PHP Code:

function listInvoice() {
error_reporting(E_ALL);
 $sql_text = "SELECT name FROM ".$this->TABLES['invoice']."";
 
    $the_array = Array();
    $handle = opendir('invoice'); 
    while (false !== ($file = readdir($handle))) { 
       if ($file != "." && $file != "..") {   /* as descripted below: these "files" will not be added to the array */
       $the_array[] = $file; 
   } 
}
closedir($handle); 
foreach ($the_array as $element) {
 $filename = '$element';

    if ($name == $element) {
       echo "$element \n" ;
    } else {
      $this->sql->Insert("INSERT INTO ".$this->TABLES['invoice']." (name) VALUES ('".$element."')");
    }
  }
  } 


This is my code. It simple gets the names of files and places those names into a variable which is able to be outputed.

The thing is, I need to be able to compare it against a database, to see if it already exists in the database :

PHP Code:

if ($name == $element) {
       echo "$element \n" ;
    } else {
      $this->sql->Insert("INSERT INTO ".$this->TABLES['invoice']." (name) VALUES ('".$element."')");
    } 


This is always a true statement. $element is the current name of the current file.

So, it checks the database to see if the filename is already in the database. If it is, not much happens, if it isnt, it stores the filename into the database. But at the moment, it seems as though its not getting any information from the database. For it is always false.

I have no idea why this is. Maybe $sql_text = "SELECT name FROM ".$this->TABLES['invoice'].""; is the incorrect way to pull the info from the database?

Originally Posted by lothop

I have no idea why this is. Maybe $sql_text = "SELECT name FROM ".$this->TABLES['invoice'].""; is the incorrect way to pull the info from the database?

I assume you have already created a connection to your database. Otherwise, you'll need to do that before you can execute any queries.

As far as I can tell, all you have done is assigned your query to a string variable. You have not actually executed the query.

Check out Example 1 on the following page from the PHP MySQL function reference. It will walk you through creating a basic database connection, running a simple select query, and outputting the results:

http://us2.php.net/mysql

In particular, look at the section for mysql_query

I hope that helps!

Mike

Yes, it already connects to the database.

Its just running the function. The thing is, I dont want to output any information. I just want to check to see if the filename already exists in the database.

It works for my other functions, but for some reason its as if it isnt getting the info out of the database.

Cheers for reply

I've also tried it this way.

PHP Code:

function listInvoice() {
error_reporting(E_ALL);
    $the_array = Array();
    $handle = opendir('invoice'); 
    while (false !== ($file = readdir($handle))) { 
       if ($file != "." && $file != "..") {   /* as descripted below: these "files" will not be added to the array */
       $the_array[] = $file; 
   } 
}
closedir($handle);
foreach ($the_array as $element) {
 $filename = '$element';

        if ($this->sql->Select("SELECT filename FROM ".$this->TABLES['invoice']." WHERE filename = '".$element."'")) {
               $this->sql->Insert("INSERT INTO ".$this->TABLES['invoice']." (filename) VALUES ('".$element."')");
               
                } else {
                  
                }
                }
  }
  } 


PHP Code:

if ($this->sql->Select("SELECT filename FROM ".$this->TABLES['invoice']." WHERE filename = '".$element."'")) { 


Looks at database to see if the filename already exists, if it doesnt,

PHP Code:

$this->sql->Insert("INSERT INTO ".$this->TABLES['invoice']." (filename) VALUES ('".$element."')"); 


puts the filename into the database. And if does it

PHP Code:

echo "$element \n" ; 


Ok, this does work, but not exactly how I wanted it.
It only selects the first database entry.

Basically, what I want it to do, is look at the database for a duplicate entry of $element.
If there is a duplicate entry, it does nothing.
If there isnt a duplicate entry, it inserts the $element

WOOHOO

Managed to get it working

PHP Code:

foreach ($the_array as $element) {
 $filename = '$element';
        $query = mysql_query("SELECT * FROM invoice WHERE filename = '".$element."'");
        $result = mysql_fetch_array($query);
               if(!$result){
                $this->sql->Insert("INSERT INTO ".$this->TABLES['invoice']." (filename) VALUES ('".$element."')");
                } else {
                echo "$element<br>" ;
                 
                }
                }
  }
  } 


Thanks for ur help .Mike