PHP - Include Code Works On One Page But Not Another
Hi all,
I am a bit of a noob when it comes to website building and php etc so please have patients with me I am learnig as I go. I have come across this problem before and cannot remember what I did to fix it but any way it is presenting its self to me again. I have this page: http://minecraftcons...a.com/index.php with an include file loading random images, however the same code on these pages: http://minecraftconsolecrafting.comxa.com/sections/build-navi.php http://minecraftconsolecrafting.comxa.com/sections/springhill-kingdom-01.php are not working. I have other websites that this method of coding is working fine on for example here is another of my websites: http://divinegardensandlandscapes.com/index.php here the include code for random image loader is working fine on all pages it appears on so what is happening when this fail takes place? Any help and or advice would be hugely appreciated, Thanks in advance. Please remeber I am a noob when answering. Thanks Edited by eGate-Network, 12 July 2014 - 10:55 PM. Similar TutorialsHi, I am a php newbie, who has a page that relies on some php scripts, and to which I am trying to add a login page written in php. I took the example from he http://www.howtodothings.com/computers-internet/how-to-make-a-login-system-for-your-website Basically it consists of adding: <? require("log.php"); ?> to the top of any page I want to protect, a log.php file which performs the actions of the form, linking to a mySQL database, and a login.php file which contains the form. I have the login working fine, but it breaks one of the PHP scripts on the page that is protected. It is an upload script, called Weaverbox, based on FancyUpload. The uploads which are handled by a file called upload.php, aren't happening. The progress shows that they are being uploaded, but nothing is uploaded, and there is no success message. As soon as I remove the code from the top of the page requiring log.php all works fine again. I think I may have to add some rules/extensions to resolve this conflict, but I don't know how to go about this. Would someone be able to help me get it sorted? Thanks Nick @requinix, In an earlier thread of mine this week, I think you - or maybe it was @Psycho - suggested putting my code which create an array of my photo files into an include file. I think the logic was that then when either my "photo-gallery.php" OR "photo-detils.php" scripts call that code, that the array will always synched because it is the same array. Ring a bell? So super newbie question, but I am rusty on includes... How does this sound...
1.) Create an "/includes/retrieve-photos.php" script?
2.) Reference that in each script, I would do this...
require_once('../../includes/retrieve-photos.php
right?
3.) To use that code, I would then just reference the populated array...
foreach($photoFiles as $photoKey => $photoValue){ }
Right?
Hi, This is fairly urgent, so I appreciate any help anybody can give me ASAP. I recently upgraded from PHP 4 to 5 and now certain "include" code is not working on my site. Here's an example: Code: [Select] // if topic selected, display the relevant content. if ($topic == 'page1') { include ("$topic.php"); } elseif ($topic == 'page2') { include ("$topic.php"); } // else wrong or no link selected. else { print "Go to the <A HREF = 'main.php'>Main page</A>"; } Let's say the above code was in a PHP file called "frontpage.php". It used to be that if a person went to "mydomain.com/frontpage.php?topic=page1" then the page called for in the above code would include the content of "page1.php" in "frontpage.php". It's no longer working. Any ideas why or how to fix it? Any better way of doing this? Thanks, Steve In my code include_once "http://myserver.com/someFile.php" does not work but include_once "/someFile.php" works I NEED to be able to full URL. Anyone know what PHP setting I need to change to get this working for me? Thanks Hi y'all. I just inherited a legacy system that is in mid-rebuild but needs to be tended until it can be completely phased out and have run into an issue I can't remember how to deal with. The file system is as such: DocumentRoot/ m/ js/ css/ In the 'm' directory I have many files that use require_once as so: require_once "m/another_file.php"; The problem is that none of them have a $_SERVER['DOCUMENT_ROOT'] or even a __DIR__ to start the path, so I'm blowing up with errors. I know this code works as it's currently in production, but I can't remember for the life of me what php.ini or .htaccess setting needs to be set to make it viable. My coworker swears he didn't have to do any magic to make it work on his system. Anybody old enough to remember? Edited June 25, 2020 by maxxdI want to include a variable (for example an object $user) in each PHP page. Previously I declared $var in an include file then included that file in every page, however it generates an error in Zend Studio and it's also not very good programming style. I want to know how I can do it any other way. I was thinking of using constants but I don't know if constants should be used for this purpose and also I do not think constants can be defined as objects. Thank you very much for your help. Hi all not been playing with php over the past year and still a noob lol i did have some code to include files based on the URL link E.G: Code: [Select] http://www.yoursite.com/index.php?page=about then i had some form of string that would access the URL so it would have checked if the url was just index.php or had ?page=somepage after it I can not seem to find any referance on the www or shift the headach i have got trying to remember the code so thought some of you wondaful people would no what i am on about and point me in the rught direction of a tutorial for the information or have the code tohand so i can referance from it thanx all Greetings to all!
