abeer Posted June 13, 2011 Share Posted June 13, 2011 I have a script which display id contet from mysql database by grabing the url value here is the script: <?php mysql_connect("localhost", "dhaka", "dhaka") or die("Connection Failed");mysql_select_db("dhaka")or die("Connection Failed");// Sanitize the page ID$id = isset($_GET['id']) ? filter_var($_GET['id'], FILTER_VALIDATE_INT) : NULL;if (is_int($id)) { $result = mysql_query("SELECT * FROM page WHERE id='$id'") or die(mysql_error()); $row = mysql_fetch_array($result); echo 'content: ' . $row['dtl'];} else { die('Possible hack attemp!');}?> it works fine.. but i have another script to edit mysql rows by text area . but i want to make it like above script which will change the id automatic by grabinng the url value <?php mysql_connect("mysql1", "splash", "splash") or die("Connection Failed"); mysql_select_db("splash")or die("Connection Failed"); if(isset($_POST['id'])) { echo $sql="UPDATE page SET dtl='".$_POST['content']."' WHERE id = ".$_POST['id']; mysql_query($sql) or die(mysql_error()); } $result = mysql_query("SELECT * from page WHERE id = 3"); $data = mysql_fetch_assoc($result); $id = $data['id']; $content = $data['dtl']; ?> <html> <head></head> <body> <form name="change_content" method="POST" action="editnow.php"> <input type="hidden" name="id" value="<?php echo $id; ?>"> <textarea name="content"><?php echo $content; ?></textarea> <input type="submit" value="change"> </form> </body> </html> 1. please help me in where ihave to make change to do this? 2. look the script in action http://www.platformsociety.org/ck.php whwn i click the change button it echos all the enterd data with row number above the text area and the text area appear without ck editor i am very new in php. please help me. Advance thanks and salute for you all Link to comment https://forums.phpfreaks.com/topic/239215-help-needed/ Share on other sites More sharing options...
RussellReal Posted June 13, 2011 Share Posted June 13, 2011 change $_POST to $_GET, $_POST is for post form data, and $_GET is for url variables. Link to comment https://forums.phpfreaks.com/topic/239215-help-needed/#findComment-1228995 Share on other sites More sharing options...
abeer Posted June 13, 2011 Author Share Posted June 13, 2011 Friend... here is the modified codes is it ok now? <?php mysql_connect("mysql1", "splash", "splash") or die("Connection Failed"); mysql_select_db("splash")or die("Connection Failed"); if(isset($_GET['id'])) {echo $sql="UPDATE page SET dtl='".$_POST['content']."' WHERE id = ".$_POST['id'];mysql_query($sql) or die(mysql_error());}$result = mysql_query("SELECT * from page WHERE id = 3");$data = mysql_fetch_assoc($result);$id = $data['id'];$content = $data['dtl'];?><html><head></head><body><form name="change_content" method="POST" action="editnow.php"><input type="hidden" name="id" value="<?php echo $id; ?>"><textarea name="content"><?php echo $content; ?></textarea><input type="submit" value="change"></form></body></html> Link to comment https://forums.phpfreaks.com/topic/239215-help-needed/#findComment-1229010 Share on other sites More sharing options...
runnerjp Posted June 13, 2011 Share Posted June 13, 2011 Where are you string ur id? Link to comment https://forums.phpfreaks.com/topic/239215-help-needed/#findComment-1229023 Share on other sites More sharing options...
abeer Posted June 13, 2011 Author Share Posted June 13, 2011 i am really new in php. please correction me where is the problem Link to comment https://forums.phpfreaks.com/topic/239215-help-needed/#findComment-1229058 Share on other sites More sharing options...
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