Create mysql database with PHP script

[Dave_Stro]

I am trying to write a php script that will enable me to create a database using an html form.

I keep getting this error:

Notice: Undefined variable: db in c:\inetpub\wwwroot\svscripts\svsitetest\mydbadmin.php on line 131
Query failed : Query was empty

This is the code:

<?php 
$conn = @mysql_connect($host,$user,$pwd);
 if($conn)
 {
   $msg = "<font color='#00aa00' size='4'><b>Welcome ".ucfirst($user)."! "
          ."You are connected to MySQL</b></font>";
 }
  else
 {
   $msg = "<font color='#ff0000' size='4'>Sorry ".ucfirst($user)."! "
         ."MySQL has No Connection.</font>";
 }

      $db = "CREATE DATABASE $newdb"; 
     $rs =  mysql_query($db) or die("Query failed : " . mysql_error()); 
     $rs1 = mysql_list_dbs($conn);
     for ($row=0; $row < mysql_num_rows($rs1); $row++)
     {
       $list .= mysql_tablename($rs1, $row)." | ";
     }
    ?>
    <p>
      <form action="<?php echo($PHP_SELF); ?>" method='post'>
       Current databases: <?php echo ($list); ?> <hr>
      Name:<input type='text' name='newdb'>
      <input type='submit' value='Create database'>
     </form>

    </p> 

register globals is turned on.

I am still learning php. I am writing this just for learning experiance. any help would be greatly apprciated.

Thank you,

Dave Strosnider

[rachel2004]

You'll need to check if a value is assigned to the variable $newdb. You can check that by doing:

echo "Value is: ".$newdb;

and seeing what you've got. In case you weren't already aware, it's a bad idea to run php with register_globals set to ON. Turn it off and get accustomed to the idea of working with it being off. Develop good habits early on.

[Dave_Stro]

Thank you rachel.

That did help. I turned the register globals off and now have the script working, but modified it a bit.

<?php 

$conn = @mysql_connect($host,$user,$pwd);
	if($conn)
	{
	  $msg = "<font color='#00aa00' size='4'><b>Welcome ".ucfirst($user)."! "
	         ."You are connected to MySQL</b></font>";
	}
  else
	{
	  $msg = "<font color='#ff0000' size='4'>Sorry ".ucfirst($user)."! "
	        ."MySQL has No Connection for you.</font>";
	}

			  if(!$_POST['newdb'])
				{
				  $rs1 = mysql_list_dbs($conn);
				  for ($row=0; $row < mysql_num_rows($rs1); $row++)
				  {
				    $_POST['list'] .= mysql_tablename($rs1, $row)." | ";
				  }

				  echo "<p><form action='".($_SERVER['PHP_SELF'])."'  method='post'>"
				       ."Current databases:".($_POST['list'])."<hr>"
				       ."Name: <input type='text' name='newdb' value='".$_POST['newdb']."'>"
			         ."<input type='submit' value='Create database'></form></p>";
				}
				else
				{
			    $db = "CREATE DATABASE {$_POST['newdb']}"; 
				  $rs =  mysql_query($db) or die("Query failed : " . mysql_error()); 
				  $rs1 = mysql_list_dbs($conn);
				  for ($row=0; $row < mysql_num_rows($rs1); $row++)
				  {
				    $_POST['list'] .= mysql_tablename($rs1, $row)." | ";
				  }

				  echo "<p><form action='".($_SERVER['PHP_SELF'])."'  method='post'>"
				       ."Current databases:".($_POST['list'])."<hr>"
				       ."Name: <input type='text' name='newdb' value='".$_POST['newdb']."'>"
			         ."<input type='submit' value='Create database'></form></p>";
				}
			?>

It works and adds the db, but I still get these errors

Notice: Undefined index: newdb in c:\inetpub\wwwroot\svscripts\svsitetest\mydbadmin.php on line 130

Notice: Undefined index: newdb in c:\inetpub\wwwroot\svscripts\svsitetest\mydbadmin.php on line 134

but once I fill out the form and hit submit it adds the new db. Just not sure why I still get the errors.

[rachel2004]

Dave, for future reference, keep in mind that when your error message specifies a line number, no one viewing your code will know what line of your code that is unless you tell us. In this case, though, your problem is pretty common for those having worked with register global = ON and now working with them off.

At any place in your code where you had been referencing $user, it will now be known to PHP as $_POST['user'] (at least in these cases from what you've posted). It looks like you started doing that, but at least in two places, you are tryng to reference $user in the ucase() function.
Make those changes and see if makes a difference for you.

Be aware, that working with the $_POST['user'] directly in your code is not a good idea as you need to do error checking before with the values. But if this is just development code as you are learning PHP, then it's okay as long as you are aware that in "real world" code, you should perform error checking.

[rachel2004]

Ooops, I posted a response, then looked at your code again and realized that it's possible that $user isn't a form variable so my previous suggestion may be in vain. If you could, let me know what lines 130 and 134 of your code are and I should be able to narrow down your problem.

