Can not get select menu to appear when in query.

Shinmeiryu

On this control panel I have two staff levels. (1 and 2 are the supposed values). On my add data form I have option menus that determine which author posted this data, which category it's going to, etc.

I have two queries where I need a stafflevel variable to be recognized from the staff table, and likewise with author for the news table. $author needs to pick up the records from the username variable/field from the staff table as that was already queried from login activation, and it does though only with the name that I'm logged in as and not the full list. The if statement, which I'm using to determine which level of staff is logged on, doesn't pick up anything and doesn't show that selection menu for author on the value 1.

Still not sharp with fetching arrays via mysql, but I'm trying. What's off with my syntax?

      $query = "SELECT * FROM news WHERE author='$author'";  

      $result = mysql_query($query) or die('error making query');  

      $row = mysql_fetch_array($result); 
      $author=$row['author']; 
      $author=$username;  

  $query2 = "SELECT * FROM staff WHERE stafflevel='$stafflevel'";  
  $result = mysql_query($query2) or die('error making query');  
  $row = mysql_fetch_array($result); 
  $stafflevel=$row['stafflevel']; 
     if ($stafflevel == 1) //is an admin, gets to select which author posted 
     {   
     echo "<select name='author'>";  
     echo "<option value='$author'>$author</option>'";  
     echo "</select>";  
     }  
     else  
     {  
     echo ""; //does not have the option to choose which author 
  //submits this data, will only be submitted as the name that's logged in 
     }

cgraz

      $query2 = "SELECT * FROM staff WHERE stafflevel='$stafflevel'";  

      $result = mysql_query($query2) or die('error making query');  

      $row = mysql_fetch_array($result); 
      $stafflevel=$row['stafflevel'];

shouldn't $stafflevel=$row['stafflevel']; be before you run your query referring to that variable? When you run your query, $stafflevel does not exist. You define it after your query. Move that line of code above $query2

Cgraz

Shinmeiryu

Placing it above '$query2 = ....' doesn't work either. Shouldn't the $stafflevel variable be created once selected from the staff table in $query2 that it would recognize it's being performed in the if statement? I just don't understand why the selection menu for that still won't show up.

cgraz

Well how can you use $stafflevel if you define it later?

Taht's the same as having

$query2 = "SELECT * FROM staff WHERE stafflevel=''";

cause $stafflevel has no value. I thought you were getting stafflevel from the first query. Anyways, what you should do is use the persons id

SELECT * FROM staff WHERE id='$id'

ANd make sure that you get have the persons ID load on that page. I don't know how your site is built or whatever, but if you're accessing this page via link, then pass their id as a variable in the url, then grab it, and run the query using their id.

You'd then extract $stafflevel for that persons id to determine what their level was, and use your if statement to display accordingly.

Cgraz

Shinmeiryu

Oh should have told you then that stafflevel does have a value...for my username being admin, I gave it the value 1.

cgraz

I guess i'm a little confused as to what you're trying to do. If you already defined stafflevel, then what do you need this for:

$stafflevel=$row['stafflevel'];

And how did you define stafflevel? I thought you wanted to select a certain person from the database, check what their staff level was, and display accordingly. But if stafflevel is already set to 1, why even run the query?

Also are you sure $stafflevel still has a value on THIS page? Have you tried echoing it? I'd try

$query2 = "SELECT * FROM staff WHERE stafflevel='$stafflevel'";  

   echo $query2 . "<br>\n"; // echo your query to see what it looks like
      $result = mysql_query($query2) or die('error making query');  

      $row = mysql_fetch_array($result);

Cgraz

ab4net

man, i think the best way to make levels:
make the users table like:
id |user |pass |level

so in the page save the level in sessions

and include another page regarding the level ..

$pg="pagelevel".$level.".php";
if ($level){
include('$pg');
}

try it ..

Shinmeiryu

I think I get what you're trying to say ab4me. Include this certain bit of code in a different php page in session regarding the level. Though I'm confused towards that direction, because I didn't think setting what should show up on a page according to the stafflevel my staff member is would be that complicated.

Maybe I should try at simplifying what I am trying to do.

Basically it's a news posting CMS, this was part of the code in the addnews.php file. I have two stafflevels: 1 = admin, and 2 = editor. The editor's rights are restricted compared to what admin can do. 1 is already listed as a value in admin, or so I think it should be as I added the SQL code in my database. I would think if stafflevel = 1 then it should show the function, and else would show something different (whatever restriction I want for the editor)

Basically, I'm trying to make that dropdown menu not show if you are an editor staff member logged on to the admin panel. The other problem is, I also seem to not be able to display the list of names except just the one I am logged in as in the dropdown list (only shows when take the echo calls out of the 'if' statement). I also want to make sure the editor level, when posting his news that it will be still posted under his logged in name, but that he can not select others.

I've searched through sites and forums for information on multiple queries and dropdown list functions, and am starting to delve deeper into the php manual w/ mysql_query. Though I still keep finding myself back at square one writing up the same block or something similar that was done before. Which I guess, is what I will have to do, until I figure it out. It seems that I still do have a long way to go in knowing all about php.

ahundiak

One monir point. If stafflevel is defined in the database as an integer then don't use 's in the query.

$query2 = "Select * from staff where stafflevel=$stafflevel";

Of course, you still need to have $stafflevel set to something before using it in a query.

ab4net

i got an idea for you its usefull ..
if u like to show them as a drop menu and you some thing to appears on level 1 and not to appear in level 2 do:

make each menu contains table and each row in table contains DIV then u can by javascript use visible:none, or display:none

regarding the level ..

think about it more, its nce and usefull ..

and the good in it, that in the future u can easily custumise the each level options, or add another level, like if u have now: Admin, Operator ... later u can add : Browser User

good luck