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PHP - Displaying Menu From Mysql By Priority
i am storing my menu in the database, i want to be able to output it by priority, heres so far what i have. I have no idea were to start.
Database dump: -- -- Table structure for table `menu` -- CREATE TABLE IF NOT EXISTS `menu` ( `menu_access_lvl` int(2) NOT NULL, `priority` int(11) NOT NULL, `name` varchar(200) NOT NULL, `comment` text NOT NULL, `location` text NOT NULL, `creator_id` varchar(255) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=latin1; -- -- Dumping data for table `menu` -- INSERT INTO `menu` (`menu_access_lvl`, `priority`, `name`, `comment`, `location`, `creator_id`) VALUES (0, 1, 'Home Page', 'Home page', 'index.php', 'admin'), (0, 3, 'Contact', 'Contact', 'index.php?PG=contact', 'admin'), (0, 2, 'Events & Meetings', 'Events & Meetings', 'index.php?PG=events', 'admin'), (0, 4, 'About', 'About', 'index.php?PG=about', 'admin'), (2, 5, 'Admin', 'Admin', 'index.php?PG=admin', 'admin'); And here is the php code displaying it //gets the role of the user if set, otherwise role = 0 if(isset($_SESSION['SESS_MEMBER_ID']))$lvl = $_SESSION['SESS_ROLE']; else $lvl = 0; // this loads the menu buttons that correspond to the users role $menuqry="SELECT * FROM menu WHERE menu_access_lvl<='$lvl'"; $menuresult=mysql_query($menuqry); while($row = mysql_fetch_array($menuresult)){ echo "<li class=\"menuitem\"><a href=\"".$row['location']."\">".$row['name']."</a></li>"; } What this currently displays: Home Contact Events & meetings About I want it to be according to priority in the menu like: Home Events & meetings Contact About Similar TutorialsHey guys im looking to do a drop down menu which is pretty basic however, in this drop down menu it uses an array of 4 values eg Orange, Apples, Bananas, Strawberry's. For this menu if a user has previously chosen an Orange which is stored on the db then then when the drop down menu is loaded it should move Orange to the top of the list, the same applies for the other values. How could I do this? I know how to do the select statement and place the values in an array. from there I dont know quite how to approach it. () Thank you. Hello everyone, So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help. A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right. Some of this probably doesnt even make sense but I'll show you the kind of things I've tried... <select name="gender" size="1" id="gender"> <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option> <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option> </select> OR <select name="gender" id="gender"> <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option> <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option> <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option> </select> OR <select name="gender" size="1" id="gender"> <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option> <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option> <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option> </select> OR <select name="gender" id="gender"> <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option> <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option> </select> Honestly man, I've got no idea. The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen. And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol. Any help would be tremendous and greatly appreciated. Thanks very much, Learner P.S Man! If I have three files such as the following, and put session_start() in each, the session cookie will be domain: "mydomain.com" and path "/". As such, the session will be shared between all three files.
http://mydomain.com/index.php http://mydomain.com/folder1/index.php http://mydomain.com/folder1/folder2/index.phpIf I want the cookie to be unique to each file, then it seems I have several options. I could use session_name() to give the cookie a unique name for each file. I could use session_set_cookie_params() to set a unique path for each (/, /folder1, /folder1/folder2), however, the index files in the children directories will also be sent the root directories session cookie, right? Will this cause problems? Should both approaches be implemented? Or is there another approach? Thanks I have a list of x,y coordinates. (drill positions) I have Three Primary x,y co-ords. (station positions) I want to assign drill positions to stations based on how close they are and the priority of the station(1, 2 or 3). I also want the option of making all stations have a priority or 1 so they divide equally.(for drill positions that are close to two stations) I have absolutely no idea how to tackle this. Is there a formula anyone is aware of? Can anyone even attempt to show me how to tackle this. I am 100% lost. i want to sort the available words by the "searched word"...is there any way ? Code: [Select] <?php $query = "SELECT * FROM ce_cour_chap WHERE nom LIKE'%$find%' ORDER BY nom"; $result = mysql_query($query); while($search = mysql_fetch_array( $result )) { echo $search['nom']." - ".$search["chid"]."<br/>"; } ?> Hello everyone
Hoping someone could lend me a hand. I have a form that takes some end-user's details and adds the date and time into a MySQL table of when the form was submitted. I wish to display that date/time + 4 days ahead using PHP.
