Making sortable output by using drop-down selection.

[ErlendAnderson]

Hello!
I’m trying to make a selection box in html that sorts the mysql table output accordingly to the selected values.
I’ve created to menu and a button for that but I have no idea what to do next.

[code]Sorteeri:

Osakond Soetusaasta IT Number Tooterühm Mudeli nimetus SN Riigivara nr Inventaari nr Maja Ruum Vastutaja Märkus

Järjekord:

Ascending
Descending

---

[/code]

PHP code that does the searching:

[php]<?php //otsi
mysql_query (“set character_set_results=‘utf8’”);
if($_SERVER[‘REQUEST_METHOD’] == “POST”){
$query = "SELECT * FROM norse5_proov WHERE 1=1 ";

		if(!empty($_POST["osakond"])){
			$query .= " AND osakond LIKE '%".mysql_real_escape_string($_POST["osakond"])."%'";
		}
		if(!empty($_POST["soetusaasta"])){
			$query .= " AND soetusaasta LIKE '%".mysql_real_escape_string($_POST["soetusaasta"])."%'";
		}
		if(!empty($_POST["it_number"])){
			$query .= " AND it_number LIKE '%".mysql_real_escape_string($_POST["it_number"])."%'";
		}
		if(!empty($_POST["tooteruhm"])){
			$query .= " AND tooteruhm LIKE '%".mysql_real_escape_string($_POST["tooteruhm"])."%'";
		}
		if(!empty($_POST["mudeli_nimetus"])){
			$query .= " AND mudeli_nimetus LIKE '%".mysql_real_escape_string($_POST["mudeli_nimetus"])."%'";
		}
		if(!empty($_POST["sn"])){
			$query .= " AND sn LIKE '%".mysql_real_escape_string($_POST["sn"])."%'";
		}
		if(!empty($_POST["riigivara_nr"])){
			$query .= " AND riigivara_nr LIKE '%".mysql_real_escape_string($_POST["riigivara_nr"])."%'";
		}
		if(!empty($_POST["inventaari_nr"])){
			$query .= " AND inventaari_nr LIKE '%".mysql_real_escape_string($_POST["inventaari_nr"])."%'";
		}
		if(!empty($_POST["maja"])){
			$query .= " AND maja LIKE '%".mysql_real_escape_string($_POST["maja"])."%'";
		}
		if(!empty($_POST["ruum"])){
			$query .= " AND ruum LIKE '%".mysql_real_escape_string($_POST["ruum"])."%'";
		}
		if(!empty($_POST["vastutaja"])){
			$query .= " AND vastutaja LIKE '%".mysql_real_escape_string($_POST["vastutaja"])."%'";
		}
		if(!empty($_POST["markus"])){
			$query .= " AND markus LIKE '%".mysql_real_escape_string($_POST["markus"])."%'";
		}
		if(!empty($_POST["id"])){
			$query .= " AND id LIKE '%".mysql_real_escape_string($_POST["id"])."%'";
		}
	
		?>[/php]

the output:

[php] $result = mysql_query($query);
while($row = mysql_fetch_assoc($result)){
echo “

”;
echo “”.$row[“osakond”]."";
echo “”.$row[“soetusaasta”]."";
echo “”.$row[“it_number”]."";
echo “”.$row[“tooteruhm”]."";
echo “”.$row[“mudeli_nimetus”]."";
echo “”.$row[“sn”]."";
echo “”.$row[“riigivara_nr”]."";
echo “”.$row[“inventaari_nr”]."";
echo “”.$row[“maja”]."";
echo “”.$row[“ruum”]."";
echo “”.$row[“vastutaja”]."";
echo “”.$row[“markus”]."";
echo “”.$row[“id”]."";
echo “”;
echo “Kustuta”;

				echo "<form method='post' action='edit.php?action=delete&id=".$row["id"]."'><input type='submit' value='Muuda'>";
				echo "</td>";
				echo "</tr>";
			}
			mysql_free_result($result);
			?>
		</table>
		<?php
	}
	?>[/php]

Any help please?

[Thiago]

Hello friend,

So, i don’t understand really do you want to do in your code, because has many columns as name like %name , why???

But, to get the value from select you can make it…

// IN php…

if(isset($_POST[‘item’])){
$query .= " AND {$_POST[‘item’]} LIKE ‘%".mysql_real_escape_string($_POST[“item”])."%’";
}

tell me more about your project…