rahulvicky00

Sorry for my bad english but i hope you would understand my query. How can i set 404 not found for database driven pages..? I am using header("HTTP/1.0 404 Not Found"); in my 404 error page and ErrorDocument 404 /404.html, whereas 404.html is my error page. i have an URL http://www.example.com/folder/my-test-for-url-95/ which is opening correctly because it has existence on database. but when i am opening http://www.example.com/folder/my-test-for-url-951/ which has existence on database is also opening without any information but i want here a 404 not found page. How can i do so. Thanks

yes i have checked but there is no error....

This seems very helpful information, but there is one confusion... where to find the php.ini file on the server. i am using window hosting for the website and using tomcat server...

When i am uploading under 2 MB file, it is successfully uploaded but when i am trying to upload bigger file then its showing failed to upload issue... my code is following....please help me to get out of this situation... <?php function UploadOne($fname) { $uploaddir = 'uploadedfiles/'; if (is_uploaded_file($fname['tmp_name'])) { $filname = basename($fname['name']); $uploadfile = $uploaddir . basename($fname['name']); if (move_uploaded_file ($fname['tmp_name'], $uploadfile)) $res = "File " . $filname . "was successfully uploaded and stored. Upload More<br>"; include ('../config.php'); mysql_query("INSERT INTO videos (id, path, name, title, content) VALUES ('$_POST[id]', 'url/$filname', '$_POST[date]', '$_POST[title]' , '$_POST[content]')"); $data = mysql_query('SELECT * FROM videos ORDER BY id DESC LIMIT 1'); $info = mysql_fetch_array($data); $res = "<strong>$info[title]</strong>," . "\n Sucessfully Uploaded. \n Upload more". "<br>" . "<a href='../videos.php'>View Videos</a>"; } else $res = "File ".$fname['name']." failed to upload."; return ($res); } ?> Thanks in advance...

Its not only images are going to store on that, there is a combination of images and text...By the way it stores the URL of the image in that column. The whole text is stored as text in the column...

My db table has columns "id" (int-11), "date(varchar-50)", "author name(varchar-100)", "text (varchar-100000)"... I am storing images + text in text field of database. Everything is going fine while i am using ckeditor to insert data into the database for that particular text field. Now when i am retrieving data from database... i am able to extract id, date, autorname successfully (i have check it while the issue came)..but when i am trying to retrieve text field which have image, that jumps to page not found error... what could be the issue... am i having problem with database or editor that i am using????

I am using CKeditor 3.2. Everything is going fine except image uploader, flash uploader and table... These content are not working.. my site is in PHP.. I have searched too much to solution on Google not found a right answer... infact, there are many results talking about filemanager folder which i was not able to find on downloaded zip. One more thing; when i am uploading images through default uploader and send it to database everything is fine, but when i am retrieving that image on some page; the image is not displaying, /"/" kind of code displaying instead of the images.... What to do to resolve this issue..... Kindly help me its very needed... Thanks in advance....

Thanks for a very quick reply but can you give me some coding, actually i am very new in PHP... A example code would be a grate help..

I need something like "The session to expire after a certain amount of inactivity? For instance, if they get up and go to lunch, when they come back they'll be logged out?"...

As my title said the question, i want to destroy my session after a particular time if user is being ideal.. currently i am using this code to define my session: session_start(); // Check, if username session is NOT set then this page will jump to login page if (!isset($_SESSION['myusername'])) { header('Location: admin.php'); }

its not working...

