PHP - Broken Pagination Script
I'm very out of practice, and trying to figure out old broken code.
From a previous search page, results are pulled from the database, and the variables are passed. At first it works, but then, when I click on the pagination links to see the next page of results, it stops working. The next page appears with no info. Here's the code, I think it has something to do with the passed variables toward the bottom of the code, but I can't figure out what to do. <code> <?php require ('databaseconnection'); $display = 2; // it's intentionally only 2 for the moment to test pagination if (isset($_GET['np'])) { $num_pages = $_GET['np']; } else { $data = "SELECT COUNT(*), `descriptors`.*, `plantae`.* FROM `descriptors` LEFT JOIN `plantae` ON (`descriptors`.`plant_id` = `plantae`.`plant_name`) WHERE `leaf_shape` LIKE '%$s1%' AND `leaf_venation` LIKE '%$s3%' AND `leaf_margin` LIKE '%$s4%'"; $result = mysql_query ($data); if (!$result) { die("Oops, my query failed. The query is: <br>$data<br>The error is:<br>".mysql_error()); } $row = mysql_fetch_array($result, MYSQL_NUM); //row 40 above seems to be where a problem is $num_records = $row[0]; if ($num_records > $display) { $num_pages = ceil ($num_records/$display); } else { $num_pages = 1; } } if (isset($_GET['s'])) { $start = $_GET['s']; } else { $start = 0; } if(isset($_GET[submitted])) { // Now collect all info into $item variable $shape = $_GET['s1']; $color = $_GET['s2']; $vein = $_GET['s3']; $margin = $_GET['s4']; // This will take all info from database where row tutorial is $item and collects it into $data variable $data = mysql_query("SELECT `descriptors`.* ,`plantae`.* FROM `descriptors` LEFT JOIN `plantae` ON (`descriptors`.`plant_id` = `plantae`.`plant_name`) WHERE `leaf_shape` LIKE '%$s1%' AND `leaf_venation` LIKE '%$s3%' AND 'leaf_margin' LIKE '%$s4%' ORDER BY `plantae`.`scientific_name` ASC LIMIT $start, $display"); //chs added this in... echo '<table align="center" cellspacing="0" cellpading-"5"> <tr> <td align="left"><b></b></td> <td align="left"><b></b></td> <td align="left"><b>Leaf margin</b></td> <td align="left"><b>Leaf venation</b></td> </tr> '; while($row = mysql_fetch_array($data)){ echo '<tr> <td align="left"> <a href="view_plant.php?id=' . $row['plant_name'] . '">View plant</a> </td> <td align="left"> </td> <td align="left">' . $row['scientific_name'] . '</td> <td align="left">' . $row['common_name'] . '</td> <td align="left">' . $row['leaf_shape'] . '</td> </tr>'; } echo '</table>'; } if ($num_pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; if ($current_page != 1) { echo '<a href="leafsearch4c.php?s=' . ($start - $display) . '&np=;' . $num_pages . '&s1=' . $s1 . '&s2=' . $s2 . '&s3=' . $s3 . '&s4=' . $s4 . '">Previous</a> '; } for ($i = 1; $i <= $num_pages; $i++) { if($i != $current_page) { echo '<a href="leafsearch4c.php?s=' . (($display * ($i - 1))) . '$np=' . $num_pages . '&s1=' . $s1 . '&s2=' . $s2 . '&s3=' . $s3 . '&s4=' . $s4 .'">' . $i . '</a>'; } else { echo $i . ' '; } } if ($current_page != $num_pages) { echo '<a href="leafsearch4c.php?s=' . ($start + $display) . '$np=' . $num_pages . '&s1=' . $s1 . '&s2=' . $s2 . '&s3=' . $s3 . '&s4=' . $s4 .'"> Next</a>'; } } //added curly ?></code> Similar TutorialsI have a very weird problem. On my website i have a script that takes random pages and displays them on the homepage. It works without a problem on its own but when i have it included in the homepage using include('webmaster fav.php'); i get this error: Warning: Cannot use a scalar value as an array in /mnt/w0210/d28/s25/b02a8bb2/www/webmaster fav.php on line 18 and this error: Warning: array_unique() [function.array-unique]: The argument should be an array in /mnt/w0210/d28/s25/b02a8bb2/www/webmaster fav.php on line 20 and this error: Warning: array_unique() [function.array-unique]: