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PHP - Avoiding Broken Links
If I have a web page located he www . company . com / how-to-repair-your-computer.html
And I decide to re-structure my website like this www . company . com / articles / how-to-repair-your-computer.html
How do I make sure that people don't search and end up at the old, now broken, link?
It seem inevitable that as a website grows, that you will want to re-organize things. What is the best way to make sure that anyone who searches or clicks on an old link - say from an email from a friend - doesn't get a 404 error?
Also, how do you avoid ruining a web pages rank on Google after you move things? (I think if the URL changes, Google makes you start all over as far as getting listed on page-1 and all of that?
Is this something I have to hande on my end, or is it a Google issue, or something else?
Similar TutorialsI am trying to build an app which will scan a site multple times, the only problem is the 403 error, how do I get around this. Searching seems to imply curl or user_agent, but can't get it working. Any suggestions? Thanks When user submits data through a textarea input and he then refreshes the page, the data will be re-submitted and inserted into the database a second time. I used to avoid this with a header redirect, but this solution is not an option to me anymore, since I need to echo out in the header.php, which I have to put before the submission area. Another weak point with the header redirect solution is that the user could go a page back and then hit refresh. I'm wondering how are other sites avoiding resubmission? When you post a comment on YouTube it will simply freeze the submit button after you've posted, I'm guessing they have achieved that with JavaScript, is that solution recommended? And which type of other techniques are people generally using to avoid resubmission on page refresh? Excuse the beginner question. Though, how do you avoid the processing of HTML when text becomes inputted, without stripping away the tags, or trimming the text in any way, simply leaving it as is? As the title says, I have a .txt file with about 30,000+ lines of data presented in a pipe delimited list, which i'm parsing and inserting into my database. Problem is, my server seems to always time out every time I try to parse the whole file at once. I'm sure naturally it would work with out any timeout errors, but i'm ensuring the data is xss clean before it's being inserted and i'm doing that on about 15 items on each line, which means i'm calling the xss clean function over 450,000 times in one execution. So my friend suggested I break the files down maybe into each file having 5,000 lines of code, which would mean i'd generate about 6 files (if I had 30,000 lines of data). I've managed to code a script that breaks the main file into several files. Now what I want to do is pass each of those files to my parser method, but i'd like to do them one by one, rather than in one execution as I want to avoid the timeout error. Any ideas? I've got a script where the client can upload pictures. The pictures are then resized, thumbnailed, and added to the database. In the process I'm trying to search the database for a duplicate file name and create a new name if necessary: Code: [Select] $userfile = 'userfile' . $i; $tmpLoc = $_FILES[$userfile]['tmp_name']; $name = $_FILES[$userfile]['name']; $error = $_FILES[$userfile]['error']; $type = $_FILES[$userfile]['type']; $temp = 'album' . $i; $album = $_POST[$temp]; if($error > 0) { echo "Error on $name: "; switch ($error) { case 1: echo "File exceeded upload_max_filesize"; break; case 2: echo "File exceeded max_file_size"; break; case 3: echo 'File only partially uploaded'; break; case 4: echo 'No file uploaded'; break; } echo "</div>"; exit; } // Check for name duplicates and deal with $query = "SELECT * FROM pictures WHERE src = $name"; $result = mysql_query($query); if($result) $dup = true; while($dup) { echo "Duplicate file name $name <br />"; $ext; if($type == 'image/gif') $ext = '.gif'; else if($type == 'image/jpeg') $ext = '.jpg'; else if($type == 'image/png') $ext = '.png'; else die("Error: Unsupported file type"); $x = 0; $name = $x . $ext; echo "Checking $name <br />"; $query = "SELECT * FROM pictures WHERE src = $name"; $result = mysql_query($query); if(!$result) { $dup = false; echo "File successfully renamed to $name to avoid duplicate <br />"; } $x++; } I don't get any errors of any sort, it just never enters the loop I have the following code:
