Blank PHP page, yet php working

nick_kirby

Hello there, I wrote this little script to create mysql databases on the fly to test something. My difficulty is that when I run the script, there is nothing on the page - it is completely blank.

I have a site that relies upon php, and that works fine, error reporting is on inside of php.ini, and I'm stumped. Here's my code:

<?php
$conn = @mysql_connect("localhost","nick","568610")
or die ("Sorry, could not connect to MySQL");

$rs1 = @mysql_create_db($_REQUEST['db']);

$rs2 = @mysql_list_dbs($conn);
for($row =0; $row < mysql_num_rows(rs2); $row++)

{ $list .= mysql_tablename($rs2, $row) . " ¦ "; }
?>

<html><head><title>Creating Databases</title></head>
<body>
<form action = "<?php echo($_SERVER['PHP_SELF']); ?>
method ="post">Current databases: <?php echo($list); ?>
<hr>Name:<input type = "text" name = "db">
<input type = "submit" value = "Create database">
</form>
</body>
</html>

saved as create_db.php. If someone could let me know where I am going wrong, what config I have missed, I would be exceptionally grateful. it appears only working code shows anything.

Many thanks,
Nick

herve

First : when debugging your code, you should delete all the '@' stuff on mysql functions
because it will show you errors if there's at least one.

There's a logical error in your script : your are trying to create a database before
getting the name of that database.

The line :

$rs1 = @mysql_create_db($_REQUEST['db']);

is executed , even if the $_REQUEST['db'] is not set (on the very first call of your script)

So you have to test first if that variable is set or not.

If it's not set, then you display your HTML form, otherwise you create the database.

Something like this :

<?php

$conn = @mysql_connect("localhost","nick","568610")
or die ("Sorry, could not connect to MySQL");

if (isset($_REQUEST['db']))
{
	// Second call to our script, we have a value in 'db'
	// Let's try creating the new database

	if ( mysql_create_db($_REQUEST['db'],$conn) )
	{
		// mysql_create_db return TRUE, which mean OK

		print "<html><head><title>Database created</title></head>";
		print "<body> Confirmation : Database created !</body></html>";
	}
	else
	{
		// mysql_create_db return FALSE, which means there's an error

		print "<html><head><title>Database creation error !</title></head>";
		print "<body> Database NOT created !";
		print "Mysql error message : ".mysql_error();
		print "</body></html>";		
	}
}
else
{
	// First call, we have only to display the HTML form

	$rs2 = @mysql_list_dbs($conn);

	$list = "";		
	for($row =0; $row < mysql_num_rows(rs2); $row++)
	{ $list .= mysql_tablename($rs2, $row) . " ¦ "; }


	print "<html><head><title>Creating Databases</title></head>";
	print "<body>";
	print "<form action = \"".$_SERVER['PHP_SELF']."\" method =\"post\">Current databases:"
	print $list;
	print "<hr>Name:<input type = \"text\" name = \"db\">";
	print "<input type = \"submit\" value = \"Create database\">";
	print "</form></body></html>";
}
?>

nick_kirby

Oh, that's useful: I am very new to this and was not sure what the @ indicated.

I thought this was a script which could create me a database? Was I confused or a bit dim?

nick_kirby

However, when I run the script, I get nothing on the page - nothing at all. All is returned is a blank, white page.

herve

Yes your script (rewritten like me or not) is supposed to create a database from PHP commands.

If you don't get anything it may be due to a PHP error, not shown for security reason.

Have you some control on that or not ? (dedicated server or mutualized hosting ?)

You can delete all the '@' in mysql command and add a

print mysql_error();

after each call

(after :

mysql_connect("localhost","nick","568610")
$rs2 = @mysql_list_dbs($conn);

)

so you will get a mysql error if there's one.

nick_kirby

Daft - I've changed my code to yours, removed the @ symbols, inserted the print mysql_error codes in place and still nothing.

I am hosting the server on a virtual machine on the system I am using. What is really stupid is that my own site, which relies upon php is wokring fine and is hosted in the same system.

rincewind456

Try this it works on my box,

<?php

$conn = mysql_connect("localhost","nick","568610")
or die (mysql_error("Sorry, could not connect to MySQL"));

if (isset($_POST['submit'])) {

$rs1 = mysql_create_db($_REQUEST['db']) or die(mysql_error());

} else {

$rs2 = mysql_list_dbs($conn)or die(mysql_error());

for($row =0; $row < mysql_num_rows($rs2); $row++)

$list='';

{
	$list .= mysql_tablename($rs2, $row) . " ¦ ";
}
?>

<html><head><title>Creating Databases</title></head>
<body>
<form action = "<?php echo($_SERVER['PHP_SELF']); ?>
method ="post">Current databases: <?php echo($list); ?>
<hr>Name:<input type = "text" name = "db">
<input type = "submit" value = "Create database">
</form>
</body>
</html>
<?php
}
?>

herve

Strange indeed :evilgrin:

Well, try to do it line by line.

Even if it sound silly, try first with a simple PHP script like :

<?PHP

print "hello world";

Does it work ? If yes, add the first two lines of my script, and so on, while it work.