[Dave_Stro]

heh...sorry didnt think about the line numbers.

they are refering to:

line 130

if(!$_POST['newdb'])

line 134

."Name: <input type='text' name='newdb' value='".$_POST['newdb']."'>"

Here is the full code with comments at the lines

<?php  

$conn = @mysql_connect($host,$user,$pwd); 
    if($conn) 
    { 
      $msg = "<font color='#00aa00' size='4'><b>Welcome ".ucfirst($user)."! " 
             ."You are connected to MySQL</b></font>"; 
    } 
  else 
    { 
      $msg = "<font color='#ff0000' size='4'>Sorry ".ucfirst($user)."! " 
            ."MySQL has No Connection for you.</font>"; 
    } 

              if(!$_POST['newdb'])  // LINE 130
                { 
                  $rs1 = mysql_list_dbs($conn); 
                  for ($row=0; $row < mysql_num_rows($rs1); $row++) 
                  { 
                    $_POST['list'] .= mysql_tablename($rs1, $row)." | "; 
                  } 

                  echo "<p><form action='".($_SERVER['PHP_SELF'])."'  method='post'>" 
                       ."Current databases:".($_POST['list'])."<hr>" 
                       ."Name: <input type='text' name='newdb' value='".$_POST['newdb']."'>" // LINE 134
                     ."<input type='submit' value='Create database'></form></p>"; 
                } 
                else 
                { 
                $db = "CREATE DATABASE {$_POST['newdb']}";  
                  $rs =  mysql_query($db) or die("Query failed : " . mysql_error());  
                  $rs1 = mysql_list_dbs($conn); 
                  for ($row=0; $row < mysql_num_rows($rs1); $row++) 
                  { 
                    $_POST['list'] .= mysql_tablename($rs1, $row)." | "; 
                  } 

                  echo "<p><form action='".($_SERVER['PHP_SELF'])."'  method='post'>" 
                       ."Current databases:".($_POST['list'])."<hr>" 
                       ."Name: <input type='text' name='newdb' value='".$_POST['newdb']."'>" 
                     ."<input type='submit' value='Create database'></form></p>"; 
                } 
            ?> 

[rachel2004]

Give this block a try:

if(!isset($_POST['newdb']))  // LINE 130
{ 
	$rs1 = mysql_list_dbs($conn); 
	$num = mysql_num_rows($rsl);
	for ($row=0; $row < $num; $row++) 
	{ 
		$list .= mysql_tablename($rs1, $row)." | "; 
	} 

echo '<p><form action="'.$_SERVER['PHP_SELF'].'"  method="post">' 
      .'Current databases:'.$list.'<hr>' 
      .'Name: <input type="text" name="newdb">' // LINE 134
      .'<input type="submit" value="Create database"></form></p>'; 
} 

I'm a bit confused where you're getting $POST['list'] from since it doesn't appear in your HTML form as an <input> field. You may want to revisit that. In the above code that I'm giving you, I used the variable $list where you had $POST['list'].

Oh, yeah, notice where I am using the single quotes and the double quotes. For consistency's sake, use double quotes for your HTML tag attributes as you see I've done. If you follow this rule in your echo statement that you're using for the HTML form, you may fix your error messages.

[Dave_Stro]

Ok. The use of the double qoutes was a huge help. It makes more sense now. It does make the code much more readable. and it will help me import html files much easier.

I also looked up isset() in the manual and I understand it checks to see if the variable is set.

I didnt fully understand the use of $POST and $GET. Turning the register globals off confused me a little, but I am figuring it out.

Ok now we have eliminated that problem another has come about.

now I get:

Notice: Undefined variable: list in c:\inetpub\wwwroot\svscripts\svsitetest\createdb.php on line 70

which is this line:

 {  

            $list .= mysql_tablename($rs1, $row)." | ";  

 }  

Thank you so much for your help. I have learned quite a bit from you today.

Now if I write it like this:

 {  

            $list = mysql_tablename($rs1, $row)." | ";  

 }  

It doesnt display the error, but it only displays the last database in the list. I could surpress the error, but I dont think that would be a good thing to do. Not sure, but it doesnt seem like that would be the best way.

Thank you for all of your help.
Dave Strosnider

[rachel2004]

Using this syntax:

.=

will concatenate (I hope I spelled that correctly) the value on the right of the operator to the variable on the left of the operator. So if you have this:

$first_name = "Dave";
$last_name =  " Strosnider";

$first_name .= $last_name; //

echo $first_name; // echos Dave Strosnider

Now, when I gave you that block to try, I made an attempt to merely convert some values to correct PHP syntax. But now that you know what .= does, you can make some further modifications.

Using $POST, $GET, $_SESSION (and there are others) may seem overwhelming right now, but you need to know them, love them, use them, live by them in PHP. I unfortunately can't explain them all right now, but the PHP manual will explain them ad nauseum. Use the keyword "superglobals" in their search engine of the php.net site and you should have some info on them.

A few more resources:
sitepoint.com (they have a 30 page tutorial on PHP and MySQL which was updated as recently as last week)
phpfreaks.com (good script respository, IMHO).

Of course, the manual has tons of user comments. Use it and and rather current book on PHP and you'll be up to speed soon enough. Best to you.

[Dave_Stro]

Rachel

For your help. Everything seems to be working well now.

I will keep reading and studying. You have given me a few other places to look over.

Thank you,
Dave Strosnider