Newbie here, I just signed up this morning. I have an issue with a code I have been using for several years with multiple files/directories. Until recently all of the files have worked flawlessly. Now I have one file in a directory that does not display the text I which I am attempting to call up and it was working until recently.
Every file with the same code as the errant one work flawlessly except for the errant one, and I have 11 files to be displayed on my website at https://TheLoveOfGod.org They are all listed under the Devotionals menu button except for one which is displayed on the home page. I have absolutely no training in web design, I am self taught. I began by hand typing html coding over 30 years ago and am now using Word Press with Elementor . I do not understand why this is happening. Can any of you offer assistance to resolve this. I thank you in advance for your assistance.
Edited September 25, 2020 by namednad more explanation Hi, I'm having a first attempt at sanitising code, but I'm not actually sure what I'm doing and how I know if it works. This is the code I have inserted, if I enter "description=re#d%widget" the description query ends so it displays everything 'red'. Not just everything 'red widget'. Code: [Select] $description = mysql_real_escape_string($description); $description = stripslashes($description); $description = htmlentities($description); return $var; $price = mysql_real_escape_string($price); $price = stripslashes($price); $price = htmlentities($price); return $var; Code: [Select] <?php ini_set('display_errors', 1); error_reporting(-1); $query = "SELECT * FROM productfeed"; if(isset($_GET['description']) && !empty($_GET['description'])) { $description = $_GET['description']; $query .= " WHERE description like '%$description%'"; } if(isset($_GET['price']) && !empty($_GET['price'])) { $price = explode('-', $_GET['price']); $lowPrice = (int)$price[0]; $highPrice = (int)$price[1]; $query .= " AND price BETWEEN $lowPrice AND $highPrice"; } $result = mysql_query($query); while($row = mysql_fetch_assoc($result)) { $id = $row['id']; $image = $row['awImage']; $link = $row['link']; $description = $row['description']; $fulldescription = $row['fulldescription']; $price = $row['price']; echo "<div class='productdisplayshell'> <div class='productdisplayoutline'> <div class='productborder'><center> <a href='$link' target='_blank'><img src='$image' width=\"95%\" /></a> </center> </div></div> <div class='productdescriptionoutline'> <div class='productdescriptionbox'> <a href='$link' target='_blank' >$description</a> </div> <div class='productfulldescriptionbox'>$fulldescription</div> </div> <div class='productpriceoutline'> <div class='productpricebox'> <center>&#163; $price</center> </div> <div class='productbuybutton'> <center><a href='$link' target='_blank' ><img src=/images/buybutton.png /></a></center> </div> </div> </div>"; } if ($_GET['description'] == $description ) { echo 'Sorry, this product is not available. Please visit our <a href="http://www.domain.co.uk">Homepage</a>.'; } ?> <?php function sanitizeString($description) { $description = mysql_real_escape_string($description); $description = stripslashes($description); $description = htmlentities($description); return $var; $price = mysql_real_escape_string($price); $price = stripslashes($price); $price = htmlentities($price); return $var; } ?> I have some code that works fine on my dev server but does not work on my production server. Dev server has PHP version 5.2.5 and production server has PHP version 5.1.6. This is the part of the code that isn't working on the prod. server: $xmlDoc=new DOMDocument(); $xmlDoc->loadXML($tmpDoc); $x=$xmlDoc->getElementsByTagName('link'); //get the q parameter from URL $q=$_GET["q"]; //lookup all links from the xml file if length of q>0 if (strlen($q)>0) { $hint=""; for($i=0; $i<($x->length); $i++) { $y=$x->item($i)->getElementsByTagName('title'); $z=$x->item($i)->getElementsByTagName('url'); if ($y->item(0)->nodeType==1) { //find a link matching the search text if (stristr($y->item(0)->childNodes->item(0)->nodeValue,$q)) { if ($hint=="") { $hint="<tr><td><a href='" . $z->item(0)->childNodes->item(0)->nodeValue . "' target='_blank'>" . $y->item(0)->childNodes->item(0)->nodeValue . "</td></tr>"; } else { $hint=$hint . "<tr><td><a href='" . $z->item(0)->childNodes->item(0)->nodeValue . "' target='_blank'>" . $y->item(0)->childNodes->item(0)->nodeValue . "</a></td></tr>"; } } } } } $tmpDoc is a variable that holds database information in xml form. It basically looks like this: $tmpDoc = $tmpDoc . "<link><title>" . $row['CustomerName'] . "****" . $row['Rep'] . "****" . $row['InstallDate'] . "****" . $row['PaidDate'] . "</title><url>accountPage.php?AccNum=" . $row['AccountNum'] . "</url></link>"; ...that is inside a while loop that loops through the rows returned by a query. Basically, as I said, the whole thing works fine on my dev server but on the production server it never makes it into the for loop so I guess the condition $i<($x->length) isn't being met. I'm at a bit of a loss here. Is there anything like the PHP version or Apache version that may cause the "->" operator to not work? The prod PHP version isn't that much older than my dev PHP version so I doubt that's the issue but it's about all I can think of. Thanks! Hi all,
Having an issue with some of my php code. The goal is to have two drop-down menus, one that pulls from "ahevents_scenarios". The second pulls the "sides" involved, however it does not go into dropdown. The other issue is that even though there are only 3 sides specified, all 5 show up on the webpage. This is in Joomla.
