PHP - Php Print Sql Into 3 Divs
Im trying to print several image url's and names from an sql database into 3 columns,
Aiming to get it to go 1,2,3 4,5,6 7,8,9 etc etc. but for some reason myne is going 1,3,5 2,4,6 etc. Dont bother mentioning that my loops do nothing, i realised that about 10 minutes ago, Any help would be appreciated. <?php include 'config.php'; mysql_connect($host, $user, $pass) or die(mysql_error()); mysql_select_db($database) or die(mysql_error()); $result = mysql_query("SELECT * FROM tracks"); echo '<div id="left_wrapper">'; for ($i=0;$i<mysql_num_rows($result);$i+=3) { $row = mysql_fetch_array($result); $id = $row['id'] + 1; echo "<img src='Thumbnails/" . $id .".gif'></img><br>"; echo $row['name']. "<br>"; } echo "</div>"; echo '<div id="middle_wrapper">'; for ($i=1;$i<mysql_num_rows($result);$i+=3) { $row = mysql_fetch_array($result); $id = $row['id'] + 1; echo "<img src='Thumbnails/" . $id .".gif'></img><br>"; echo $row['name']. "<br>"; } echo "</div>"; echo '<div id="right_wrapper">'; for ($i=2;$i<mysql_num_rows($result);$i+=3) { $row = mysql_fetch_array($result); $id = $row['id'] + 1; echo "<img src='Thumbnails/" . $id .".gif'></img><br>"; echo $row['name']. "<br>"; } echo "</div>"; ?> i know each of those loops does nothing, But you can see where they are meant to do, Each div is aligned to different positions, first loop is left, 2nd is center, 3rd is right. Similar TutorialsAre either of these OK depending on circumstance or stupid things I should never do? Is one or the other sensible/idiotic? Is there an obvious 'better' way?
I think all the php tags opening and closing within the DIVs is untidy but it's static HTML therefore if there's a LOT of it it's easier on the server right/wrong? On the other hand, maybe all the open/close of php tags defeats the object. I'm applying for a web design apprenticeship (I know, I need one) so I'm trying to appear less stupid as much as make nice php. Many thanks in advance, I know I'll end up better off by asking here <?php // php in here ?> <div class='content header'> <?php include("includes/header.php"); ?> </div> <div class='content content1'> <?php include("includes/intro.php"); ?> </div> <div class='content content2'> <?php include("includes/quals.php"); ?> </div> <div class='content content3'> <?php include("includes/experience.php"); ?> </div> <div class='content content4'> <?php include("includes/hobbies-and-interests.php"); ?> </div> <?php // More php here ?>Or this one: <?php print("<div class='content header'>"); include("includes/header.php"); print("</div> \r\n <div class='content content1'>"); include("includes/intro.php"); print("</div> \r\n <div class='content content2'>"); include("includes/quals.php"); print("</div> \r\n <div class='content content3'>"); include("includes/experience.php"); print("</div> \r\n <div class='content content4'>"); include("includes/hobbies-and-interests.php"); print("</div> \r\n"); ?>test.zip 1.5KB 0 downloads Hello, How to customize print page in php ? I had a page, but I need to print it like an invoice look page. Thanks in advance Code: [Select] $query ="SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'" ; $result = mysql_query($query); $row = mysql_fetch_array($result); $oneID = $row[0]; [code] If I then echo "$oneID" why does it not print anything? $result echos resource7 Hello.
i am totally new to php and just started to learn now. i just dont understand why the following code is not printing the username that i enter on the page.
Please note that the code itself is saved with the name "basicForm.php".
Thanks.
<html> I am trying to get the code at the bottom of the script to print just once during the loop but it either doesn't print at all or repeats with the loop im am using if (!$i++) to print once and i works the first time i use it. foreach($uploadFilename as $key => $myvar) { if (!$i++) print "<!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\" /> <title>Upload Complete....</title> </head> <body> <body onload=\"document.forms.formname.submit);\"> <form id=\"formname\" name=\"form1\" method=\"post\" action=\"reg5.php\">\n"; echo "<input type=\"hidden\" name=\"image$key\" value=\""; echo end(explode('/',$myvar)); echo "\">\n"; if (!$i++) print "</form> </body> </html>\n"; } Hi, What is the best way to print the local date and time, my website have customers from 4 or 5 countries. Saudi - UTC+3 Dubai - UTC+4 India - UTC+5.30 how to print the local time for them Thanks, This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=346728.0 can someone please give me some guidance on how to do this please I am wanting to create a status updating type application on my site and i have this idea in my head i want it to retrive and print the last 3 posts (max ids) made by the user if someone could give me some example code please and i can hopefully work from that. Thanks James This topic has been moved to Installation in Windows. http://www.phpfreaks.com/forums/index.php?topic=351150.0 Hi, I am trying to get the form field to echo a php variable. The problem is that the form is 'Printed' via php. print "<td width=\"300\" valign=\"top\"><input type=\"text\" name=\"usr\" value=\"i.e. JBloggs \" onfocus=\"if(!this._haschanged){this.value=''};this._haschanged=true;\" tabindex=\"1\"></td>"; Instead of the value being i.e. JBloggs have it echo the $name variable. Any input would be appreciated. Hello guys and gals, I am pretty green to PHP! I have an empty array that that I am trying to put images into. The thing is I have a certain file name in the folder, I want to exclude