parse error expecting "(" in line 93

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  • drgnhiker
    New Member
    • Jan 2010
    • 5

    parse error expecting "(" in line 93

    And I seem to be missing the error...can't find it and the statements in one section duplicate other sections with no error....any help would be greatly appreciated: the following is the entire code...(Yes I know it is from a book but I got to learn it all from someplace :).....thanks in advance for any and all help and advice.
    Code:
    <?php
    $db = mysql_connect('localhost', 'bp6am', '*****') or
        die ('Unable to connect. Check your connection parameters.');
    mysql_select_db('moviesite', $db) or die(mysql_error($db));
    
    if ($_GET['action'] == 'edit') {
    //retrieve the record's information
       $query = 'SELECT
           movie_name, movie_type, movie_leadactor, movie_director
         FROM
           movie
         WHERE
           movie_id =  ' . $_GET['id'];
      $result = mysql_query($query, $db) or die(mysql_error($db));
      extract(mysql_fetch_assoc($result));
    } else{
      //set values to blank
      $movie_name = '';
      $movie_type = 0;
      $movie_year = date('Y');
      $movie_leadactor = 0;
      $movie_director = 0;
    }
    ?>
    <html>
     <head>
      <title><?php echo ucfirst($_GET['action']); ?> MovieAdd Movie</title>
     </head>
      <body>
       <form action="commit.php?action=<?php echo $_GET['action']; ?>&type=movie" method="post">
        <table>
         <tr>
          <td>Movie Name</td>
          <td><input type="text" name="movie_name" value="<?php echo $movie_name; ?>"/></td>
         </tr><tr>
          <td>Movie Type</td>
          <td><select name="movie_type">
    <?php
    //select movie type information  line 39
    $query = 'SELECT
           movietype_id, movietype_label
         FROM
           movietype
         ORDER BY
           movietype_label';
    $result = mysql_query($query, $db) or die(mysql_error($db));
    
    //populate the select options with the results
    while ($row = mysql_fetch_assoc($result)) {
         foreach ($row as $value){
             if ($row['movietype_id'] == $movie_type) {
                 echo '<option value="' . $row['movieytpe_id'] .'" selected="selected">';
             } else {
                 echo '<option value="' . $row['movietype_id'] . '">';
    
             }
             echo $row['movietype_label'] . '</option>';
             }
    }
    ?>
          </select></td>
         </tr><tr>
          <td>Movie Year</td>
          <td><select name="movie_year">
    <?php
    // populate the select options with years line 66
    for ($yr = date("Y"); $yr >= 1960; $yr--) {
         if ($yr == $movie_year) {
             echo '<option value="' . $yr . '" selected="selected">' . $yr . '</option>';
         } else {
             echo '<option value="' . $yr . '">' . $yr . '</option>';
         }
    }
    ?>
         </select></td>
        </tr><tr>
         <td>Lead Actor</td>
         <td><select name="movie_leadactor">
    <?php
    //select actor records
    $query = 'SELECT
           people_id, people_fullname
        FROM
           people
        WHERE
           people_isactor = 1
        ORDER BY
           people_fullname';
    $result = mysql_query($query, $db) or die(mysql_error($db));
    
    //populate the select options with the results line 91
    while ($row = mysql_fetch_assoc($result)) {
          for each ($row as $value) {
              if ($row['people_id'] == $movie_leadactor) {
              echo '<option value="' . $row['people_id'] . '" selected="selected">';
          } else {
              echo '<option value="' . $row['people_id'] . '">';   
          }
          echo $row['people_fullname'] . '</option>';
       }
    }
    ?>
          </select></td>
         </tr><tr>
         <td>Director</td>
         <td><select name="movie_director">
    <?php
    
    //select director records line 109
    $query = 'SELECT
           people_id, people_fullname
         FROM
           people
         WHERE
           people_isdirector = 1
         ORDER BY
           people_fullname';
    $result = mysql_query($query, $db) or die(mysql_error($db));
    
    //populate the select options with the results
    while ($row = mysql_fetch_assoc($result)) {
         foreach ($row as $value) {
            if ($row['people_id'] == $movie_director) {
                echo '<option value="' . $row['people_id'] . '" selected="selected">';
            } else {
                echo '<option value="' . $row['people_id'] . '">';
            }
            echo $row['people_fullname'] . '</option>';
         }
    }
    ?>
          </select></td>
         </tr><tr>
          <td colspan="2" style="text-align: center;">
    <?php 
    if ($_['action'] == 'edit') {
          echo '<input type="hidden value="' . $_GET['id'[ . '" name="movie_id" />';
    }
    ?>
           <input type="submit" name="submit" value="<?php echo ucfirst($_GET['action']); ?>" />
          </td>
         </tr>
        </table>
       </form>
      </body>
    </html>
    Last edited by Atli; Feb 12 '10, 03:56 PM. Reason: Removed the MySQL password from the code.
  • zorgi
    Recognized Expert Contributor
    • Mar 2008
    • 431

    #2
    I didn't try to understand the logic of your code, so this might not be the only problem with it, but line 93:
    Code:
    for each ($row as $value)
    should be

    Code:
    foreach ($row as $value)

    Comment

    • drgnhiker
      New Member
      • Jan 2010
      • 5

      #3
      Oh my gosh.......spac e removed....repl aced a missing comma and all is fine in the world.....Thank s you so much

      Comment

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