I believe the MySLQ DATE_ADD should do the trick quite nicely. In fact plumbing the following statement into phpMyAdmin gives me exactly the results I requi
SELECT DATE_ADD(`datetime`,INTERVAL 4 DAY) FROM `faults` WHERE fault_id = '51';However, just having a pig of a time getting this displayed using PHP. I'm sure this is elementary so forgive me. Here's what I had in mind but is no working. Can someone please point me in the right direction:
<?php
$date_query = mysqli_query($con, "SELECT DATE_ADD(`datetime`,INTERVAL 4 DAY) FROM `faults` WHERE fault_id = '51'");
while($row = mysqli_fetch_assoc($date_query)){
echo $row ['datetime'];
}
?>
Many thanks for your help and advice.Ok... I need some help - I want to show a players balance in a game beside there name (Balance is in mysql database)I can do that but... - I also want to show if there online or offline at the same time( This is stored in a different database) I have the code which says whether they are online or offline <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> Here is the page: www.scswc.com/Offline_Users.php displaying that But I want to Create something like this: Nocvacraft Players |Online| Name:Player Balance:$20 |Offline| Name:Player Balance:$15 Here is what I have tried: <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); $query2 =mysql_query("SELECT * FROM iBalances WHERE player = $users"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query) $rows2 = mysql_fetch_array($query2)): $users = $rows['name']; $balance = $rows2['balance']; echo "<font color='black'>|Online|<br><font color='green'>Name:$usersBalance:$balance</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> I know I am trying to use a variable before it is been set - but if I don't how I have tried this as well... <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>Name:$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; $query = mysql_query("SELECT * FROM iBalances WHERE player = $users"); WHILE($rows = mysql_fetch_array($query)): $balance = $rows['balance']; echo "<font color='red'>$users $balance</font><br>"; endwhile; // ?> Can you use variables in mysql_query()?Is that why it isn't working? This is my first php script so if I need to give you more information for you to help me just tell me Thanks Here is database pictures iBalances users_online Guys, i need your help,i have one table employee with empno=number,image=blob, images uploaded sucessfully and insert into database but when i am trying to retrieve records image are not displaying instead of it some encrypted form shows please help me Greetings,
My current code logs into a database, opens a table named randomproverb, randomly selects 1 proverb phrase, and then SHOULD display the proverb in the footer of my web page.
As of right now, the best I can do is get it to display "Array", but not the text proverb... this code below actually causes my whole footer to not even show up.
Please help!
<?php
include("inc_connect.php"); //Connects to the database, does work properly, already tested
$Proverb = "randomproverb";
$SQLproverb = "SELECT * FROM $Proverb ORDER BY RAND() LIMIT 1";
$QueryResult = @mysql_query($SQLproverb, $DBConnect);
while (($Row = mysql_fetch_assoc($QueryResult)) !== FALSE) {
echo "<p style = 'text-align:center'>" . {$Row[proverb]} . "</p>\n";
}
$SQLString = "UPDATE randomproverb SET display_count = display_count + 1 WHERE proverb = $QueryResult[]";
$QueryResult = @mysql_query($SQLstring, $DBConnect);
...
