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PHP - Getting Errors On Code I Have Used Before (page View Count)
Hey everyone,
So here is my problem. I have some code to display the amount of views that page has got. In this case it is the thread in my forums section. I have used the same code to show how many people have views a certain persons profile page and that works fine but when I use it on my forum thread page I get this error. Quote You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE id='20' LIMIT 1' at line 1 Here is the section of code: Code: [Select] <?php $thread_id = preg_replace('#[^0-9]#i', '', $_GET['id']); $getThreadViews = mysql_query("SELECT view_count FROM forum_posts WHERE id='$thread_id' LIMIT 1") or die (mysql_error()); $row = mysql_fetch_assoc($getThreadViews); $counter = $row['view_count']; if($counter == 0){ $counter = 1; $startCounter = mysql_query("INSERT INTO forum_posts (view_count) VALUES ('$counter') WHERE id='$thread_id' LIMIT 1") or die (mysql_error()); } $threadViews = $counter+1; $appendCounter = mysql_query("UPDATE forum_posts SET view_count='$view_count' WHERE id='$thread_id'") or die (mysql_error()); ?> I have checked that there are no spelling errors so just wanted to show it to a fresh pair of eyes because its really starting to annoy me. Thanks in advance for any help. Similar TutorialsAlright guys i have an idea on what am going to do here, although i am unsure that it's the best way to go about it. I am creating a blog and i need to count how many views an article gets in order to display the top 10 most viewed articles. Here is my plan... 1. Add an additional field to my blog posts table called viewed 2. On the view article page have an update method that adds one to the current number on the the coloum viewed 3. When displaying the 10 most viewed articles query it by the heightest number and limet to 5 Hello all, i have a feeling im doing something wrong, but i have no idea what. being server-side code, php should not show when you 'view page source' in your browser (or rather it should display as html), correct ? why, then am i seeing php code when i view page source? see attached image I have a strange problem. When a guest visits my contact-user.php page, they get a message telling them the must login before viewing the page. After the guest logs in, they view the same page and it tells them they have to login again (keeps on looping). But if they manually refresh that page with the "you must be logged in" message, it recognizes the login and lets them in. How can I get this page to immediately recognize that the user is logged in and not require them to refresh the page manually? Here is my code for contact-user.php <?php session_start(); header("Cache-Control: private, max-age=10800, pre-check=10800"); header("Pragma: private"); header("Expires: " . date(DATE_RFC822,strtotime("+2 day"))); include("connection.php"); mysql_select_db("database"); if (isset($_SESSION['username'])) { ******** MY HTML PAGE CONTENT ******** } else { echo "<meta http-equiv='REFRESH' content='2;url=http://www.mysite.com/login.php'> <center><font color='#EE0000'><p>You must be logged in before negotiating. You will now be redirect to the login page.</p></font></center>"; } ?> Here is my code for login.php script: <?php include("connection.php"); mysql_select_db("database"); session_start(); if(isset($_POST['login'])){ $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $tUnixTime = time(); $sGMTMySqlString = gmdate("Y-m-d H:i:s", $tUnixTime); if (!$username || !$password) { print "Please fill out all fields."; exit; } $logres = mysql_num_rows(mysql_query("SELECT * FROM members WHERE username = '$username' and password = '$password'")); if ($logres <= 0) { print "Login failed. If you have not already, please signup. Otherwise, check your spelling and login again."; exit; } else { $_SESSION['username'] = $username; if (isset($_SESSION)) { echo'You are now logging in'; mysql_query("UPDATE members SET activity = '$sGMTMySqlString' WHERE username = '$username'"); } else { echo "You are not logged in!"; } echo'<html><head><meta http-equiv="REFRESH" content="1;url=http://www.mysite.com/members/' . $_SESSION['username'] . '/"></head><body></body></html>'; exit; } } ?