Drop Down Results

[kiwis]

I have a sql qyery which i want to add the results to a drop down menu.

<?php
$link = mysql_connect (localhost, user, pass);
mysql_select_db (db12816e);
$result = mysql_query ("SELECT name From table");
while ($row = mysql_fetch_array($result));
{
?>

<select NAME="venue" size="1">
<option value="something i guess">results</option>


how do i get the results in the drop down menu???

[kows]

try this:

<select name="select">
<?
  $qry = mysql_query("SELECT name FROM table"); 
  while($arr = mysql_fetch_array($qry)){
    echo "  <option value='$arr[name']>$arr[name]\n";
  }
?>
</select>

[kiwis]

Originally posted by kows
try this:

<select name="select">
<?
  $qry = mysql_query("SELECT name FROM table"); 
  while($arr = mysql_fetch_array($qry)){
    echo "  <option value='$arr[name']>$arr[name]\n";
  }
?>
</select>

I get this error

Parse error: parse error, expecting `']'' in /home/httpd/vhosts/site.com/httpdocs/history.php on line 86

[NoteMe]

I used this the other day....worked fine for me:

$resultat1 = mysql_query("SELECT sNavn FROM Sted",$db);

                            while($arr1 = mysql_fetch_array($resultat1)){
                                $sNavn = stripslashes($arr1["sNavn"]);
                                    echo "  <option value=\"$arr1[sNavn]\">$arr1[sNavn]\n";
                              } 

[kows]

typing stuff up on the fly makes for syntax errors, you should've been able to fix it yourself.

<select name="select"> 
<? 
  $qry = mysql_query("SELECT name FROM table");  
  while($arr = mysql_fetch_array($qry)){ 
    echo "  <option value='$arr[name]'>$arr[name]\n"; 
  } 
?> 
</select>

[NoteMe]

I had to look twice to see that one.......well I guess both ways work...

[kiwis]

sweet got it working... problem is the form is
<form method="POST" action="gallery.php">

<select size="1" name="cats" style="border-style: solid; border-width: 1">


  <option selected>-- Please select a category --</option>
<?php
while ($row = mysql_fetch_array($result)) {
?>
  <option value="<?php echo $row["cat_id"];?>"><font face="Verdana" size="1"><?php echo $row["cat_name"];?></option>
<?php
}
?>
  </select>

and the gallery.php script

$result = mysql_query ("SELECT * From gallery_cats where cat_id = $cat_id"); 

this is not a vaild search option... how do i fix this??

[NoteMe]

Originally posted by kiwis
sweet got it working... problem is the form is
<form method="POST" action="gallery.php">

<select size="1" name="cats" style="border-style: solid; border-width: 1">


  <option selected>-- Please select a category --</option>
<?php
while ($row = mysql_fetch_array($result)) {
?>
  <option value="<?php echo $row["cat_id"];?>"><font face="Verdana" size="1"><?php echo $row["cat_name"];?></option>
<?php
}
?>
  </select>

and the gallery.php script

$result = mysql_query ("SELECT * From gallery_cats where cat_id = $cat_id"); 

this is not a vaild search option... how do i fix this??

If gallery_cats are the table and cat_id is an atribute, then try this:

$result = mysql_query ("SELECT * From gallery_cats where cat_id = ".$cat_id.""); 

[kows]

I suggest always putting your database search values within single quotes.

$result = mysql_query("SELECT * FROM gallery_cats WHERE cat_id='$cat_id'"); 

And, also, try using <?=$var;?> to echo your variables instead of <?php echo $var;?>. It just makes it look nice/saves time.