?>
Hey guys have been trying to get this script to work for a while now, i am new to php and mysql so i am sure i am missing something simple. I have DB setup and need to pull data based on the key item code and get the following I want to get the fields item_code description allergy_statement useable_units region_availability order_lead_time ingredients for item_code 12-100 LITERALLY 12-100, no range, but like i said before i am really new to php and mysql. I have 1187 items that when a user clicks a link in search results it takes them to the product details page for that item code All that data is in my database just can't figure out how to get it out of the database. Is this even the right script to achieve that result. here is the code to get the data from database Code: [Select] <?php require_once('includes/mysql_connect_nfacts_ro.php'); $query = "SELECT item_code, description, allergy_statement, useable_units, region_availability, order_lead_time, ingredients " . "FROM products " . "WHERE item_code = '12-100' "; $resuts = mysql_query($query) or die(mysql_error()); ?> And need to display the data like so : Code: [Select] <td width="715" align="center" valign="top"> <h1>Product Details</h1> <h3>DISPLAY description HERE</h3> <table width="420" border="0"> <td class="ingreg"> </td> </table> <h5>Item Number</h5> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY ITEM_CODE HERE</td> </tr> </table> <h3>Ingredients:</h3> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY INGREDIENTS HERE</td> </tr> </table> <h4>Allergy Statement:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Allergy Statement HERE</td> </tr> </table> <h4>Useable Units Per Package:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Useable Units Per Package HERE</td> </tr> </table> <h4>Region Availability: </h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY ITEM_CODE HERE</td> </tr> </table> <h4>Order Lead Time:</h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY order lead time HERE</td> </tr> </table> <p> </p> <div align="right"></div></td> </tr> </table> how do i get data in database to display where i need it to? Can any one shine some light on this So i pull some records out of a mysql table and i want to display them in 5 even columns. I'm not entirely sure how to do the math & logic to accomplish this. The pull is simple $qry = "SELECT DIST_PART_NUM FROM $tablename"; $sql = mysql_query($qry) or die(mysql_error()); while($res = mysql_fetch_assoc($sql)) { // CREATE 5 even columns here. } so let's say i just retrieved 5,000 part numbers, i'd like to display then in a table of 5 columns with 1000 records per column. This is easy math, but i need the script to automatically figure out the #'s. Also the tricky part is that i dont want to display the part numbers like so 11111 22222 33333 44444 55555 66666 77777 88888 99999 00000 but rather 11111 44444 77777 22222 55555 88888 33333 66666 99999 00000 the remainder if there is one can go in the last column or whatever is easier. I'd tried googling this, but it's not easy to phrase what i'm looking for. Thanks for the help. PS: I'm not looking to copy and paste code, if possible please explain your way so that i can learn the logic. Hi Guys, I am a complete novice as you will soon notice. Can anyone suggest what I am doing wrong with this code. When I run the query in phpmyadmin it produces the correct answer. However when I try to output on my site with php it returns the result "Array". I am guessing I have oversimplified somewhere, aint got a clue how though Code: [Select] <?php include("configure.php"); // To grab the DB info $dbh = mysql_connect ("localhost", DB_SERVER_USERNAME, DB_SERVER_PASSWORD) or die ('<BR> - Could not connect to the database because: '.mysql_error()); mysql_select_db (DB_DATABASE, $dbh) or die(mysql_error( )); $query = "SELECT `options_values_price` FROM `rain_products_attributes` WHERE `products_id` = 526 AND `options_id` = 3 AND `options_values_id` = 3"; $result = mysql_query($query); if (!$result) { $message = "Error! Invalid Query: ".mysql_error()."\n Original Query: ".$query; die($message); } while($row = mysql_fetch_array($result)) { echo $row; } mysql_close(); ?> I am querying my database to show the visit statistics for a particular week and it shows the number of visits for the countries, but does not display the country name.
I have proved that the MySQL works by going into phpMyAdmin and pasting the query into SQL query tab, replacing the POST with 1, for week 1.
I can't see why it is not displying the country.
Here is the code:
<?php
include('connect_visits.php');
doDB7();
$WVisit_data="SELECT WeekNo15.WNo, WeekNo15.WCom, Countries.Country, Countries.CID, ctryvisits15.CVisits
FROM ctryvisits15
LEFT JOIN Countries
ON ctryvisits15.country = Countries.CID
LEFT JOIN WeekNo15
ON ctryvisits15.WNo = WeekNo15.WNo
WHERE ctryvisits15.WNo = '{$_POST['WeekNo']}'
ORDER BY ctryvisits15.CVisits DESC";
$WVisit_data_res = mysqli_query($mysqli, $WVisit_data) or die(mysqli_error($mysqli));
$display_block ="
<table width=\"20%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\" BGCOLOR=\"white\" >
<tr>
<th>Country</th>
<th>Visits</>
</tr>";
while ($WV_info = mysqli_fetch_array($WVisit_data_res)){
$Ctry = $WV_info['country'];
$Visits = $WV_info['CVisits'];
//add to display
$display_block .="
<tr>
<td width=\"10%\" valign=\"top\">".$Ctry."<br/></td>
<td width=\"5%\" valign=\"top\">".$Visits."<br/></td>
";
}
mysqli_free_result($WVisit_data_res);
mysqli_close($mysqli);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<!--
Design by Free CSS Templates
http://www.freecsstemplates.org
Released for free under a Creative Commons Attribution 2.5 License
Name : Yosemite
Description: A two-column, fixed-width design with dark color scheme.