> Hi Everyone, I have an ecommerce store that I have done some coding for, and each category has some coding to create pages to display items in that category. For some reason, I am getting a couple of "ghost" pages on the end of a couple categories. For example, look at this page: http://www.autismcommunitystore.com/view.php?category=4 Pages 5 & 6 are blank - page 4 has the last of the items in the category. I have a table called prods_to_cats which connects the products and the categories (a many to many relationship). I'm posting the code here - could someone just let me know if they spot something that looks like trouble? Thank you in advance for your time! Code: [Select] <? $category=$_REQUEST['category']; $showcategory=$category; include 'connect.php'; $thiscategory = mysql_query("SELECT * FROM category WHERE category_id=$showcategory", $dbh); $thisrow = mysql_fetch_array($thiscategory); $this_des=$thisrow["category_description"]; $this_text=$thisrow["category_text"]; $titlekey="Products listed in $this_des at Autism Community Store 877-422-5932"; // If current page number, use it // if not, set one! if(!isset($_GET['page'])){ $page = 1; } else { $page = $_GET['page']; } // Define the number of results per page $max_results = 40; // Figure out the limit for the query based // on the current page number. $from = (($page * $max_results) - $max_results); echo '<div id="product">'; $result = mysql_query("SELECT DISTINCT product.id, product.* FROM prods_to_cats INNER JOIN product ON prods_to_cats.id=product.id WHERE prods_to_cats.category_id=$showcategory ORDER BY product.title ASC LIMIT $from, $max_results", $dbh); $numresult = mysql_num_rows($result); if($numresult==0) { echo "<p style='padding-left: 50px'>There are no products listed in this category.</p>\n"; } else { $thiscategory = mysql_query("SELECT * FROM category WHERE category_id=$showcategory", $dbh); $thisrow = mysql_fetch_array($thiscategory); $this_des=$thisrow["category_description"]; echo "<table width='100%'><tr>"; echo "<div style='position:relative; left:150px;'><font size='2'>Return To: <strong><a href='http://www.autismcommunitystore.com'>Home </a></strong></font></div><br/>"; echo "<div style='position:relative ; width:100%' align='center'>"; echo "<br />"; echo "<img src='http://www.watchesandthings.com/autism/images/suppliesHeader.gif'>"; echo "</div>"; echo "<h2>"; echo "<div style='padding-right:20px ; position:relative ; width;100%; text-align:right;'>"; echo "<div style='position:relative ; width;80%; text-align:center; left:20%;'>"; echo $this_des; echo "</div>"; echo "</div>"; echo "</h2>"; echo "<div style='position:relative ; padding-left:30px ; padding-right:30px ; '>"; echo "<p>"; echo "$this_text </p>"; echo "</div>"; echo "</tr>"; $count = 1; $column = 1; //initialize column 1 or 2 //-----------------------code for drop down menu---------------------------------------- include 'browserChecker.php'; /* echo $browser; echo "<br/>"; echo $browser_version; echo "<br/>"; $main_version=$browser_version[0]; echo $main_version; */ if ($browser=='IE' && $browser_version<7){ echo '<br />'; //make version 6 compatible select box if ($category==1 || $category==24){ }else{ ?> <center> <div id="hiddendiv" style="display:;"> <form method="get" enctype="multipart/form-data" action="http://www.autismcommunitystore.com/viewsubcategory.php"> <select id="myselectbox" name="subcategory" > <? $sql="SELECT * FROM subcategory ORDER BY sub_description ASC"; $result=mysql_query($sql, $dbh); while($myrow=mysql_fetch_array($result)) { if ($myrow['category_id']==$category){ $myrow['sub_description'] ?> <option name="subcategory" value="<? echo $myrow['sub_id'] ?>" ><? echo $myrow['sub_description'] ?> </option> <? }//close if for category check }//close drop down while loop ?