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pull data from DB to a drop down list ..

Posted on 2002-05-09
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Last Modified: 2011-04-14
Hi,

I'm a ASP programmer.. and I"m very new in php.
May I know how to pull data from a table and output it in a drop down list ??

<select name=glass>
<?php
   $db = mysql_connect("localhost", "root", "");
   mysql_select_db("Optics",$db);
   $result=mysql_query("select glass from Glass");

// from here how to continue ??

?>
</select>


I found that php got no EOF() function .. how to use loop to display the data in a drop down list ?? Please advice..

Thanks
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Question by:suelow
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7 Comments
 
LVL 4

Expert Comment

by:lokeshv
ID: 6998507
after that

<?whiler($row=mysql_fetch_array($qid)){?>

<option value="<?=$row[field1]?>"><?=$row[field2]?></option>

<?}?>
</select> //end of ur select


Hope this helps...

LK
0
 
LVL 32

Expert Comment

by:Batalf
ID: 6998601
Adjust lokeshv code to your like this:

<select name=glass>
<?php
  $db = mysql_connect("localhost", "root", "");
  mysql_select_db("Optics",$db);
  $result=mysql_query("select glass from Glass");
  // Data found in the resultset?
  if($info = mysql_fetch_array($result))
  {
     // Loop through each row in the resultset.
     do
     {
           echo "<option value=\"".$info["glass"]."\">".$info["glass"]."</option>\n";

     }while($info = mysql_fetch_array($result));
  }
?>
</select>
0
 
LVL 1

Accepted Solution

by:
Rajko earned 200 total points
ID: 6998612
<html>
<head>
<title></title>
</head>
<body>
<?php
$host = "localhost";
$user = "root";
$pass = "0";
$dbase = "your_database";
$db = mysql_connect($host, $user, $pass) or die ("verbindung fehlgeschlagen");
      mysql_select_db($dbase, $db) or die ("verbindung zur datenbank fehlgeschlagen");

$sql = "select * from test_table";
$result = mysql_query($sql,$db);
?>
<select name="list">
<?php
if($result){
 while($row = mysql_fetch_array($result)){
  echo "<option value=\"".$row['column_name']."\">".$row['column_name']."</option>\n";
 }
}
?>
</select>
</body>
</html>

it's you write the tag attruibute in dopple qoutes und you define your php as <?php ?> not <? ?>
xml use the same tag! <?xml ?>. the parser can't know whta language is mean, xml or php.
0
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LVL 32

Expert Comment

by:Batalf
ID: 6999152
Rajko: You should NOT propose answer on something that does the same as the comment posted previously. Now, you have locked the question!

Why not just post comments and let the user asking the question evaluates them and then accept the one he likes best as an answer???

Batalf
0
 
LVL 4

Expert Comment

by:lokeshv
ID: 7007425
strange....

btw wat did u find in his code ?

Lk
0
 
LVL 32

Expert Comment

by:Batalf
ID: 7007856
I agree with you lokeshv. The only result of this question is as far as I can see that Rajko could continue his annoying habit of proposing answers on question whene he shouldn't.
0
 
LVL 4

Expert Comment

by:lokeshv
ID: 7007869
Yeah Baltaf yor are right.

this is really annoying somone comes and directly post the answer when already two comments have been posted into the quesation and his answer is nothing diferent..


Lk
0

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