Version : 1.0
Released : 20091106
-->
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta name="keywords" content="" />
<meta name="description" content="" />
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<title>1066 Cards 4U - Stats for country</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" />
</head>
<body>
<div id="wrapper">
<div id="menu">
<ul>
<li class="current_page_item"><a href="index.php">Home</a></li>
<li><a href="Links.html">Links</a></li>
<li><a href="Verse_Menu.html">Verses</a></li>
<li><a href="Techniques.html">Techniques</a></li>
<li><a href="blog.php">Blog</a></li>
<li><a href="Gallery.html">Gallery</a></li>
<li><a href="contact.html">Contact</a></li>
<li><a href="AboutUs.html">About Us</a></li>
<li><a href="stats1.html">Stats</a></li>
</ul>
</div><!-- end #menu -->
<div id="header">
<div id="logo">
<h1><a href="http://www.1066cards4u.co.uk">1066 Cards 4U</a></h1>
</div><!-- end #wrapper -->
</div><!-- end #header -->
<div id="page">
<div id="page-bgtop">
<div id="page-bgbtm">
<div id="content">
<h3>Statistics for Week Commencing <? echo $WkCom; ?> in 2015</h3>
<div id="table">
<?php echo $display_block; ?></div>
</div><!-- end #content -->
</body>
</html>
Can you help please?
Edited by rocky48, 07 January 2015 - 07:33 AM. I am currently looking to insert images into a database and then display them in php. I am thinking of storing the images as a VARCHAR data type but what would be a more suitable type? Hi guys. I'm trying to build a games site for a friend. Currently, the front-end (HTML/CSS) of the site is done. Now, I want to make a way for him to easily add games to the site. Ideally, I'd like to make a database with the following columns: ID, Name, Category, Link, Thumbnail_Link. So, those would be the ID, name of the game, the category, a link to the game, and a link to the 50x50 thumbnail image respectively. Then, using PHP, I'd like to call the first x number (not sure what it will be yet, let's say 50) and make format it as a grid in the following way: There's the thumbnail image followed by the game name, and they're all a clickable link to the game URL. Is this possible? How would I go about doing this? I've already set up a database for a login module to the site, so each page has already opened a connection to the MySQL database. However, I've only ever done basic PHP for mail forms and am otherwise extremely new to it, and am especially new to MySQL. Could anyone walk me through how to do this or even give me a quick example script to work off of? Thanks, any of your time is greatly appreciated! Hi, I am new to php but not to programming. I created a script on a windows platform which connects to the mysql database and returns the results of a table. A very basic script which I wrote to simply test my connection worked. The script works fine on my windows machine but not on my new mac. On the mac it simply does not display any records at all. I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine. The Mac has mysql (with data) and apache running for php. Please could someone help as I have no idea what to do now? Script below: Code: [Select] $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'root'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'test'; mysql_select_db($dbname); mysql_select_db("test", $conn); $result = mysql_query("SELECT * FROM new_table"); while($row = mysql_fetch_array($result)) { echo $row['test1'] . " " . $row['test2'] . " " . $row['test3']; echo "<br />"; } mysql_close($con); I want to display my pictures I stored in Mysql in a 4 column, 2 row table WITH pagination. Here's the code I use to display the data currently: Code: [Select] //your username $username = "username"; //your password $password = "password"; //your mySQL server $host = "host"; //The name of the database your table is in $database = "database"; //connect, but if there is a problem, display an error message telling why there is a problem $conn = mysql_connect($host,$username,$password) or die("Error connecting to Database!<br>" . mysql_error()); //Choose the database from your mySQL server, but if there is a problem, display an error telling why $db = mysql_select_db($database) or die("Cannot select database!