Here is my code...I think I've looked over this way too much and just can't find my errors anymore.
<?php if (! $my->id) { // shouldn't even get here since menu is viewable only through logged in status print "<b>Sorry, you must be registered and logged into this site to continue</b>\n"; } else { // include standard functions and db info require_once($_SERVER['DOCUMENT_ROOT'] . "/inc/admin_functions.php"); require_once($_SERVER['DOCUMENT_ROOT'] . "/inc/db_admin_functions.php"); // which scenario we doing? -- replace with table lookup in future opendb('ahevents_scenarios'); if(!isset($_POST['action'])) { // display form $sqlstr2 = "select id, name from scenarios order by name"; $result2 = mysql_query($sqlstr2) or die("Error: ".mysql_error()."<br>Query: $sqlstr2"); // display form ?> <form name="edit_reg" method="post" action="<? echo $_SERVER[ 'REQUEST_URI' ]; ?>" <table> <tr> <td>Select Scenario: </td> <td> <select name="id"> <? while ( $row = mysql_fetch_array($result2)) { printf("<option value=\"%s\">%s</option>\n",$row['id'], $row['name']); } ?> </td></tr> <tr> <td>Which Side?</td> <td><select name="side"> <option value="side_1">side_1</option> <option value="side_2" >side_2</option> <option value="side_3" >side_3</option> <option value="all" selected>All</option> </select> </td></tr> </table> <p> <input type="submit" name="action" value="Get Registrants"> <input type="reset" value="Reset"> <input type="button" name="" value="Back" onClick="history.back()"> </form>The goal is to download the registrants into a .csv file, which it does, however due to not being able to select a "side", the .csv is blank with only headers, no content. Thanks for your time Rob Is there some way of a php include which is visible on every page (footer), itself having two php includes, but only one of which would be visible, then on the user refreshing the page or on visiting another page on the site, that include would be replaced with another php include? Preferably done without any js. Any help appreciated. Chris I am trying to force home landing page text onto this wordpress themes (weblizar Enigma) standard page (which doesnt seem to allow for this even with settings/reading/static/home page option selected wordpress settings)
The landing page just seems to be wigitised.
In a nutshell, I want to have a "page" of text on home page below banner slider and instead of "BLOG" and blog text. Hope this makes sense.
Not getting help from theme author and I have been pointed towards .phpfreaks
This is the website below....