$fp = fopen(“path_to_file”, ‘a’); flock($fp, LOCK_EX); fwrite($fp, $string); flock($fp, LOCK_UN); fclose($fp);If I try to lock the file in two different places at the same time, this will cause a race condition. How can I prevent this? I know in Java, for example, it has a concurrent library which contains reentrant lock, which basically tries to get the lock and if can't waits. What can I do in PHP? I have questions but I also have some good info to share about putting your software in the cloud. The situation: Some of you may have read about the nightmare stories. A developer had a infinite loop in his code that ran all night. This code did things inefficiently that devoured CPU in each iteration. This developer, greatly skilled, opened his email the next morning and saw a email bill from his cloud provider totaling for $75K. True story. Most cloud providers let you define CPU usage thresholds that, when breached, send you a warning but these thresholds, if I understand them, are per account. It would seem the claim that cloud resources are available in whatever amounts you need, CPU, disk space, enough RAM to never have to wait on a page faults, etc... the claim that the cloud provides you with infinitely elastic resources in an "all you can eat for one price" contract smells just a like a little like 💩 I did cloud development for many years with ServiceNow starting when it was a help desk and I watched it evolve into one of the best cloud development platforms out there. At one customer site I installed and managed it it out of the cloud and saw its insides and I can tell you its core code is not so terribly efficient. IMHO the cloud DOES take away 90% of a developer's worries about app performance.. If you call ServiceNow tech support and your problem is diagnosed as a performance issue with your code the first thing they will ask you is "did you follow the developer best practices"? They will politely say "sorry. Here's a link to them implying "fix your code". Questions: PHP Functions that devour CPU and where there is a better way? What PHP functions or code techniques waste CPU? I am using similar_text and it does the job but it is slow. Better way? What is the best way to measure CPU used by a PHP script or by a particular code module, defined as a set of related functions that fulfill a common purpose or by a single line of code? The purpose being to identify inefficient modules of code and improve them and even if the code is damn near perfect then at least can know what code modules are the most expensive. CPU killin users (and developers too), how can they be identified? I need to store data on cumulative CPU usage for any of the above and compare it with the free amount they give you and warn CPU hogs before they breach a threshold and generate $75K bills that were not in the plan. Any info you have on avoiding surprise $75K CPU bills from a cloud provider are welcome
It would seem the I get a php error...Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given $sql = "SELECT `pm_messages`.`conversation_id` FROM `pm_messages` GROUP BY `pm_messages`.`conversation_id` WHERE `pm_messages`.`conversation_id` = ${conversation['id']}"; $result = mysql_query($sql); $replies = mysql_fetch_assoc($result); I copied the code for password_hash at php.net:
<?php
/**
* In this case, we want to increase the default cost for BCRYPT to 12.
* Note that we also switched to BCRYPT, which will always be 60 characters.
*/
$options = [
'cost' => 12,
];
echo password_hash("rasmuslerdorf", PASSWORD_BCRYPT, $options);
?>
and changed it for use in my login page:
$options = ['cost' => 12,]; $user = mysqli_real_escape_string($db_link,$_GET['username']); $pass = password_hash($_GET['password'], PASSWORD_BCRYPT, $options); but my page keeps saying invalid user/pass. Upon echoing the $pass I find that the result changes EACH time. so I created a test page that runs the code from php.net (verbatim code) 20x and I got: [pre]
$2y$10$Nlf0J520viR4C5jd3nIdd.6M3OMKACx503Jm3PiXDYZIs.13XAheq [/pre] Is password_hash broken? or am I mistaken to think that it's supposed to return the same output everytime fror the same input? Edited March 17, 2019 by Karaethontypos corrected Much of the PHP documentation is broken into very small pages. I find that this makes it very difficult to use. Does anyone else find this? I have a hackish but useful program which takes the PHP single-file documentation and splits it into one page per extension, ensuring that links between pages work correctly. It works quite well. I wonder if anyone else would find this useful? If so, then I should have time in the next few weeks to clean it up and make it publicly available. this doesnt work and ive spent ages trying to figure it out
its the bit with else
<?php
//CORS header
header("Access-Control-Allow-Origin: *");
//Capture parameter
$create = $_POST['create'];
$fuser = $_POST['fuser'];
if (!file_exists("uploads/$fuser/$create"));
{
if ($f = fopen("uploads/$fuser/$create", 'w')) {
fwrite($f, 1);
fclose($f);
echo 'OK';
}
}
else
{
$f = fopen("uploads/$fuser/$create", 'w')
fwrite($f, 5);
fclose($f);
echo 'FAIL' ;
}
?>
this bit does work below, its until i try to do else if , or else
<?php
//CORS header
header("Access-Control-Allow-Origin: *");
//Capture parameter
$create = $_POST['create'];
$fuser = $_POST['fuser'];
if (!file_exists("uploads/$fuser/$create"));
{
if ($f = fopen("uploads/$fuser/$create", 'w')) {
fwrite($f, 1);
fclose($f);
echo 'OK';
}
}
