lancey10 Posted February 20, 2006 Share Posted February 20, 2006 Ok so i made a a script with 2 files, db.php, and index.php...... here is my script:db.php:[code]<?php// Database variables$host = 'db304.perfora.net';$user = 'dbo154983898';$pass = 'Q439a.gw';$dbname = 'db1549838989';// Connect to Mysql Database$mysql = mysql_connect($host,$user,$pass);$db = mysql_select_db($dbname,$mysql);//grab rveiews varsfunction reviews($reviews) { $get_reviews = mysql_fetch_array(mysql_query("select * from reviews")); $reviews = $get_reviews[$reviews]; return $reviews;}?>[/code]index.php:[code]<?php include("db.php");access();echo ('<h4>Submit a Review</h4>');if($_POST['options']){ $title = $_POST['title']; $posted = $_POST['posted']; $story = $_POST['story']; $picturepath = $_POST['picturepath']; $quality = $_POST['quality']; $effect = $_POST['effect']; $overall = $_POST['overall']; $options = "UPDATE reviews SET title='$title',posted='$posted',story='$story',picturepath='$picturepath',quality='$quality',effect='$effect',overall='$overall'"; $update_options = mysql_query($options); echo("Reviews Submited.<br /><br />"); } $config = mysql_fetch_array(mysql_query("select * from reviews")); $title = $config['title']; $posted = $config['posted']; $story = $config['story']; $picturepath = $config['picturepath']; $quality = $config['quality']; $effect = $config['effect']; $overall = $config['overall'];; ?><form action="<? echo $_SERVER['PHP_SELF']; ?>" method="post"><table cellspacing="1" cellpadding="4" border="0" bgcolor="#f2f2f2"><tr> <td width="95">Title: </td><td width="352" style="border-right: 10px solid <? echo $bg1; ?>;"><input type="text" name="bg1" value="<? echo $title; ?>" size="20" /></td></tr><tr> <td>Posted By : </td><td style="border-right: 10px solid <? echo $bg2; ?>;"><input type="text" name="bg2" value="<? echo $posted; ?>" size="20" /></td></tr><tr> <td valign="top">Story: </td><td style="border-right: 10px solid <? echo $text; ?>;"><textarea name="text" cols="50" rows="8"><? echo $story; ?></textarea></td></tr><tr> <td>Picture Url: </td><td><input type="text" name="size" value="<? echo $picturepath; ?>" size="20" /></td></tr><tr> <td> </td> <td><b>Ratings:</b></td></tr><tr> <td>Picture Quality: </td> <td><label> <select name="select"> <option value="<? echo $quality; ?>">1</option> <option value="<? echo $quality; ?>">2</option> <option value="<? echo $quality; ?>">3</option> <option value="<? echo $quality; ?>">4</option> <option value="<? echo $quality; ?>">5</option> </select> </label></td></tr><tr> <td>Effects:</td> <td><select name="select2"> <option value="<? echo $effect; ?>">1</option> <option value="<? echo $effect; ?>">2</option> <option value="<? echo $effect; ?>">3</option> <option value="<? echo $effect; ?>">4</option> <option value="<? echo $effect; ?>">5</option> </select></td></tr><tr> <td>Overall:</td> <td><select name="select3"> <option value="<? echo $overall; ?>">1</option> <option value="<? echo $overall; ?>">2</option> <option value="<? echo $overall; ?>">3</option> <option value="<? echo $overall; ?>">4</option> <option value="<? echo $overall; ?>">5</option> </select></td></tr></table><br /><br /><input type="submit" name="options" value="Submit" /></form>[/code]now everytime i clciiksubmit, it isnt updating the mysql database, how do i go by fixing that? is there a problem wuith my code? Quote Link to comment Share on other sites More sharing options...
fenway Posted February 20, 2006 Share Posted February 20, 2006 First, don't post entire scripts. Second, your UPDATE statement has no where clause, which means you're updating all the records in the table. Third, I don't see any mysql_error() calls, so how can you determine what the problem is? Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.