http://www.cloudchasing.co.uk
Thankyou in advance guys
Hello, I'm new on this forum and i have not seen the rules to post a code, so i will try to explain my best : I have learned the basics of php during the past few days and i have a php code that is supposed to read parameters in several fields of a Mysql database and to return those parameters as variables in an array. (well that's how i understand it... please correct me if i'm wrong) So here the code and the part that doesn't seem to work properly (colored in red) : //I have changed the following values which are confidential... $DBName = "MyDatabase"; $DBHostName = "MyMysqlServer"; $DBUserName = "MyUsername"; $DBPassword = "MyPassword"; $Table = "MyTable"; //The fields which are in MyTable : // MemberID SMALLINT // MemberName VARCHAR(20) // MemberPassword VARCHAR(20) // MemberEmailAddress VARCHAR(50) // MemberDateTimeInscription VARCHAR(19) //Reading member parameters echo"<br>Reading the parameters of a member."; echo"<br>Defining the parameters of the member to read."; $CurrentName = "Ashley"; $CurrentPassword = "65hl3y"; $CurrentEmailAddress = "Ashley@HisDomain.com"; echo"<br>Trying to start a connection with the Mysql server."; mysql_connect($DBHostName,$DBUserName,$DBPassword) OR DIE(mysql_error()); echo"<br>Selecting the table."; mysql_select_db($DBName) OR DIE(mysql_error()); echo"<br>Searching for the fields corresponding to the CurrentName."; $Query = "SELECT * FROM ".$Table." WHERE MemberName = '".$CurrentName."'"; $Result = mysql_query($Query) or die(mysql_error()); echo"<br>Returning the parameters stored in the fields."; while($Row = mysql_fetch_array($Result,MYSQL_ASSOC)){ /////This is the start of the part that does not seem to work properly. $MemberId = $row["MemberId"]; $MemberName = $row["MemberName"]; $MemberPassword = $row["MemberPassword"]; $MemberEmailAddress = $row["MemberEmailAddress"]; $MemberDateTimeInscription = $row["MemberDateTimeInscription"]; echo"<br>MemberId : ".$MemberId; echo"<br>MemberName : ".$MemberName; echo"<br>MemberPassword : ".$MemberPassword; echo"<br>MemberEmailAddress : ".$MemberEmailAddress; echo"<br>MemberDateTimeInscription : ".$MemberDateTimeInscription; /////This is the end of the part that does not seem to work properly. } echo"<br>Ending the connection with the Mysql server."; mysql_close(); All of echo are here for debug, i'm a beginner with php. This is what the php page shows : Reading the parameters of a member. Defining the parameters of the member to read. Trying to start a connection with the Mysql server. Selecting the table. Searching for the fields corresponding to the CurrentName. Returning the parameters stored in the fields. MemberId : MemberName : MemberPassword : MemberEmailAddress : MemberDateTimeInscription : Ending the connection with the Mysql server. I don't understand why the variables $MemberId, $MemberName, $MemberPassword, $MemberEmailAddress, $MemberDateTimeInscription are empty. Your advices are welcome, Thanks, Dear php freaks, In my test file the following code works perfectly: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <script language="javascript" type="text/javascript"> function dropdownlist(listindex) { document.formname.subcategory.options.length = 0; switch (listindex) { case "1" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); break; case "2" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); break; case "3" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); break; case "4" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); break; case "5" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); document.formname.subcategory.options[6]=new Option("5","5"); break; case "6" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); document.formname.subcategory.options[6]=new Option("5","5"); document.formname.subcategory.options[7]=new Option("6","6"); break; case "7" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); document.formname.subcategory.options[6]=new Option("5","5"); document.formname.subcategory.options[7]=new Option("6","6"); document.formname.subcategory.options[8]=new Option("7","7"); break; case "8" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); document.formname.subcategory.options[6]=new Option("5","5"); document.formname.subcategory.options[7]=new Option("6","6"); document.formname.subcategory.options[8]=new Option("7","7"); document.formname.subcategory.options[9]=new Option("8","8"); break; case "9" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); document.formname.subcategory.options[6]=new Option("5","5"); document.formname.subcategory.options[7]=new Option("6","6"); document.formname.subcategory.options[8]=new Option("7","7"); document.formname.subcategory.options[9]=new Option("8","8"); document.formname.subcategory.options[10]=new Option("9","9"); break; case "10" : document.formname.subcategory.options[0]=new Option("Select kinderen thuis",""); document.formname.subcategory.options[1]=new Option("0","0"); document.formname.subcategory.options[2]=new Option("1","1"); document.formname.subcategory.options[3]=new Option("2","2"); document.formname.subcategory.options[4]=new Option("3","3"); document.formname.subcategory.options[5]=new Option("4","4"); document.formname.subcategory.options[6]=new Option("5","5"); document.formname.subcategory.options[7]=new Option("6","6"); document.formname.subcategory.options[8]=new Option("7","7"); document.formname.subcategory.options[9]=new Option("8","8"); document.formname.subcategory.options[10]=new Option("9","9"); document.formname.subcategory.options[11]=new Option("10","10"); break; } return true; } </script> </head> <title>Dynamic Drop Down List</title> <body> <form id="formname" name="formname" method="post" action="submitform.asp" > <table width="50%" border="0" cellspacing="0" cellpadding="5"> <tr> <td width="41%" align="right" valign="middle">Aantal kinderen :</td> <td width="59%" align="left" valign="middle"><select