[RESOLVED] data not showing in the option box

alighouston

hi i need some help with this code trying to get data from database and show it in option box
thanks



<HTML>
<HEAD>
 <TITLE>New Document</TITLE>
</HEAD>
<BODY>
<select name="employ_name" style="width:150px" >

<?php

$connect = mysql_connect ("localhost", "some data", "some") or die ("lost") ;


mysql_select_db('data base',$connect) ;

$employ = "SELECT employ_name FROM empoly";
$result = mysql_query($employ)
        or die ("lost: " . mysql_error());
while ($row = mysql_fetch_assoc($result)) {
           $data=$row['data'] ;
      echo '<option value="'. $data ['employ_name'] . '">' .
       '</option>' . "\r\n";
      }





  ?>



   </select>


</body>
</html>

HalfaBee

The query is not returning data it is returning employ_name

           $data=$row['employ_name'] ;

alighouston

changed that still no luck

<HTML>
<HEAD>
 <TITLE>New Document</TITLE>
</HEAD>
<BODY>
<select name="employ_name" style="width:150px" >

<?php

$connect = mysql_connect ("localhost", "some data", "some user") or die ("lost") ;


mysql_select_db('some data',$connect) ;

$employ = "SELECT employ_name FROM empoly";
$result = mysql_query($employ)
        or die ("lost: " . mysql_error());
while ($row = mysql_fetch_assoc($result)) {
           $data=$row['employ_name'] ;
      echo '<option value="'. $data ['employ_name'] . '">' .
       '</option>' . "\r\n";
      }





  ?>



   </select>


</body>
</html>

HalfaBee

$data is the variable you assigned the info to, not $data ['employ_name']

alighouston

the code is not complete its a form employ_name will be link to drop info at the right place on form processing file

alighouston

http://www.bella.firenetsolution.com/addtime.php here is the link there are 3 records in the database u will get the idea it shows there are 3 records but they never show

HalfaBee

heh, my html is crap 😉

      echo '<option value="'. $data . '">' .
       '$data</option>' . "\r\n";

http://www.w3schools.com/TAGS/tag_select.asp

Pikachu2000

Your query reads:

$employ = "SELECT employ_name FROM empoly";

I believe it should read . . . FROM employ, no? Unless you actually named the table empoly, that is.

alighouston

when i put
echo '<option value="'. $data . '">' .
'$data</option>' . "\r\n";

it fill out the 3 records coming from database with word $data and treat it as text take a look its funny

HalfaBee

<sigh>

"$data</option>"

I tend not to use single quotes.</sigh>

alighouston

now its giving me word array that means that the array thats coming from database don't have any place to temp hold it

HalfaBee

use print_r( ) to see what is happening to your variables, its called debugging 😉

alighouston

same results as before with print_r still saying array

Pikachu2000

alighouston wrote:
same results as before with print_r still saying array

Can you repost your edited code, so we can see what changes you've made to it?

alighouston


<HTML>
<HEAD>
 <TITLE>New Document</TITLE>
</HEAD>
<BODY>
<select name="employ_name" style="width:150px" >

<?php

$connect = mysql_connect ("localhost", "some data", "password") or die ("lost") ;


mysql_select_db('some data',$connect) ;

$employ = "SELECT employ_name FROM empoly";
$result = mysql_query($employ)
        or die ("lost: " . mysql_error());
while ($row = mysql_fetch_array($result)) {
       print_r ('<option value="'. $row ['employ_name'] . '">' .
       "$row </option>" . "\r\n");





      }





  ?>



   </select>


</body>
</html>

HalfaBee

    while ($row = mysql_fetch_array($result)) {
           print_r ('<option value="'. $row ['employ_name'] . '">' .
           "{$row['employ_name]} </option>" . "\r\n");
     }

That should work.

alighouston

Warning: Unexpected character in input: ''' (ASCII=39) state=1 in public_html/bella/addtime.php on line 25

alighouston

and fixed it was was missing ' after employ_name

thanks alot guys