my coding is OK, but it seems like PHP or MySQL is having issues here. I'm not sure what to do though because we have other PHP/MySQL applications working perfectly on this same server.

**Justin-Credible** · 04-21-2005, 05:08 PM

I tried the tutorial on a different server and it worked no problem. I guess ISS is locked down too tightly on the first server.

Has anyone had this problem before?

**Gaffer Sports** · 04-21-2005, 10:30 PM

IIS can be a right pain in the behind, simply because the owner did not think of PHP.

I have had experiences of Dedicated servers and they are great for certain things and a nightmare for others. If you go for a host that offers a Windows package, make sure they know how to configure it or you can have plenty of downtime.

Likewise, if you decide to dedicate your time to a server, make sure you read up on it first as it is not as simple as linux.

Steve.

**Justin-Credible** · 04-22-2005, 04:29 PM

Both of the servers are my own, I host my own stuff.

I ran the ISS lockdown wizard to unlock all the restrictions that I had put in place (which unfortunately deleted all of my virtual directories

). Anyways after I got everything back up and running I got the PHP fixed. There are still some weird things happening, but I'm getting closer.

**merlin** · 04-22-2005, 04:39 PM

I might be right out to lunch but I've had this error before and found if I put a "@" in front of the mysql_fetch_array it corrected the problem. On the other hand I could be completely missing the point and not helping at all.

You could try this:

PHP Code:


			
while ( $row = @mysql_fetch_array($result) ) {

**leavethisplace** · 04-22-2005, 10:34 PM

mysql_fetch_array() is an array anyway isn't it? So putting the @ there isn't going to be much difference?

Oberon needed here!