> </select> </center> <div id="go_subcat"> <input type="submit" value=" GO " style="background-color: #E4DFF3; color: #6F5079; border: 2px solid #6F5079"> </div> </form> </div> <? }//end check category }else{ //make normal select box if ($category==1 || $category==24){ }else{ ?><div style="z-index:2"><br /><form method="get" enctype="multipart/form-data" action="http://www.autismcommunitystore.com/viewsubcategory.php"> <div align="center" style='z-index:5'> <select name="subcategory" size="1" style="background-color: #ffffff; color: #6F5079; border: 2px solid #6F5079"> <? $sql="SELECT * FROM subcategory ORDER BY sub_description ASC"; $result=mysql_query($sql, $dbh); while($myrow=mysql_fetch_array($result)) { if ($myrow['category_id']==$category){ ?> <option name="subcategory" value="<? echo $myrow['sub_id'] ?>" ><? echo $myrow['sub_description'] ?> </option> <? }//close if for category check }//close drop down while loop ?> </select> <input type="submit" value=" GO " style="background-color: #E4DFF3; color: #6F5079; border: 2px solid #6F5079"> </div> </form> </div> <? } }//end browser checker if //-----------------------end code for drop down menu---------------------------------------- //reste variables $result = mysql_query("SELECT DISTINCT product.id, product.* FROM prods_to_cats INNER JOIN product ON prods_to_cats.id=product.id WHERE prods_to_cats.category_id=$showcategory ORDER BY product.title ASC LIMIT $from, $max_results", $dbh); echo "<TABLE cellspacing='20'>"; while($myrow = mysql_fetch_array($result)) { extract($myrow); $location=$image; $price=$sale_price; $item=$myrow["id"]; $stockstatus=$stat; if($stat=='IN STOCK') { $stockstatus=""; } else { $stockstatus="--Item Coming Soon--"; } if($price>0) { $printout="<table border=0 height=70 width=250><tr><td width='100px' align='center' ><a href=http://www.autismcommunitystore.com/item/$item/><img src=http://www.watchesandthings.com/autism/files/thumbnails/$location border=0 alt=$product_id></a></td><td><a href=http://www.autismcommunitystore.com/item/$item/>$title</a><br><span class=prod-dollar>$$price</span><br>$stockstatus</td></tr></table>"; } else { $printout="<table border=0 height=70 width=250><tr><td width='100px' align='center' ><a href=http://www.autismcommunitystore.com/item/$item/><img src=http://www.watchesandthings.com/autism/files/thumbnails/$location border=0 alt=$product_id></a></td><td><a href=http://www.autismcommunitystore.com/item/$item/>$title<br /><span class=prod-dollar>Click for Pricing</span></a><br>$stockstatus</td></tr></table>"; } if ($column=="1") { echo "<tr><td>$printout</td>"; // first column } else { echo "<td>$printout</td></tr>"; // 2nd column } // end of columns $count += 1; // this is a modulus operator it gets the remainder of the equation $column = $count % 2; } // end of while loop echo "</table>"; ; // Figure out the total number of results in DB: $total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM prods_to_cats WHERE category_id=$showcategory"),0); //$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM product"),0); // Figure out the total number of pages. Always round up using ceil() $total_pages = ceil($total_results / $max_results); echo "<p><center>Pages<br />"; // Build Previous Link if($page > 1){ $prev = ($page - 1); echo "<a href='http://www.watchesandthings.com/autism/view.php?category=$showcategory&page=$prev'><<Previous</a> "; } for($i = 1; $i <= $total_pages; $i++){ if(($page) == $i){ echo "$i "; } else { echo "<a href='http://www.watchesandthings.com/autism/view.php?category=$showcategory&page=$i'>$i</a> "; } } // Build Next Link if($page < $total_pages){ $next = ($page + 1); echo "<a href='http://www.watchesandthings.com/autism/view.php?category=$showcategory&page=$next'>Next>></a>"; } echo "</center>"; }//end of else> echo "</table>"; echo "</div>"; ?> Advance thank you. Can you help please. The error..... Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in C:\wamp\www\test_dabase.php on line 24 code. Code: [Select] <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES"; //select show and show. $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())"; } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_assoc($mysql_select2)){ //show it. echo $data; } } ?