help<html> <?php $id = $_GET['id']; $dbusername="web148-matt"; $dbpassword="matt"; $dbdatabase="web148-matt"; mysql_connect(localhost,$dbusername,$dbpassword); @mysql_select_db($dbdatabase) or die( "Unable to select database"); mysql_query("UPDATE count SET clicks=clicks+1 WHERE id='$id'"); $sql = mysql_query("SELECT link FROM count WHERE id='$id'"); $fetch = mysql_fetch_row($sql); $result = mysql_query("SELECT * FROM count"); while($row = mysql_fetch_array($result)) { echo "<a href=" .$row['link']. ">Link</a>"; } ?> <a href='http://www.google.com'>Google</a> <a href='/index.php?id=2'>link2</a> </html> So on my website I have a basic if statement that checks some arguments to see if a user can add another user as a friend. Well I had gotten that part down and for the longest time other people on my website have been able to use the feature. Now all of a sudden the if statement doesn't work? Why? Here is the statement: if ($privacy['privacy']['who_can_add'] == '1' AND $zext->user['id'] != '0' AND $zext->user['id'] != $u['id']) { $add_friend = $u['add_friend']; } of course if I put $add_friend outside the if statement, the button appears. How can a statment work one day but not the other? Is it an issue with my server? dump of $privacy: Code: [Select] $ => Array (4) ( | ['hide_o_status'] = Integer(1) 0 | ['who_can_view'] = Integer(1) 1 | ['who_can_add'] = Integer(1) 1 | ['who_can_contact'] = Integer(1) 1 ) dump of $zext->user['id']: Code: [Select] $ = String(2) "10" dump of $u['id']: Code: [Select] $ = String(2) "4" it all has correct information and the if statement has not been changed from before when it had worked and outputted $add_friend all day long. it worked until last night, i don't know what happened or why, php version has not been changed or anything. if anybody has any ideas on what's going on help would be much appreciated. Thanks, Matt. Ok mixing javascript with php.... im having bugs . I basically want to replace any broken image links with a picture "noimage.gif" in the images folder. I tried this code but am getting the error: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom.php on line 78 This is the javascript header - all seems fine: <script language="JavaScript" type="text/javascript"> function ImgError(source){ source.src = "/images/noimage.gif"; source.onerror = ""; return true; } </script> this is the code thats erroring... is it the way ive written in the code into the IMG tag? while($row = mysql_fetch_array($result)){ echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100% BORDER=0>"; echo "<TR />"; echo "<TD WIDTH=30% VALIGN=TOP />"; echo " <A HREF='images/".$row['photo']."' target=_blank><IMG SRC='images/".$row['photo']."' width=186 height=155 border=0 onerror="ImgError(this);" /></A> "; echo "<br />"; echo "</TD>"; echo "<TD WIDTH=10 VALIGN=TOP />"; Hi I am a php learner and I am having some problems while loading images from templates. I will explain everything so hope you can understand my question.. My Folder Structure WEB SITE NAME - index.php [default landing page] + Images [images folder] + css [css folder] + templates [templates folder] |-- header.inc.php [header template] |-- footer.inc.php [footer template] + includes [folder for all classes and variables] + js [folder for all js files] + contact-us [this is a folder] |--index.php [this is the file inside the contact-us folder] + about-us [this is a folder] |-- index.php [this is the file inside the about-us folder] This is the header.inc.php file without some html markups Quote <title>Web Site Name</title> <link href="css/reset.css" rel="stylesheet" type="text/css" /> <link href="css/default.css" rel="stylesheet" type="text/css" /> <script type="text/javascript" src="js/jquery-1.4.4.min.js"></script> </head> <body> <img src="images/phpmadeeasy.jpg" width="200" height="70" alt="php made easy logo" /> </body> ------------------------------------------------------------------------------ This is a sample main landing index.php page [i added codes only where i get into problems] Quote <?php include_once('includes/header.inc.php') ?> <div id="mainWrapper"> <img src="images/phpbigbanner.jpg" width="200" height="70" alt="php big banner" /> </div> <?php include_once('includes/footer.inc.php') ?> ------------------------------------------------------------------------------ As you can guess, default index.php file works fine... load both logo and banner images but... this is the index.php file located under the about-us folder Quote <?php include_once('includes/header.inc.php') ?> <div id="mainWrapper"> <img src="../images/phpbigbanner.jpg" width="200" height="70" alt="php big banner" /> </div> <?php include_once('includes/footer.inc.php') ?> ------------------------------------------------------------------------------ Here only the banner image load because the template file still show the logo image path as "images/phpmadeeasy.jpg" instead of "../images/phpmadeeasy.jpg" so is there any way me to define the default image folder as a variable so i can use that variable to load images from any directory level Example: Quote <?php echo $images; ?>images/phpmadeeasy.jpg" <script type="text/javascript" src="<?php echo $js; ?>/jquery-1.4.4.min.js" <link href="<?php echo $css; ?