select list with all results

[searls03]

I have never tried this before, but I must now.....how do I make a select list that will display all results from the table pictures but I only need the row "event".  make sense?  please help? 

[ManiacDan]

You need to get your terminology straight, and be a little more clear.  A "row" and  "result" are the same thing.

 

Are you saying you want a PHP script that selects a single COLUMN (not row) from a database table and spits out a select box with every value from that column?

 

What have you tried so far?  Do you know how to query a database and/or print HTML from PHP?

 

-Dan

[searls03]

sorry, here is the code that pulls things from database(ignore $content please:

<?php
include_once "connect_to_mysql.php";

// if no id is specified, list the available articles


   $query = "SELECT image, event, name, id, site FROM pictures where id='".$_GET['id']."'";
   $result = mysql_query($query) or die('Error : ' . mysql_error());
  
   // create the article list

   while($row = mysql_fetch_array($result, MYSQL_NUM))
   {
      list($image, $event, $name, $id, $site) = $row;
  
      $content .= "<input name=\"check[]\" type=\"checkbox\" value=value='".$row['userid']."' /><li><img src='$image'/></a></li>";
   }
   ?>

 

When I referred to row, I was referring to the table column from the database, not the result.  and yes, I would like it to select from one column and spit all it out.  sorry for confusion, I forget that people cant read my mind.  haha.

[lordshoa]

$row = ("SELECT `event`FROM `pictures` WHERE id='.$id.';

print_r($row);

[mikesta707]

$row = ("SELECT `event`FROM `pictures` WHERE id='.$id.';

print_r($row);

 

what is this? This really doesn't have anything to do with OP's question, and in fact has a syntax error in it. Even if you did have correct syntax, all this would do is display that $row is a string, and print the contents of the string?

 

Anyways, for OP, printing out a select box with the info you want is fairly easy. I'll give an example

 

$sql = "SELECT something FROM somewhere WHERE somecolumn='somevalue'";
$res = mysql_query($sql);
echo "<select name='someName' ... >";//print out begining select tag once
while($row = mysql_fetch_array($res)){
echo "<option value='".$row['column_you_want']."' >".$row['column_you_want']."</option>";//print each option for every row
}
echo "</select>";//print out closing select tag

 

Hope this helps