> looking for a way to view the PHP source code of any PHP page on the web. Looking for a way to obtain PHP source code in order to help me with a certain issue. Advise. Hi, I'm looking for a code to view the date of the past weekdays. Forexample, if it's thursday, I want it to view the dates for wednesday, tuesday, monday, Friday before. Thank you in advance. when i right click to view page source i just saw the javascript calling php as source file , while on page there is table, i need to read that table in database , is there nay way to do that, file_get contents dont work on it its just take javascript not the page table I've got another funny one... I've noticed that sometimes on some of my pages (all of which are generated by PHP), when i click view source, the source it shows me is not from the page i'm looking at. This isn't a problem that affects me in any way (not yet anyway), it's just a bit odd. I've noticed it happens a little more often in chrome than any other browser. Has anyone else noticed anything like this before? I'm intrigued to know the cause. The only real google response was this: http://www.mail-archive.com/debian-bugs-dist@lists.debian.org/msg144609.html and i haven't seen another issue similar on here. Looke on the PHP manual for the "count" function and it has confused me. it seems to count arrays more than anything. i have a small site and i wanted to display how many items are contained within a Database table. for example i want it to count the amount of users are in the user table. and i then need to echo that amount + 100. Can someone help me with the code. i have already included a database connect script within the header of the page with includes etc. Help would be appreciated. but i would like to know how the code works else i will never learn. I wrote the code below as a way of deleting books from a database. The variables sent to this piece of code come from the page before it, through checkboxes with names corresponding to books, for example the page may have 3 checkboxes with the names 3, 4 and 5. If the user was to select checkbox 4, the variable 4 would be sent through post to this piece of code. The code below selects all of the books from the database in the users school, and then cycles through it, checking whether a book should be deleted by checking whether the post value for that book has been set, eg. if book three has been sent, isset($_POST[$temp]) should return a true, and thus the book is deleted from the database via the mysql_query. The code however will not run, currently I am getting Parse error: syntax error, unexpected T_STRING in /home/textexch/public_html/home/exchange/deletebooks.php on line 81, but i fear there are other problems. Does anyone have any advice as to how to do this better? Code: [Select] $result = mysql_query("SELECT * FROM `books` WHERE School ='".$_COOKIE['School']."'"); while($row = mysql_fetch_array($result)){ $temp = $row['BookID']; if(isset($_POST[$temp])){ mysql_query("DELETE FROM books WHERE BookID = '$Delete'); } } if (mysql_affected_rows() == 0) { echo "sorry didn't work"; } else { echo "Books successfully deleted. Return home <a href='../'>here</a>"; } Hi all . I have a script , which will display the records from sql . If user wish to view more details of the record , they can click view . Code: [Select] echo "<td>$objResult[Message] </td>"; echo "<td>$objResult[mylocaltime] </td>"; echo '<td><a href="historydetails.php?id=' . $objResult[Groupnumber] . '">View</a></td>'; Then it will link to a page which will display more details according to the Groupnumber . Which will display 2 records per page . Code: [Select] <? $groupnumber = $_GET['id']; $strSQL = "SELECT * FROM esp WHERE Groupnumber='$groupnumber'"; $objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]"); $Num_Rows = mysql_num_rows($objQuery); $Per_Page = 2; // Per Page $Page = $_GET["Page"]; if(!