The argument should be an array in /mnt/w0210/d28/s25/b02a8bb2/www/webmaster fav.php on line 21 and this error: Warning: Cannot use a scalar value as an array in /mnt/w0210/d28/s25/b02a8bb2/www/webmaster fav.php on line 23 I am kindof new to php but i think its a good script and it works without errors when it isn't included on the homepage it works. what could be the problem? heres the code: <?php $directory = "/mnt/w0210/d28/s25/b02a8bb2/www/data/"; //the list of pages i want to be random on the site $directory = (!strstr($directory,"*") || $directory =="./" ) ? $directory."*" : $directory; //Checks if the wildcard operator is present, and if not it adds it by default at the end; $files = glob($directory); //Yes, it was that easy to get all the files; $size=sizeof($files); for($i=0;$i<sizeof($files) ; $i++){ //Loop through the files and adds to array; $fp = fopen($files[$i],"r"); $contents[$i]=fgets($fp,999); fclose($fp); } for($x=0;$x<15;$x++){ $numb[$x]=rand(1, sizeof($files)); } $x=count($numb)-count(array_unique($numb)); $num = array_unique ($numb); for($q=0;$x<15;$x++){ $numb[$x]=rand(1, sizeof($files)); } //$imploded = implode(" ", $contents); //get rid of spaces //$newcontent=explode("~", $imploded); // sort into chucks so i can display the data. for($i=0;$i<15; $i++){ $number=$num[$i]; if($contents[$number]==""||$contents[$number]==" "||$contents[$number]==null){ } else{ echo "<li>"; $replacedcontent=str_replace(' ', '-',$contents[$number]); echo "<br/><a href='games/$replacedcontent'>"; $newrcontent=str_replace('-', ' ',$replacedcontent); echo "<img src='$newrcontent.jpg' border='2'></img>"; echo "<br/>$newrcontent</a></li>"; } } ?> i copied and pasted it from many sites examples so thats why some comments are weird... but basically it gets all the data files. reads the title and puts them in an array, then chooses some random ones and puts them in with their image so they can be displayed on the homepage. Good morning, First, I want to thank this forum for all the help provided to me. I am new to Php programming, and honestly, I am still not good at reading codes. I am somehow able to understand how the code is written, but I am still learning, and I know myself, I still have lots to learn. I was able to finish now my very php project, again, thanks to this site and to other forum sites I have visited, and for my friends who helped as well. The web application I made is working fine now, but as I look into it, I know that once the database is populated with information, I will be required to put up pagination on my search page. And so, I looked online for pagination tutorials. I have found some, but I have to admit, it is still hard for me to comprehend the scripts I found. Just today, I found a script that I think would be a very good implementation to the web application I made. The problem is, I am not well familiar with adopting pre-made codes to the one I did, plus the fact that I can't well understand how the code is written. Can anyone help me implement the code, and somehow provide comments on the pre-made code I found in the web? Thank you in advanced for your responses. Here's my code: <link href="add_client.css" rel="stylesheet" type="text/css"> <?PHP include("dbconnection.php"); $query = "SELECT * FROM records"; $result = array(); if(isset($_POST["btnSearch"])) { $query .= " WHERE last_name LIKE '%".$_POST["search"]."%' OR first_name LIKE '%".$_POST["search"]."%'OR territory LIKE '%".$_POST["search"]."%'OR job_title LIKE '%".$_POST["search"]."%'OR title LIKE '%".$_POST["search"]."%'OR employer LIKE '%".$_POST["search"]."%' ORDER BY territory ASC" ; $result = mysql_query($query, $connection) or die(mysql_error()); } ?