Simple Search from Database Fails.. HELP HELP!!

[aragorn911]

Hello,
im trying to search some song lyrics from my database and display in the webpage. But somehow it fails..

Code: Select all

<form metod="POST" action="test.php">
      
       
        <label> Select Language:
        <select name="Language">
            <option value="English">English</option>
            <option value="Sinhala">Sinhala</option>
            </select>
        </label>
           
       
        <label>Search Song: <input type="text" name="criteria" /> </label>
        <input type="submit" value="Search" />
       
        </form>
        <br />

       
        <?php
      
      if(isset($_POST['submitted'])){
      
      $Language = $_POST['Language'];
      $criteria = $_POST['criteria'];
      $query = "SELECT * FROM songs WHERE category ='$criteria'";
      $result = mysql_query($query) or die();
      
      echo "<table>";
      echo "<tr> <th>Title</th> <th>Lyrics</th> </tr>";
            
      while($row = mysql_fetch_array($result)){
      
      echo "<tr><td>";
      echo $row['Title'];
      echo "</td><td>";
      echo $row['Lyrics'];
      echo "</td></tr>";
      }      
            
      echo "</table>";   


      }
      
      
      ?>

PLEASE tell me what I did wrong.. for some reason it never goes into the if statement..

[macek]

Code: Select all

$query = "SELECT * FROM songs WHERE category ='" . mysql_real_escape_string($criteria) . "'";


because, if you enter: " it's nice " that the final query is:

SELECT * FROM songs WHERE category = 'it's nice'

but better could be
WHERE category LIKE '" . mysql_real_escape_string($criteria) . "'

[aragorn911]

My problem is not returning search results.. its just that the POST thing doesn't work.. i think..
if i put an else{} for the if and echo something, the page will always display it... even if i press the submit button..

plss Help!!

[macek]

Code: Select all

<input type="submit" value="Search" name="someVeryNiceName />
...
if(isset($_POST['someVeryNiceName'])) {

}