$_GET["Page"]) { $Page=1; } $Prev_Page = $Page-1; $Next_Page = $Page+1; $Page_Start = (($Per_Page*$Page)-$Per_Page); if($Num_Rows<=$Per_Page) { $Num_Pages =1; } else if(($Num_Rows % $Per_Page)==0) { $Num_Pages =($Num_Rows/$Per_Page) ; } else { $Num_Pages =($Num_Rows/$Per_Page)+1; $Num_Pages = (int)$Num_Pages; } $strSQL .=" order by id DESC LIMIT $Page_Start , $Per_Page"; $objQuery = mysql_query($strSQL); ?> But when I click next to view the third record , then the table will become blank . Is it because when I click to next page , the $groupnumber is no longer getting the data from $_GET['id'] ? Or is there any other problem? Thanks for every reply . hope someone can help me out here, stuck on this page... This page shows the details of the listing they had just clicked on on the previous page. this part works If the person logged in is the person who is looking at the page, the email section does not show. this part works However, the part I cannot get to work, is when the logged in user types their email message and clicks send email. at this point the details of listing stop showing please SOMEONE HELP ME thanks in advance, i hope Code: [Select] <?php // Start the session require_once('startsession.php'); // Insert the page header $page_title = 'Listing Details'; require_once('header.php'); require_once('connectvars.php'); // Show the navigation menu require_once('navmenu.php'); if(isset($_POST['submit'])){ $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // Retrieve the user data from MySQL $query1 = 'SELECT email FROM ob_user WHERE username="' . $user_name . '"'; $data1 = mysqli_query($dbc, $query1); $row1 = mysqli_fetch_array($data1); $email = $row1['email']; $to = $email; $subject = 'OurBazzar response to' . $subject . ''; $body = $_POST['email']; $headers = 'From: caleb.jordan.flax@gmail.com'; if (mail($to, $subject, $body, $headers)) { echo("<p>Message sent!</p>"); } else { echo("<p>Message delivery failed...</p>"); } end(); } if (isset($_GET['listing_id']) && isset($_GET['subject']) && isset($_GET['description']) && isset($_GET['price']) && isset($_GET['date']) && isset($_GET['city']) && isset($_GET['state']) && isset($_GET['user_id']) && isset($_GET['category_id'])) { // Grab the score data from the GET $listing_id = $_GET['listing_id']; $title = $_GET['subject']; $description = $_GET['description']; $price = $_GET['price']; $date = $_GET['date']; $city = $_GET['city']; $state = $_GET['state']; $userid = $_GET['user_id']; $categoryid = $_GET['category_id']; } // Connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // Retrieve the user data from MySQL $query = 'SELECT username FROM ob_user WHERE user_id="' . $userid . '"'; $data = mysqli_query($dbc, $query); $row = mysqli_fetch_array($data); $query1 = 'SELECT name FROM ob_category WHERE category_id="' . $categoryid . '"'; $data1 = mysqli_query($dbc, $query1); $row1 = mysqli_fetch_array($data1); $user_name = $row['username']; $category_name = $row1['name']; echo '<table>'; echo '<tr><td colspan="2"><strong>' . $title . '</strong></td></tr>'; echo '<tr><td>Created by: </td><td>' . $user_name . '</td></tr>'; echo '<tr><td>Date created: </td><td>' . $date . '</td></tr>'; echo '<tr><td>Location: </td><td>' . $city . ', ' . $state . '</td></tr>'; echo '<tr><td>Category: </td><td>' . $category_name . '</td></tr>'; echo '<tr><td>Description: </td><td>' . $description . '</td></tr>'; echo '</table>'; if ((isset($_SESSION['username'])) && ($_SESSION['username'] != $user_name)) { ?> <br /> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset> <legend>Send Email to <?php echo $user_name; ?></legend> <label for="email">Message: </label> <input type="text" name="email" /><br /> <input type="submit" value="Send Email" name="submit"> </fieldset> <?php } require_once('footer.php'); ?