Help with IF statement

Sharif

Here is my code block:

if ($id == '')
{
echo ("");
}
else 
{
echo '<table width="630" border="0" align="center" cellpadding="0" cellspacing="0" class="bodyTEXT">
				<tr valign="middle"> 
					<td align="center"><a href="http://forums.onlyonxbox.net"><img src="http://forums.onlyonxbox.net/templates/subSilver/images/folder.gif" alt="Discuss In The Forums" border="0"></a>&nbsp;<a href="http://"forums.onlyonxbox.net" class="bodynav">Discuss 
						In The Forums</a>&nbsp;&nbsp;|&nbsp; <a href="printnews.php"><img src="images/mini/print.jpg" alt="Printable Version" border="0"></a> 
						<a href="printnews.php" class="bodynav">Printer 
						Friendly Version</a>&nbsp;&nbsp;|&nbsp; <a href="send.php"><img src="images/mini/email.gif" alt="Send To A Friend" border="0"></a> 
						<a href="send.php" class="bodynav">Send To A Friend</a></td>
				</tr>
			</table>';
}
?>

What this code does is that IF there is no $id given in the URL, it does not output the page options ("discuss in forums"... etc.). How can I make it so that IF the $id is not specified, OR does not exist in the database, it does not output anything?

leatherback

if(!isset($id))
  {}
  else
    {
     if($id == "")
        {
         }
         else
              {run query to database, and check wether input exists}
     }

Sharif

Okay, I dont really get that.. can you please put more detail in it? Sorry.. but not too familiar with a lot of stuff.

stolzyboy

if (!$id)
{
//when there isn't an id available
}
else
{
//output whatever you want if id is available
}

kevinsequeira

hope this answers your question

if(isSet($id)){//if $id is specified
     if ($id == ''){
	     	echo ("id is blank");
     }else{
	 		// checking if id is present in the database.
			$host = "your_host_name";
			$user = "your_user_name";
			$password = "you_password";
			$database = "your_database_name";
			$table = "your_table_name";
			mysql_connect($host,$user,$password);
			mysql_select_db($database);
	 		$query = "select * from $table where id=$id";
			$result = mysql_query($result);
			if(mysql_num_rows($result)==1){// if equal to 1 that means id is present in the database
          	echo '<table width="630" border="0" align="center" cellpadding="0" cellspacing="0" class="bodyTEXT">
                          <tr valign="middle"> 
                              <td align="center"><a href="http://forums.onlyonxbox.net"><img src="http://forums.onlyonxbox.net/templates/subSilver/images/folder.gif" alt="Discuss In The Forums" border="0"></a>&nbsp;<a href="http://"forums.onlyonxbox.net" class="bodynav">Discuss 

                              In The Forums</a>&nbsp;&nbsp;|&nbsp; <a href="printnews.php"><img src="images/mini/print.jpg" alt="Printable Version" border="0"></a> 
                              <a href="printnews.php" class="bodynav">Printer 

                              Friendly Version</a>&nbsp;&nbsp;|&nbsp; <a href="send.php"><img src="images/mini/email.gif" alt="Send To A Friend" border="0"></a> 
                              <a href="send.php" class="bodynav">Send To A Friend</a></td>

                      </tr>

                  </table>';
 		}else{//id not present
			echo "id is not present in the database";
		}
		mysql_close();
 }
}else{//id not specified
	echo "id is not specified";
}
?>

reg
kevin

Sharif

Alright guys,

I don't know if I explained it right but this is what I am trying to do:

If the value for the variable ID in the url does not exist in the database, I want to display a certain message... if there is no id given, I want to display nothing (echo ''). AGAIN, if the variable is wrong/doesn't exist in the database.. to display one error message, and if it isn't specified, then display nothing...

All I'll be outputting if everything is right... is just some HTML, nothing from the database. Am I being clear now?

superwormy

I'm sure you can figure th rest out youself...

if ($_GET['id']) {

// run query to see if id is VALID here ( ie in database or whatever )

if ($id_is_valid) {

print (" stuff to do if id is valid ");

} else {

print (" id is not valid ");

}

} else {

print (" no id specified ");

}

Sharif

A few questions:

Sharif

A few questions:

How does $_GET['id'] check the database? What is $id_is_valid?

kevinsequeira

link to this page like <a href="test.php?id=1">link 1</a>//id specified
link to this page like <a href="test.php">link 2</a>// no id specified
// test.php

if(isSet($id)){//if $id is specified
     if ($id == ''){
	     echo ("id is blank");
     }else{
          // checking if id is present in the database.
          $host = "your_host_name";
          $user = "your_user_name";
          $password = "you_password";
          $database = "your_database_name";
          $table = "your_table_name";
          mysql_connect($host,$user,$password);
          mysql_select_db($database);
          $query = "select * from $table where id=$id";
          $result = mysql_query($result);
          if(mysql_num_rows($result)==1){// if equal to 1 that means id is present in the database
    	      echo "id is present in the database";
          }else{//id not present
     	      echo "id is not present in the database";
				}
		mysql_close();
		}
}else{//id not specified
	echo "id is not specified";
}

?>

is this clear enough ???

reg
kevin

stolzyboy

$_GET['id'] is the better way for $id, and that is just getting the id
for you, once you have that, you run a query on the database to
see if the id is there or is correct, if it isn't then you can use
mysql_num_rows on your query result and do

if (mysql_num_rows($result) != 0)
{
//put your html here
}
else
{
//this is when the id is not there or is incorrect
}

kevinsequeira

thats right $_GET['id'] is a better way than using $id

reg
kevin