djp1988 Posted April 13, 2008 Share Posted April 13, 2008 Hello, I have a problem with my user rating system: require_once ('mysql_connect.php');$userid = $_SESSION['user_id']; // Make the user id a variable$query = "SELECT member_id, trip_id, note FROM note_trips WHERE member_id = $userid AND trip_id = 3"; // query if there is a result for this member id on this trip (this page)$result = mysql_query ($query, $dbc);$row = mysql_fetch_array ($result, MYSQL_BOTH);if($row){ //if there is a result to display then he has already voted on this page echo 'You have already voted'; }else{ if(isset($_GET['n'])){ // otherwise if he has not voted and the $_get has a value, get the value and set it in a variable $note = (int) $_GET['n']; if ($note <6) { $query = "INSERT INTO note_trips(note_id,trip_id,member_id,note) VALUES('',3,$userid,$note)"; // and insert that value along with the predefined id of the trip (page) and the member id } } echo '<form action="test.php" method="get" id="note" > // this is the vote form <select name="note" onchange="MM_jumpMenu(\'parent\',this,0)"> <option value="">Rate this Report</option>'; $query = "SELECT id_note FROM note_name ORDER BY id_note ASC"; $result = mysql_query ($query, $dbc); while ($row = mysql_fetch_array ($result, MYSQL_BOTH)) { echo"<option name=\"note\" value=\"test.php?n={$row['id_note']}\">{$row['id_note']}</option>\n"; } echo '</select></form>'; } Link to comment Share on other sites More sharing options...
Synook Posted April 14, 2008 Share Posted April 14, 2008 Err... what sort of problem?By the way, I think you'd find the mysql_num_rows() function useful. Link to comment Share on other sites More sharing options...
djp1988 Posted April 21, 2008 Author Share Posted April 21, 2008 I have used your suggested mysql_num_rows() to do something else, thanks ! Link to comment Share on other sites More sharing options...
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