> <table width="760" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td><table width="760" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="199" align="center" valign="top"><a href="login.html"><img src="invent-asia.gif" alt="" width="152" height="58" border="0" /></a> <script type="text/javascript" src="menu.js"></script></td> <td width="176" align="right" valign="bottom"><a href="main.php"><img src="Home.jpg" width="104" height="20" border="0"/></a></td> <td width="130" align="right" valign="bottom"><img src="View.jpg" width="104" height="20" border="0"/></td> <td width="146" align="right" valign="bottom"><a href="add_client.php"><img src="Add.jpg" width="104" height="20" border="0"/></a></td> <td width="109" align="right" valign="bottom"> </td> </tr> </table></td> </tr> <tr> <td><table width="760" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="200" height="3" bgcolor="#1B1C78"><img src="images/topspacerblue.gif" alt="" width="1" height="3" /></td> <td width="560" bgcolor="#0076CC"><img src="images/topspacerlblue.gif" alt="" width="1" height="3" /></td> </tr> </table></td> </tr> <tr> <td height="553" align="center" valign="top" bgcolor="#F3FAFE"><br /> <form name="form" action="view_client.php" method="post"> <br /> <table width="351" border="0"> <tr> <td width="137" align="left" valign="middle">SEARCH RECORD:</td> <td width="144" align="center" valign="middle"><input type="text" name="search" /></td> <td width="56" align="left" valign="middle"><input type="submit" name="btnSearch" value="Search" /></td> </tr> </table> <br /> <table border="0" cellpadding="3" cellspacing="1" bordercolor="38619E" > <tr> <th width="80" align="center" bgcolor="#E0E8F3">Territory</th> <th width="330" align="center" bgcolor="#E0E8F3">Employer</th> <th width="160" align="center" bgcolor="#E0E8F3">Name</th> <th width="80" align="center" valign="middle" bgcolor="#E0E8F3"> </th> </tr> <?php if($result) { for($i=0; $i<mysql_num_rows($result); $i++) { $id = trim(mysql_result($result, $i, "id")); $territory = trim(mysql_result($result, $i, "territory")); $employer = trim(mysql_result($result, $i, "employer")); $first_name = trim(mysql_result($result, $i, "first_name")); $last_name = trim(mysql_result($result, $i, "last_name")); echo "<tr>"; echo "<td>".$territory."</td>"; echo "<td>".$employer."</td>"; echo "<td>".$last_name.", ".$first_name."</td>"; echo "<td><a href='edit_client.php?id=".$id."'>edit</a> | <a href='delete_client.php?id=".$id."'>delete</a> </td>"; echo "</tr>"; } } ?> </table> <p><br /> </p> </form> <p> </p></td> </tr> <tr> <td height="38"><table width="760" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="200" height="35" align="center" bgcolor="#1B1C78" class=white><img src="images/topspacerblue.gif" alt="" width="1" height="3" /> <a href="disclaimer.html"><font color="#FFFFFF">Legal Disclaimer</font></a> </td> <td width="560" align="center" bgcolor="#0076CC" class=white><img src="images/topspacerlblue.gif" alt="" width="1" height="3" /> Copyright © 2006 - 2010 InventAsia Limited. All rights reserved. </td> </tr> </table></td> </tr> </table> And here is the link of the pagination script I found: http://www.phpeasycode.com/pagination/ I'd rather have the pagination style of the first one. Note: I am not trying to be spoon fed here, but really in this case, I have tried what I think should work, but Sorry guys. (The way my code is written, you should see that I am a noob xD) Hello All! I was hoping one of you phpFreaks could help me out I have the following code that produces links from a database. Certain sets of results have over 3000 links. I can currently set the amount of links to display per page but that's it. I would like to be able to truncate the links to all the pages using pagination. Any help would be greatly appreciated! Here is the code: <?php echo "Results<br><HR WIDTH=100% SIZE=1><br>"; mysql_connect("mysite.com", "mysite_mysite", "mysitepassword") or