The best way to reference...

[guyfromfl]

Im creating a payroll program for a local business to clock in and out.  The customer wants a drop box with the 9 employees that work for him, then the employee enters a pin as their password.

 

The problem is he wants the drop box to be formated like this:

Shmoe, Joe

 

where the lastname and first name are different rows in the db but I made them into 1 string when i populated the <select>. I would like to return the value of the empId I have in the table but I don't want to display the number-- just the name.  Is there anyway to reference empId or should i just use str_getcsv or a function like that get the employee's row from the db?

 

Thanks

[karatekid36]

Are you saying that you do not want the empId in the pull down menu?

[guyfromfl]

right he only wants lastName, firstName in the drop box, but its not really a good idea to reference the different employees by only their name....

 

kinda what i wanted was for each entry in the drop box it would say their name and then some how associate their employee number with out showing it in the drop box.

 

 

[MadTechie]

erm..

drop down menus can "display" the name but have a value of the "id"

 

i use this all the time..

[guyfromfl]

damn it. i just tried adding that tag...thats what i wanted

 

thanks

[karatekid36]

Okay here is the code that will create the pull down menu for you.  This will get the menu to pull the name from the table.  You obviously need to replace the right database info into the query.  I was having a similar issue with this because I also did not want to refer to people by their name.  I wanted to use their ID number.  I think I have thought of a way around it, but let me know if you want to know about that.

 

  <?php
       $query = "SELECT empID, CONCAT_WS(' ', last_name, first_name) AS name FROM emp ORDER BY last_name, first_name ASC";
  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) {
      while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        echo "<option value=\"{$row['user_id']}\">{$row['name']}</option>\n";
      }
    } else {
      echo "No results found";
    }
  } else {
    echo "Query failed<br />$query<br />". mysql_error();
  }
  
?>

[guyfromfl]

i like your code better 

 

thanks alot!!

 

[guyfromfl]

so how do you get the employee # on the other page?

[karatekid36]

The # is the value in the pull down menu.  You can submit it to a another table or pass the variable in any way you want.

[guyfromfl]

according to HTML i have:

 

<select>
<option value=0>mom, your</select>
<option value=1>mom, my</select>
</select>

(where 0 & 1 are the empId [primary keys from my MySQL db])

then the form is post...

 

so in formPostFile.php:

 

$mom = $_post['value']

 

 

? i cant get php to return any thing from the html tags. the problem im having is just passing the empId from 1.php to 2.php and still associating the employee names with the empId....I know how to $_POST and $_GET but not when the identifier is a tag within an <option>...

 

thanks for your help.

 

 

sorry i went to the pub for a few so that might not make sense..

[Barand]

Your select needs a name

 

<select name='employee'>
<option value=0>mom, your</select>
<option value=1>mom, my</select>
</select>

 

When form is posted,

 

$emp = $_POST['employee'];