>reset.css" rel="stylesheet" type="text/css" Thanks........... Hi I am a php learner and I am having some problems while loading images from templates. I will explain everything so hope you can understand my question.. My Folder Structure WEB SITE NAME - index.php [default landing page] + Images [images folder] + css [css folder] + templates [templates folder] |-- header.inc.php [header template] |-- footer.inc.php [footer template] + includes [folder for all classes and variables] + js [folder for all js files] + contact-us [this is a folder] |--index.php [this is the file inside the contact-us folder] + about-us [this is a folder] |-- index.php [this is the file inside the about-us folder] This is the header.inc.php file [just a example to let you understand my problem] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "URL/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="URL/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Web Site Name</title> <link href="css/reset.css" rel="stylesheet" type="text/css" /> <link href="css/default.css" rel="stylesheet" type="text/css" /> <script type="text/javascript" src="js/jquery-1.4.4.min.js"></script> </head> <body> <img src="images/phpmadeeasy.jpg" width="200" height="70" alt="php made easy logo" /> </body> ------------------------------------------------------------------------------ This is a sample main landing index.php page [i added codes only where i get into problems] <?php include_once('includes/header.inc.php') ?> <div id="mainWrapper"> <img src="images/phpbigbanner.jpg" width="200" height="70" alt="php big banner" /> </div> <?php include_once('includes/footer.inc.php') ?> ------------------------------------------------------------------------------ As you can guess, default index.php file works fine... load both logo and banner images but... this is the index.php file located under the about-us folder <?php include_once('includes/header.inc.php') ?> <div id="mainWrapper"> <img src="../images/phpbigbanner.jpg" width="200" height="70" alt="php big banner" /> </div> <?php include_once('includes/footer.inc.php') ?> ------------------------------------------------------------------------------ Here only the banner image load because the template file still show the logo image path as "images/phpmadeeasy.jpg" instead of "../images/phpmadeeasy.jpg" so is there any way me to define the default image folder as a variable so i can use that variable to load images from any directory level Example: <?php echo $images; ?>images/phpmadeeasy.jpg" <script type="text/javascript" src="<?php echo $js; ?>/jquery-1.4.4.min.js" <link href="<?php echo $css; ?>reset.css" rel="stylesheet" type="text/css" Thanks........... $compQ = "SELECT companies.companyid, companies.companyname, companies.companylogo, companies.companyoccupation, companies.industry, eQuestions.capitalrequested FROM companies LEFT JOIN eQuestions ON companies.companyid = eQuestions.companyid"; This is not displaying data correctly. I'm assuming eQuestions.capitalrequested is not in the correct spot? OK, hello everyone from your newest newbie - to this forum anyway. First thing to say is I do not claim to be the best php coder there is - as you will see when you look at my code ! Secondly, I have scoured/googled many sources to try and understand an answer to my problem without success. So, please first look at this page - http://www.thepearsons-ws.co.uk/php/MetMonthly.php If you pick March 2011 say you see a list of data presented on the same page. This is the effect I want to get to with a GD graph. If you now try my first attempt - http://www.thepearsons-ws.co.uk/php/raingraph3monthlyselect.php - and pick March 2011 and Submit AND click the link you get a graph on a new page. OK, but now what I want - I want the image on that same first page. So, I have the attached code which produces this - http://www.thepearsons-ws.co.uk/php/raingraph3monthlyselect2.php I have hacked the code around a bit but basically it's - PHP - form and data selection HTML - form with pull downs PHP - display graph. The image is broken. If I remove the HTML block completely it produces output but of course I can not vary the selection. Any small element of HTML here destroys the image - no whitespace or such - just <html> is enough. So, any clues on how to correctly structure this code would be greatly appreciated. Regards Phil I recently changed hosts, now my image uploader which used to work fine doesn't work. $indeximage = $_FILES['indeximage']; if($indeximage) { $indeximagename = basename($_FILES['indeximage']['name']); $indeximagenew = $_SERVER['DOCUMENT_ROOT'] . '/images/uploaded/index/' . $indeximagename; if (!file_exists($indeximagenew)) { if ((move_uploaded_file($_FILES['indeximage']['tmp_name'], $indeximagenew)) === true) { echo 'Index Image uploaded to this address '; echo 'http://www.address.co.uk/images/uploaded/index/'; echo $indeximagename; echo '<br />'; }else { echo 'Unable to move Index Image into the right folder.'; } } } It now echos Unable to move Index Image into the right folder. I tried putting: ini_set("display_errors", "1"); error_reporting(E_ALL); at the beginning. Before the upload it reads: Notice: Undefined index: images in /var/www/vhosts/huhmagazine.co.uk/httpdocs/admin/images.php on line 58 Notice: Undefined index: indeximage in /var/www/vhosts/huhmagazine.co.uk/httpdocs/admin/images.php on line 62 After it reads: Warning: move_uploaded_file(): Unable to move '/tmp/phpTd67fh' to '/var/www/vhosts/huhmagazine.co.uk/httpdocs/images/uploaded/index/calidewitt.jpg' in /var/www/vhosts/huhmagazine.co.uk/httpdocs/admin/images.php on line 71 [ m ]printf[/ m]produces a link to
php.net/<span>printf
Unless you use nobbc tags, then it works fine ???
Edited by Barand, 24 November 2014 - 02:25 PM. |
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