PHP-Problem :)

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BongoKarl

Unregistered

PHP-Problem :)

Sunday, December 14th 2003, 12:34am

Hi !

Source code

$sql_1="SELECT
           Name,
           PASSWORD
         FROM
           Users";
$result_1= MYSQL_QUERY($sql_1) OR die(mysql_error());

if(mysql_num_rows($result_1))  {
   while($row_1 = mysql_fetch_assoc($result_1)) 
 {

   if($_SESSION['Login_Name']==$row1['Name'] AND $_SESSION['Login_Password']==$row1['PASSWORD']) 
  {

  echo("<br><a href='index.php?action=addnews'>Add News</a><br><br>");

  }

 }
 }
 else
  {
   echo("Zum Newsposten bitte einloggen :)");
  }

Fehler :
Notice: Undefined index: Login_Name in i:\webserver\gm\news.php on line 21

Notice: Undefined variable: row1 in i:\webserver\gm\news.php on line 21

Notice: Undefined index: Login_Password in i:\webserver\gm\news.php on line 21

Notice: Undefined variable: row1 in i:\webserver\gm\news.php on line 21

und abfrage geht nich. "Add News" wird angezeigt

PAB

God

Posts: 4,636

Re: PHP-Problem :)

Sunday, December 14th 2003, 12:37am

while($row_1 = mysql_fetch_assoc($result_1))

if($_SESSION['Login_Name']==$row1['Name']...

BongoKarl

Unregistered

Re: PHP-Problem :)

Sunday, December 14th 2003, 12:39am

Quoted from "PAB"

while($row_1 = mysql_fetch_assoc($result_1))

if($_SESSION['Login_Name']==$row1['Name']...

haja stimmt doch , oder ?
das sagt
$row1 is das array der MYSQL Datenbank

also is $row1['Name'] die Spalte "Name" aus der Datenbank ... oder is daran was falsch ? mir gins mehr darum ,dass wenn es net definiert ist , müsste er ja das "else" ausgeben ..

PAB

God

Posts: 4,636

Re: PHP-Problem :)

Sunday, December 14th 2003, 12:44am

Du packst die Daten ausser DB in ein Array namens $row_1 und willst sie später aus nem Array $row1 auslesen,
welches laut der Fehlermeldung " Notice: Undefined variable: row1 in i:\webserver\gm\news.php on line 21" nicht existiert.

BongoKarl

Unregistered