name="category" id="category" onchange="javascript: dropdownlist(this.options[this.selectedIndex].value);"> <option value="">------</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> <option value="6">6</option> <option value="7">7</option> <option value="8">8</option> <option value="9">9</option> <option value="10">10</option> </select></td> </tr> <tr> <td align="right" valign="middle">Waarvan thuiswonend : </td> <td align="left" valign="middle"><script type="text/javascript" language="JavaScript"> document.write('<select name="subcategory"><option value="">-----</option></select>') </script> <noscript><select name="subcategory" id="subcategory" > <option value="">Select Sub-Category</option> </select> </noscript></td> </tr> </table> </form> </body> </html> When i want to implement it in my signup page, it doesnt do anything. I've checked it a thousand times, i double checked that also the -formname of my signup page is 'formname' -the element name stays 'subcategory' I really need this to be implemented. Where can be the problem? What should i look for? Hi, I have some code which works but when I created a function and call this same code it doesn't. The error I get is as follows: Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/fhlinux010/l/languageschoolsuk.com/user/htdocs/admin/email.php on line 42 Error retrieving schools The code function CreateSchoolCheckboxes() { echo '<div style="height:400px;width:400px;font:16px/26px Georgia, Garamond, Serif;overflow:scroll;">'; $querySchools = "SELECT * FROM school"; $result = mysql_query($querySchools, $conn) or die ("Error retrieving schools ".mysql_error()); while($row = mysql_fetch_array($result)) { $schoolname = $row['name']; echo '<input type="checkbox" name="school" value="'.$schoolname.'">'; echo $schoolname . '<br>'; } echo '</div>'; } Im sure that this is probably something simple but any suggestions would be much appreciated. Thanks, Joe I have made a bit of code that queries the database for articles published within the last year. <?php echo $numposts = $wpdb->get_var("SELECT COUNT(*) FROM $wpdb->posts WHERE post_status = 'publish' AND post_type = 'post' AND 'post_date' > '" . date("Y") . "-01-01-01 00:00:00'"); if (0 < $numposts) $numposts = number_format($numposts); ?> I really don't even know how it works and need to because I want to figure out how to make it also get articles published within the last day, month and week. I suppose if I understood what needs changing I might be able to work with it but so far everything I've tried either returns the exact same number or nothing at all. It's for Wordpress just to note. Any ideas? I have 2 websites, a companies main website written in PHP and a web application they have purchased written in aspx (main site) www.example.com (web app) www.webapp.com/login.aspx?CompanyName=example The client wants people to be able to login from their main website which then redirect to the webapp (which will be styled like their branding) without the customer realising they have left the main site. I thought i may be able to achieve this with a simple include from a page within the main site such as; Code: [Select] <?php // www.example.com/software include"http://www.webapp.com/login.aspx?CompanyName=example"; ?> This pulls in the login page fine, however when you try to login it does not redirect, it gives a 404 error as it trys to open this URL http://www.example.com/software/login.aspx?CompanyName=example Is there a way of pulling in the necessary html from the aspx site to provide a login box in my clients main site but then have it redirect to the correct aspx page (on the aspx site)? I have thought about using an iframe, but that wont redirect to the webapp upon login, but just keep everything withing the iframe thanks for any advice... Hi, Please bear with me, I have no real experience with PHP but am using it for the first time in a page for maintenance purposes. One of the things I am trying to do is include a random page from within a certain folder (folder: 'modules/did-you-know'). Inside this folder there are currently three files named in this format 'did-you-know-###.php' - where ### is the page number. I have no problems including the named page individually, but it is when randomising it where it causes errors. The code I am using is the following: Code: [Select] <?php $i=0; $myDirectory = dir("modules/did-you-know"); while($file=$myDirectory->read()) { $array[$i]=$file; $i++; } $myDirectory->close(); $num = count($array); $random = rand(0, $num); include "$array[$random]"; ?> When I load the page, I get numerous errors, as follows: Warning: include(..) [function.include]: failed to open stream: Operation not permitted in /home/{USERNAME}/public_html/test2/index.php on line 103 Warning: include(..) [function.include]: failed to open stream: Operation not permitted in /home/{USERNAME}/public_html/test2/index.php on line 103 Warning: include() [function.include]: Failed opening '..' for inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/{USERNAME}/public_html/test2/index.php on line 103 Warning: include(did-you-know-002.php) [function.include]: failed to open stream: No such file or directory in /home/{USERNAME}/public_html/test2/index.php on line 103 Warning: include(did-you-know-002.php) [function.include]: failed to open stream: No such file or directory in /home/{USERNAME}/public_html/test2/index.php on line 103 Warning: include() [function.include]: Failed opening 'did-you-know-002.php' for inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/{USERNAME}/public_html/test2/index.php on line 103 Now, it's clear from the above that it is (at least sometimes) reading the files from within the 'did-you-know' directory, so not sure why I am getting these errors. (Especially the last one, as that would suggest an error to do with permissions, would it not?) Line 103 is the "include" line. Any help would be appreciated. Thanks |