I am trying to make an edit page to edit any of my post...so i designed a manage post page manage-posts.php with the given code: <?php echo '<form name="frmMain" action="del1.php" method="post" OnSubmit="return onDelete();">'; $objConnect = mysql_connect("hostname","username","password") or die(mysql_error()); $objDB = mysql_select_db("dbname"); $strSQL = "SELECT * FROM text"; $objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]"); echo '<table width="600" border="1">'; echo '<tr>'; echo '<th width="91"> <div align="center">ID</div></th>'; echo '<th width="91"> <div align="center">Date</div></th>'; echo '<th width="91"> <div align="center">Title</div></th>'; echo '<th width="91"> <div align="center">Author</div></th>'; echo '<th width="30"> <div align="center">Edit</div></th>'; echo '<th width="30"> <div align="center">Select</div></th>'; echo '</tr>'; while($objResult = mysql_fetch_array($objQuery)) { ?> <tr> <td><?=$objResult["id"];?></td> <td><?=$objResult["date"];?></td> <td><?=$objResult["title"];?></td> <td><div align="center"><?=$objResult["author"];?></div></td> <td align="center"><a href="edit.php?NewsID=<?php echo $objResult["id"];?>" name="edit">Edit</a></td> <td align="center"><input type="checkbox" name="chkDel[]" value="<?=$objResult["id"];?>"></td> <input type="hidden" name="id" value="<?=$objResult["id"];?>" /> </tr> <? } echo '</table>'; echo '<input type="submit" name="btnDelete" value="Delete">'; echo '</form>'; and i designed another page edit.php to perform deletion of that particular post with the following code " <?php $objConnect = mysql_connect("hostname","username","pass") or die(mysql_error()); $objDB = mysql_select_db("dbname"); $strSQL = "SELECT * FROM text"; $objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]"); $objResult = mysql_fetch_array($objQuery); ?> <input type="hidden" name="id" value="<?=$objResult["id"];?>" /> Title : <br /><input type="text" name="title" size="100" maxlength="100" value="<?=$objResult["title"];?>"/> <br /> Date : <br /><input type="text" name="date" size="20" maxlength="12" id="TextBox" value="<?=$objResult["date"];?>"/> <br /> Author : <br /><input type="text" name="author" size="20" maxlength="100" value="<?=$objResult["author"];?>"/> <br /> <br /> <input type="submit" value="Submit" name="submit" /> </p> </form></fieldset></div> But the problem is that i am not able to get that particular post every time whenever i clicked on the respective post's edit link.. i suppose there is any issue in calling the id from the mysql... kindly suggest solutions... thanks in advance...

I am putting title in my URL whenever a new auto generated page is generates, so if i use single word tile then it is fine but when there is more that two words in the title it shows space in every word and therefor in title too. how can i add this hyphen or any other special char to my URL so it would look like? i.e. If i use title to MY FIRST WEB PAGE.php, what should i do so i can make it like MY-FIRST-WEB-PAGE.php

Thank You for your reply..i have tried it but its not working...creating a page as "Resource id #3.php"

Hi Mates, I want to know how to create auto generated pages in submit button with the existing templates...Wordpress users may easily understand what actually i mean...i am using the code that is creating a page but not getting the desired page name and giving a blank page that is without any formatting... <?php $content = <<<EOL <head> </head> <body> New page EOL; include ('config.php'); mysql_select_db("$db_name"); $result = mysql_query("SELECT * FROM text ORDER BY ID") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { echo $row['title']; echo "<br />"; } $content .= <<<EOL </body> </html> EOL; $file = '$result' . '.php'; $open = fopen($file, "w"); fwrite($open, $content); fclose($open); ?>

Thanks for your reply... My code is looks like: <html> <head> <script type="text/javascript" src="/jwplayer/jwplayer.js"> </script> </head> <body> <?php $link = mysql_connect("localhost", "dbname", "password"); if(!$link) { die("Connection Lost" . mysql_error()); } mysql_select_db ("635177_vik", $link); $data = mysql_query('SELECT * FROM vik ORDER BY id DESC LIMIT 1'); $info = mysql_fetch_array($data); echo "<video class="jwplayer" src=".$info['path']." width="252" height="250"></video>"; mysql_close($link); ?> </body> </html> it showing an error :: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /srv/disk3/635177/www/indiavsaustralia.in/vup/acha/flash.php on line 16

Thanks for your reply... but the problem is still.. i am clearing my scenario here. i have uploaded a video in my sites in a specified folder and stored the URL in database. now i want to retrieve that particular URL to my video player with the help of "video tag". it requires "src attribute" in that i want to retrieve the URL so every time when the new video have been uploaded that video will run in the video player.. please send some suggestions...

Hi Mates, I am very new to PHP so this forums too. i have stored a URL in database as VARCHAR, now i want to retrieve that URL in src attribute..is it possible in php to retrieve that URL in src so i can get the latest URL running with the tag? if yes then how to get that.