die(mysql_error()); mysql_select_db("user_database") or die(mysql_error()); if (isset($_GET['pagenum'])) { $pagenum = $_GET['pagenum']; } else { $pagenum = 1; } $rpg = $pagenum; $data = mysql_query("SELECT * FROM mydb WHERE mfg='golf'") or die(mysql_error()); $rows = mysql_num_rows($data); $page_rows = 80; $last = ceil($rows/$page_rows); if ($pagenum < 1) { $pagenum = 1; } elseif ($pagenum > $last) { $pagenum = $last; } $max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows; $data_p = mysql_query("SELECT * FROM mydb WHERE mfg='golf' $max") or die(mysql_error()); while($info = mysql_fetch_array( $data_p )) { $page=$info['link']; echo "<a href='$page' class='blue1'>". $info['title'] . "</a><br>"; } echo "<p>"; if ($pagenum > $last) {$pagenum=1;} else { echo "<HR WIDTH=100% SIZE=1><br>"; $pagenum = 1; while ($pagenum <= $last) { $next = $pagenum; echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$next' class='g5'>$pagenum</a> "; $pagenum = $pagenum+1; } Echo "<br><br>Page $rpg of $last <br>"; } ?> Simple insertion of pagination would be GREAT! Thanks guys.... So i have this page that pretty much displays all the members in the database, and its split into pages using a pagination php script. The problem is that i want the ID number displayed next to it, however the database record id jumps since some records have been deleted. So while there were a total of 1000 records 10 have been removed so instead of showing the accurate ID number of 995 on the last record it displays the number 1000 .... I've tried just assigning a number to each record using the following code... $get_members = "SELECT * FROM members ORDER BY ID ASC LIMIT $rowsperpage OFFSET $offset"; $run_mem_query = mysql_query($get_members) or die(mysql_error()); $club_id = 1; while($member = mysql_fetch_assoc($run_mem_query)){ ?> <li><strong><?php echo $club_id++ ;?>.</strong><?php echo $member['username'] ;?> - <span><?php echo $member['state'] ;?></span></li> <?php } The problem with that is that every time i click to go to the next page the number sequence refreshed back to 1 and starts over.... i've been trying to find a way around his, but can't....does anyone have any solutions to this??? Here's the script that controls the pagination <?php if ($totalpages > 1){ /*********** Start the pagination links ********/ echo "<p>"; // range of num links to show $range = 6; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a> "; } // end if if($result > 0){ // loop to show links to range of pages around current page for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $totalpages)) { // if we're on current page... if ($x == $currentpage) { // 'highlight' it but don't make a link echo " <b>$x</b> "; // if not current page... } else { // make it a link echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a> "; } // end else } // end if } // end for }else{ echo "<a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>1</a>"; } // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'> Next </a> "; // echo forward link for lastpage echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'> Last </a> "; } // end if echo "</p>\n"; /****** end build pagination links ******/ } ?> Hey guys,
I'm using the following pagination script from a website I found. I'ts basically a pagination script that takes data from a MySQL database and displays a certain number of sets before placing the next set on a new page.
This script works great if I substitute a static value for the data I want returned but If the request is stored in a $POST variable like from an HTML form the data refuses to paginate. It will render the first set of data but will not paginate the rest and only shows blank pages
I'm thinking it has something to do with the $POST variable not being set or carrying over to each page but I'm not sure.
I truly appreciate the help!