> I have a form that is passing the User to following code. The code below is just ported from another site that I created, which works extremely well. I've had to change the datatable, database connection and some of the variables for this site, but it's otherwise the same. I've triple checked the variables. The datatable is accurate. It doesn't appear to be passing variable from the Form with the exception of "content", and I'm getting the following error: Quote You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'year='', position='', content='Test', ppg='', rp' at line 6 Query: INSERT INTO players SET playerFirst='', playerLast='', feet='', inches='' year='', position='', content='Test', ppg='', rpg='', apg='', spg='', bpg='', fgp='', ftp='', status='' What am I missing? Code: [Select] <?php include('db.php'); $playerFirst = $_POST['playerFirst']; $playerLast = $_POST['playerLast']; $feet = $_POST['feet']; $inches = $_POST['inches']; $year = $_POST['year']; $position = $_POST['position']; $content = $_POST['content']; $ppg = $_POST['ppg']; $rpg = $_POST['rpg']; $apg = $_POST['apg']; $spg = $_POST['spg']; $bpg = $_POST['bpg']; $fgp = $_POST['fgp']; $ftp = $_POST['ftp']; $status = $_POST['status']; //if(isSet($_POST['playerFirst']['playerLast']['feet']['inches']['year']['status'])) //{ /* search for existing row */ $sql = "SELECT msg_id FROM players WHERE playerFirst='".mysql_real_escape_string($playerFirst)."' AND playerLast='".mysql_real_escape_string($playerLast)."'"; if(!$result = mysql_query($sql)) { die(mysql_error()."<br />Query: ".$sql); } if(mysql_num_rows($result)) { $row = mysql_fetch_assoc($result); /* update existing row */ $sql = "UPDATE players SET feet='".mysql_real_escape_string($feet)."', inches='".mysql_real_escape_string($inches)."' year='".mysql_real_escape_string($year)."', position='".mysql_real_escape_string($position)."', content='".$content."', ppg='".$ppg."', rpg='".$rpg."', apg='".$apg."', spg='".$spg."', bpg='".$bpg."', fgp='".$fgp."', ftp='".$ftp."', status='".$status."', WHERE msg_id='".$row['msg_id']."'"; if(!$result = mysql_query($sql)) { die(mysql_error()."<br />Query: ".$sql); } } else { /* insert new row */ $sql = "INSERT INTO players SET playerFirst='".mysql_real_escape_string($playerFirst)."', playerLast='".mysql_real_escape_string($playerLast)."', feet='".mysql_real_escape_string($feet)."', inches='".mysql_real_escape_string($inches)."' year='".mysql_real_escape_string($year)."', position='".mysql_real_escape_string($position)."', content='".$content."', ppg='".$ppg."', rpg='".$rpg."', apg='".$apg."', spg='".$spg."', bpg='".$bpg."', fgp='".$fgp."', ftp='".$ftp."', status='".$status."'"; if(!$result = mysql_query($sql)) { die(mysql_error()."<br />Query: ".$sql); } } Hi, can any see is the are mistakes in this for me. Thanks echo'<img src="skinFiles/'.$skin['thumb_name'].'"class="skinImage" onclick="changeSkin(\'skinFiles/'.$skin["css_name"].'\')" />'; The following works, but I don't want to have to write 50 query statements for 50 states. How can I just write one query (using a variable I presume) and then echo the variable count each time for each state? I'm just trying to list the total number of rows for each state, that exist in the db. <? $result = mysql_query("SELECT * FROM mytable WHERE source = 'alabama'") or die(mysql_error()); $num_rows = mysql_num_rows($result); ?> href="/state/alabama/">alabama</a> (<?php echo("$num_rows");?>) <br> <? $result = mysql_query("SELECT * FROM mytable WHERE source = 'alaska'") or die(mysql_error()); $num_rows = mysql_num_rows($result); ?> href="/state/alaska/">alaska</a> (<?php echo("$num_rows");?>) <br> The above would give output such as: alabama (122) alaska (212) Hi all ! I just passed my code through an analyzer and it showed that a lot of it was not following best practices. Some examples are below: 1.Direct use of $_SERVER Superglobal detected.
if($_SERVER['REQUEST_METHOD']==="POST"){
if(!isset($_SESSION)) sess_start();
if(isset($_SESSION['timeout'])){
$_SESSION['user']=$user; 2. Direct use of $_POST Superglobal detected.
if(isset($_POST['submit']) && $_POST['submit'] ==='Logoff'){
$_POST = array(); $usertype = fcheckRecruiter($_POST['usertype']); and many more like these concerning the use of SUPERGLOBALS. 3. Discouraged functions : header(), session_unset(), mysqli_close(), session destroy() & require_once to name a few besides a lot of other common php functions.