May be the catch is outside the script itself (or not, this test will tell us)

I have not test the code so may be there's some stupid mistake inside
(c) me 🙂

herve

For instance, there's a missing ';' at the end of the line :

print "<form action = \"".$_SERVER['PHP_SELF']."\" method =\"post\">Current databases:"

nick_kirby

I've been fiddling and wondered exactly that - if the issue was not with the code proper, but with the server.

I configured hosts to point to another vhost on apache, and lo, that reverts to my first web site, not the one it should.

Now, why, despite Apache being great, can it not understand that I want my sites to be referenced by name? What switch do I make inside of httpd.conf or httpd-vhosts.conf to force this?

herve

Big subject 🙂

That's a bit far away for a PHP forum but, well, show us the relevent parts of your apache config file ?

Are you specifying two differents directories for the two virtuals hosts ?

nick_kirby

I am using XAMPP, however there appear to be major holes in it, adn frankly parts of it just don't work properly. The trouble is I would expect to get those sort of errors: missing; on line X, but I don't. I get nothing. Sites aren't even resolving properly, so there are some serious holes here.

If you guys can think of anything overnight, I'd be most grateful.

Here goes:

NameVirtualHost 192.168.1.4:80

<VirtualHost *:80>
ServerName www.niknet.homeip.net
ServerAdmin nick.kirby@freeuk.com
DocumentRoot "C:/WWW/niknet"
#ErrorLog C:/WWW/niknet/logs/error.log
#CustomLog C:/WWW/niknet/logs/access.log combined
</VirtualHost>

<VirtualHost *:80>
ServerName php
ServerAdmin bob.kirby@freeuk.com
DocumentRoot "C:/WWW/php"
</VirtualHost>

<VirtualHost *:80>
ServerName www.niknet.co.uk
ServerAdmin nick.kirby@freeuk.com
DocumentRoot C:/WWW/niknet.co.uk
ErrorLog C:/WWW/niknet.co.uk/logs/error.log
CustomLog C:/WWW/niknet.co.uk/logs/access.log combined
</VirtualHost>

<VirtualHost *:80>
ServerName www.emma-search.homeip.net
ServerAdmin nick.kirby@freeuk.com
DocumentRoot C:/WWW/emma-search
ErrorLog C:/WWW/emma-search/logs/error.log
CustomLog C:/WWW/emma-search/logs/access.log combined
</VirtualHost>

Houdini

Have you run phpinfo()? On the server as localhost, and www.emma-search.homeip.net, www.niknet.co.uk, also www.niknet.homeip.net. Simply create

 <?php phpinfo()?>

name it phpinfo.php put it in your htdocs folder (directory) and see if that runs on all the hosts especially the localhost, if not then your php is not working.

herve

This paragraph is a bit strange :

<VirtualHost *:80>
ServerName php
ServerAdmin bob.kirby@freeuk.com
DocumentRoot "C:/WWW/php"
</VirtualHost>

because it means that "php" is the Fully Qualified Domain Name of your virtual host,
but depending on how your DNS resolv is set, it will not work.

(it will try to find a host on Internet with "php" as a full name which will fail
so the web server may fall back to the default server)

Also I think that the

NameVirtualHost 192.168.1.4:80

and all the

<VirtualHost *:80>

lines should be the same.

Either

NameVirtualHost 192.168.1.4:80

<VirtualHost 192.168.1.4>
...
<VirtualHost 192.168.1.4>
...
<VirtualHost 192.168.1.4>
...

NameVirtualHost *:80

<VirtualHost >
...
<VirtualHost >
...
<VirtualHost *>
...

Cf : http://httpd.apache.org/docs/1.3/vhosts/name-based.html

nick_kirby

Hello - I nabled UseCanonicalName to On, then revamped the referenced IP address as above to *:80.

Also Apache.org recommends setting the virtual host as the first site hosted, so I added that in, and lo! Everything flew up and worked!

However, the script returns this error:

Call to undefined function mysql_create_db() in C:\WWW\php\create_db.php on line 11

	if ( mysql_create_db($_REQUEST['db'],$conn) )

Any thoughts most welcome.

bradgrafelman

If you're using PHP 5, I don't believe they have the MySQL extension supported now? Only the [man]MySQLi[/man]? Do a [man]phpinfo/man as suggested above to find out more information about MySQL support and PHP version unmber.

nick_kirby

OK, got to PHP info -

mysql.allow_persistent On On
mysql.connect_timeout 60 60
mysql.default_host no value no value
mysql.default_password no value no value
mysql.default_port no value no value
mysql.default_socket no value no value
mysql.default_user no value no value
mysql.max_links Unlimited Unlimited
mysql.max_persistent Unlimited Unlimited
mysql.trace_mode Off Off

What should these settigns be? What do they change/affect?

My username is a clone of the root user. Not sure if that's pertinent.

Many thanks,

Nick

nick_kirby

As an aside, I have found many scripts that allow for data to be entered into a table, but I have not come across one that lets a user change an entered value:

For example, I am writing a target tracking database for school, so I can log where the children are through each term relative to their previous scores, running reports on those failing below, and also monitoring the gifted and talented groups.

Basically my question is this: I enter a target of 4a for Alex, and a week later I want to change that. How would I go about changing that value? I was thinking of soemthing like

where firstname == alex and lastname == smith and year ==3 show database fields available fr editing, or provide dialogs to enter different values?