that file. This is what I have tried, any advice would be appreciated! Code: [Select] $thumbImg[] = array(); foreach (glob($DImg) as $PImg) { if (!is_file("thumbnail.jpg")) { $thumbImg[] = "<img src=\"pathtoimage\">"; } } Later on the page I am printing it out with this. It is still including the thumbnail.jpg image. Thank you in advance!! Code: [Select] for ($i=0; $i<count($thumbImg); $i++) print $thumbImg[$i]; I currently have a search page on my site that prints the products but it prints the products more than once if its in more than one category I have tried getting distinct item in my SQL. But this doesnt work so im trying an if statement that if there is more than one specific result then to just print this once. I was wondering if anyone had any ideas of how to do this using an if statement I just dont know how to go about just printing the result just once if its greater than 1. The code is below to make it clearer. $searchterm = $_POST['searchterm']; trim ($searchterm); /*check if search term was entered*/ if (!$searchterm){ echo 'Please enter a search term.'; echo $searchterm; } /*add slashes to search term*/ if (!get_magic_quotes_gpc()) { $searchterm = addslashes($searchterm); } /*query the database*/ $query = "SELECT * from (products LEFT JOIN categories_products_link ON products.prod_id = categories_products_link.prod_id) LEFT JOIN categories ON categories_products_link.cat_id = categories.cat_id WHERE prod_title LIKE '%" . $searchterm . "%' ORDER BY cat_title, prod_title"; $result = mysql_query($query); /*number of rows found*/ $num_results = mysql_num_rows($result); echo '<p><h1>Search Results: '.$num_results.'</h1></p><br />'; /*loops through results*/ for ($i=0; $i <$num_results; $i++) { $num_found = $i + 1; $row = mysql_fetch_assoc($result); echo "$num_found. "?><a href="store-<?php echo $row['cat_id'];?>-<?php echo $row['prod_id']; ?>/<?php echo seo_makeSafeURI($row['prod_title']); ?>.html"><strong><?php echo $row['prod_title']; ?></strong></a> <br /> Hi all this might be a bit of a novice question, but if anyone knows the following i would be pleased and my eyes will too I am looking for a way to print on paper or pdf (en print the pdf in the end) all the stringfunctions on php.net (http://www.php.net/manual/en/ref.strings.php) If anyone knows a way to do so i would love to hear it, because after a few days behind a computer my eyes are about to go on strike. I have the following code but I cannot get it to print the results of the array into my web page. Help would be so appreciated!! Code: [Select] $sql = "SELECT RegName, SireID, DamID FROM pedigrees WHERE ID="; $DoggieIDQ = $sql . $values["ID"]; $DoggieIDR = db_query($DoggieIDQ,$conn); $DoggieID = $_GET['ID']; $Depth = 1; $DepthLimit=8; function getParents($DoggieID, $Depth, $DepthLimit) { $Sire = getSire($DoggieID); $Dam = getDAM($DoggieID); $Depth++; if ($Depth == $DepthLimit) { return array($Sire, $Dam); } else { # Now get grandparents too $SireParents = getParents($Sire, $Depth, $DepthLimit); $DamParents = getParents($Dam, $Depth, $DepthLimit); return array($Sire, $Dam, $SireParents, $DamParents); } } Is there a way to dynamically print the url of a web page once it loads? If so, how? This is for metadata purposes. Thanks! Hi Guys I want to knw if its possible to send job to print with php. example, in my database I have a table called Letters which stores letters, while I loop through the table, I want to print each letter off, any examples available? Thank you hihi, so I have the following, except it goes all the way up to server 400. Is there a way to make it print after each echo? As it is right now it will not print the entire list until all 400 servers are done echo "Server 01: " . count($server01->listaccts()) . " / 130" . "<br />" ; echo "Server 02: " . count($server02->listaccts()) . " / 130" . "<br />" ; echo "Server 03: " . count($server03->listaccts()) . " / 130" . "<br />" ; echo "Server 04: " . count($server04->listaccts()) . " / 130" . "<br />" ; echo "Server 05: " . count($server05->listaccts()) . " / 130" . "<br />" ; echo "Server 06: " . count($server06->listaccts()) . " / 130" . "<br />" ; echo "Server 07: " . count($server07->listaccts()) . " / 130" . "<br />" ; echo "Server 08: " . count($server08->listaccts()) . " / 130" . "<br />" ; thanks! I am using function to insert into database. But the primary key is automatic and I used Quote $_SESSION['Tes_ID'] = mysql_insert_id(); to retrieve this. But now that I use function method. I am not sure how to retrieve the primary key on to the next page. Code: [Select] $value = modulesql($postVar1, $postVar2, $SessionVar1, $SessionVar2); $_SESSION['Tes_ID'] = mysql_insert_id(); echo $value, $_SESSION['Tes_ID']; Code: [Select] <?php function modulesql($Tes_Name, $Tes_Description, $Use_ID, $Sub_ID){ $con = OpenConnection(); mysql_select_db("examination", $con); $module = ("INSERT INTO test (`Tes_Name`, `Tes_Description`, `Use_ID`, `Sub_ID`) VALUES ($Tes_Name, $Tes_Description, $Use_ID, $Sub_ID)") or die('Cannot Execute:'. mysql_error()); CloseConnection($con); return $module; } ?> Have I lost you with my question?? I spent some time "Googling" today but all I found was how to create PDFs on the fly. What I wonder is if it is possible to print text on an existing PDF. And if so, could I get some pointer where to obtain this knowledge? Simply put. I have a PDF form that is not yet filled out. I would like to be able to look up an address and other information in a database, and use that information to print on the existing PDF file. Any help is much appreciated. |