header ("Location: donepage.php");
session_unset(); mysqli_close($link); session_destroy(); Well the question is obviously how to tackle these. The surprising part though is that prior to checking the code by an analyzer, I had no clue, like many other coders on this forum perhaps, especially the newbies, that my code was flawed or at least not following the best practices. I never found a single piece of code on the net, in examples, even in examples in the PHP manual that showed the correct usage of these as per best practices. The most surprising of these were of course the SUPERGLOBALS since they are used everywhere and by almost everybody. Googling the internet shows that hardly anyone is clear about these. People are debating on the direct usage of suberglobals where they are used for checking the existence of the variable. So it's all very moot and very grey it seems. Then there are common functions some of which i mentioned above. For example how would I reset the super global $_POST if not by setting it to a blank array? $_POST = array(); Why are these functions, enlisted above, being discouraged from use and what and how should the alternate functions be used ? How to achieve the same functionality in an alternate way? For the use of superglobals I found that it's proposed to use the filters or filter functions to sanatise or validate the input. If i recall correct, Guru Jacques strongly advised against sanitizing any user input. While I can understand validation of user input, sanitization of it seems to be wrong ?? I would be very grateful if someone can shed some light on these very basic and important questions and provide, if possible, some examples of the correct method of using these in code. Thanks all ! Can you see any problems with the code below? It doesn't work. What I am trying to do is quite straightforward. The links on my site look like this: go.php?id=1&url=usa go.php?id=1&url=uk go.php?id=2&url=usa go.php?id=2&url=uk When a visitor clicks on one of these links, I want to grab the values of the id and url parameters, count the click, and redirect the visitor. Here's the error that I get when I try the code: Quote Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\sites\go.php on line 18 Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\sites\go.php:18) in C:\xampp\htdocs\sites\go.php on line 19 <?php $conn = mysql_connect('localhost','username', 'password') or trigger_error("SQL", E_USER_ERROR); mysql_select_db('database', $conn) or trigger_error("SQL", E_USER_ERROR); $id = intval($_GET['id']); $url = intval($_GET['url']); mysql_query("UPDATE urls SET click_counter = click_counter+1 WHERE id=$id"); if ($url=="usa") $href = "SELECT usa FROM urls WHERE id=$id"; elseif ($url=="uk") $href = "SELECT uk FROM urls WHERE id=$id"; elseif ($url=="aus") $href = "SELECT aus FROM urls WHERE id=$id"; elseif ($url=="can") $href = "SELECT can FROM urls WHERE id=$id"; else $href = "SELECT int FROM urls WHERE id=$id"; $qry = mysql_query($href); list($href)=mysql_fetch_row($qry); header("Location:$href"); mysql_close($conn); ?> Hello- I have a social network that has a blogs and questions section. There are community pages for each one ( blogs/questions ) On each of those pages. the first ten blogs/questions are displayed. To the right of those are a list of clickable categories that sort the pages according to their category. I have pagination implemented after each ten. I am doing a mod_rewrite for pretty urls on my dev and have everything working correctly. The problem is, after hitting next the first time 'page=' is blank and just reloads the first (defaulted page) but after a second click it becomes 'page=10' as it should and then the third click 'page=20" as it should. So basically I am trying to get it to go to 'page=10" after one click, not two. Here is the code for the pagination: Code: [Select] <div id="all_page_turn"> <ul> <?php if($totalBlogs > 10 && $_GET['page'] >= 10) { ?> <li class="PreviousPageBlog round_10px"> <a href="/blogs/?cat=<?php if(isset($_GET['cat'])) { echo $_GET['cat'];} ?>&sort=<?php if(isset($_GET['sort'])) { echo $_GET['sort'];} ?>&page=<?php if(isset($_GET['page'])) { echo ($_GET['page'] - 10);} ?>">Previous Page</a> </li> <?php } ?> <?php if($totalBlogs > 10 && $_GET['page'] < ($totalBlogs-10)) { ?> <li class="NextPageBlog round_10px"> <a href="/blogs/?cat=<?php if(isset($_GET['cat'])) { echo $_GET['cat'];} ?>&sort=<?php if(isset($_GET['sort'])) { echo $_GET['sort'];} ?>&page=<?php if(isset($_GET['page'])) { echo ($_GET['page'] + 10);} ?>">Next Page</a> </li> <?php } ?> </ul> </div> </div>and here is the defaulted category link: Code: [Select] <div id="RightBlogs"> <div id="search_blogs_future" class="round_10px"> <form name="searchBlogs" action="/blogs" method="get"> <input type="text" name="BlogSearch" class="text" value="<?php if(empty($_GET['BlogSearch'])) { echo "Search Blogs"; }else{ echo $_GET['BlogSearch'];} ?>" onclick="clearify(this);" /> <input type="submit" name="subBlogSearch" value="Search" /> </form> </div> <div class='<?php if(empty($_GET['cat']) || $_GET['cat'] == "All") { echo "all_blog_cats_Highlighted"; }else{ echo "all_blog_cats_unHighlighted"; } ?> round_10px'> <a href='/blogs/?cat=All'> All </a> </div>Here is a screen shot of the page, as I'm on my dev so I can't provide a link. you can't see the "next" button, but it's there on the bottom, and then the categories are on the right. 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