<br>" . mysql_error()); // find out how many rows are in the table $sql = "SELECT COUNT(*) FROM myphotos"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); $r = mysql_fetch_row($result); $numrows = $r[0]; // number of rows to show per page $rowsperpage = 4; // find out total pages $totalpages = ceil($numrows / $rowsperpage); // get the current page or set a default if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) { // cast var as int $currentpage = (int) $_GET['currentpage']; } else { // default page num $currentpage = 1; } // end if // if current page is greater than total pages... if ($currentpage > $totalpages) { // set current page to last page $currentpage = $totalpages; } // end if // if current page is less than first page... if ($currentpage < 1) { // set current page to first page $currentpage = 1; } // end if // the offset of the list, based on current page $offset = ($currentpage - 1) * $rowsperpage; // get the info from the db $sql = "SELECT * FROM myphotos ORDER BY id ASC LIMIT $offset, $rowsperpage"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); /****** build the pagination links ******/ // range of num links to show $range = 3; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> "; } // end if // loop to show links to range of pages around current page for ($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 == $currentpage) { // 'highlight' it but don't make a link echo " <span class=\"paginationDown\"><b>$x</b></span> "; // if not current page... } else { // make it a link echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> "; } // end else } // end if } // end for // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> "; // echo forward link for lastpage echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br> "; } // end if /****** end build pagination links ******/ // while there are rows to be fetched... while ($list = mysql_fetch_assoc($result)) { extract ($list); // echo data $url = $list['url'];; $title = $list['title']; $description = $list['description']; echo("$title<br><a rel=\"example_group\" title=\"$description\" href=\"$url\"><img src=\"$url\" alt=\"\" width=\"\" height=\"\" class=\"gallery_images\" /></a>"); } // end while /****** build the pagination links ******/ // range of num links to show $range = 3; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> "; } // end if // loop to show links to range of pages around current page for ($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 == $currentpage) { // 'highlight' it but don't make a link echo " <span class=\"paginationDown\"><b>$x</b></span> "; // if not current page... } else { // make it a link echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> "; } // end else } // end if } // end for // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> "; // echo forward link for lastpage echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br>"; } // end if /****** end build pagination links ******/ The above code just displays the pictures vertically. Here's this code live: http://www.djsmiley.net/gallery/albums/my_photos.php (you can't view the page in IE) Now how would I make it so that the first 8 images in the database display in a 4x2 table, etc.? Hi, I am trying to array a mysql tables data into a php table. Not having luck... <?php include('dbconnect.php') ?> <?php // Make a MySQL Connection $query = "SELECT * FROM cars"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo $row['CarName']. " - ". $row['CarTitle']; echo "<br />"; echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100%>"; echo "<TR />"; echo "<TD />"; echo $row['CarName'].; echo "</TD>"; echo "<TD />"; echo $row['CarTitle'].; echo "</TD>"; echo "</TR>"; echo "</TABLE>"; } ?> Error says: Parse error: syntax error, unexpected ';' in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/viewcars.php on line 69 echo </TD /> is line 69. Please help Ian Hi I have got results being displayed after clicking the search button in a form on my home page but it brings up all the results which is ok but how do I get onlt the results a user searches for for example a location or property type etc as its for a property website The coding is below for the results page Also sorry how do I add a background image to the php page, I tried using css but wouldn't work Code: [Select] <style type="text/css"> body {background-image:url('images/greybgone.png');} </style> <?php mysql_connect ("2up2downhomes.com.mysql", "2up2downhomes_c","mD8GsJKQ") or die (mysql_error()); mysql_select_db ("2up2downhomes_c"); echo $_POST['term']; $sql = mysql_query("select * from properties where typeProperty like '%$term%' or location like '%$term%'"); while ($row = mysql_fetch_array($sql)){ echo 'Type of Property: '.$row['typeProperty']; echo '<br/> Number of Bedrooms: '.$row['bedrooms']; echo '<br/> Number of Bathrooms: '.$row['bathrooms']; echo '<br/> Garden: '.$row['garden']; echo '<br/> Description: '.$row['description']; echo '<br/> Price: '.$row['price']; echo '<br/> Location: '.$row['location']; echo '<br/> Image: '.$row['image']; echo '<br/><br/>'; } ?> |
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