<?php /* Place code to connect to your DB here. */ include('config.php'); // include your code to connect to DB. $tbl_name=""; //your table name // How many adjacent pages should be shown on each side? $adjacents = 3; /* First get total number of rows in data table. If you have a WHERE clause in your query, make sure you mirror it here. */ $query = "SELECT COUNT(*) as num FROM $tbl_name"; $total_pages = mysql_fetch_array(mysql_query($query)); $total_pages = $total_pages[num]; /* Setup vars for query. */ $targetpage = "filename.php"; //your file name (the name of this file) $limit = 2; //how many items to show per page $page = $_GET['page']; if($page) $start = ($page - 1) * $limit; //first item to display on this page else $start = 0; //if no page var is given, set start to 0 /* Get data. */ $sql = "SELECT column_name FROM $tbl_name LIMIT $start, $limit"; $result = mysql_query($sql); /* Setup page vars for display. */ if ($page == 0) $page = 1; //if no page var is given, default to 1. $prev = $page - 1; //previous page is page - 1 $next = $page + 1; //next page is page + 1 $lastpage = ceil($total_pages/$limit); //lastpage is = total pages / items per page, rounded up. $lpm1 = $lastpage - 1; //last page minus 1 /* Now we apply our rules and draw the pagination object. We're actually saving the code to a variable in case we want to draw it more than once. */ $pagination = ""; if($lastpage > 1) { $pagination .= "<div class=\"pagination\">"; //previous button if ($page > 1) $pagination.= "<a href=\"$targetpage?page=$prev\">� previous</a>"; else $pagination.= "<span class=\"disabled\">� previous</span>"; //pages if ($lastpage < 7 + ($adjacents * 2)) //not enough pages to bother breaking it up { for ($counter = 1; $counter <= $lastpage; $counter++) { if ($counter == $page) $pagination.= "<span class=\"current\">$counter</span>"; else $pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>"; } } elseif($lastpage > 5 + ($adjacents * 2)) //enough pages to hide some { //close to beginning; only hide later pages if($page < 1 + ($adjacents * 2)) { for ($counter = 1; $counter < 4 + ($adjacents * 2); $counter++) { if ($counter == $page) $pagination.= "<span class=\"current\">$counter</span>"; else $pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>"; } $pagination.= "..."; $pagination.= "<a href=\"$targetpage?page=$lpm1\">$lpm1</a>"; $pagination.= "<a href=\"$targetpage?page=$lastpage\">$lastpage</a>"; } //in middle; hide some front and some back elseif($lastpage - ($adjacents * 2) > $page && $page > ($adjacents * 2)) { $pagination.= "<a href=\"$targetpage?page=1\">1</a>"; $pagination.= "<a href=\"$targetpage?page=2\">2</a>"; $pagination.= "..."; for ($counter = $page - $adjacents; $counter <= $page + $adjacents; $counter++) { if ($counter == $page) $pagination.= "<span class=\"current\">$counter</span>"; else $pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>"; } $pagination.= "..."; $pagination.= "<a href=\"$targetpage?page=$lpm1\">$lpm1</a>"; $pagination.= "<a href=\"$targetpage?page=$lastpage\">$lastpage</a>"; } //close to end; only hide early pages else { $pagination.= "<a href=\"$targetpage?page=1\">1</a>"; $pagination.= "<a href=\"$targetpage?page=2\">2</a>"; $pagination.= "..."; for ($counter = $lastpage - (2 + ($adjacents * 2)); $counter <= $lastpage; $counter++) { if ($counter == $page) $pagination.= "<span class=\"current\">$counter</span>"; else $pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>"; } } } //next button if ($page < $counter - 1) $pagination.= "<a href=\"$targetpage?page=$next\">next �</a>"; else $pagination.= "<span class=\"disabled\">next �</span>"; $pagination.= "</div>\n"; } ?> <?php while($row = mysql_fetch_array($result)) { // Your while loop here } ?> <?=$pagination?> 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); 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 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. 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 />"; 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 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........... 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........... 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 [ 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. $compQ = "SELECT companies.companyid, companies.companyname, companies.companylogo, companies.companyoccupation, companies.industry, eQuestions.capitalrequested FROM companies LEFT JOIN eQuestions ON companies.companyid = eQuestions.companyid"; This is not displaying data correctly. I'm assuming eQuestions.capitalrequested is not in the correct spot? If I have a web page located he www . company . com / how-to-repair-your-computer.html
And I decide to re-structure my website like this www . company . com / articles / how-to-repair-your-computer.html
How do I make sure that people don't search and end up at the old, now broken, link?
It seem inevitable that as a website grows, that you will want to re-organize things. What is the best way to make sure that anyone who searches or clicks on an old link - say from an email from a friend - doesn't get a 404 error?
Also, how do you avoid ruining a web pages rank on Google after you move things? (I think if the URL changes, Google makes you start all over as far as getting listed on page-1 and all of that?
Is this something I have to hande on my end, or is it a Google issue, or something else?
So on my website I have a basic if statement that checks some arguments to see if a user can add another user as a friend. Well I had gotten that part down and for the longest time other people on my website have been able to use the feature. Now all of a sudden the if statement doesn't work? Why? Here is the statement: if ($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. |