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PHP - Displaying Multiple Row Stored Data In A Single <td>
I have stored supplier information in a table and same form user will select multiple products for that supplier and that will be stored in a seperate table against that supplier id. Now when i would like to display all supplier information with muliple products in a single line. Now all the information will be displayed as many time as the products are. How to display all the products in a same line.
Now my display is like this
if(isset($_POST['submit']) && ($_POST['vendor']!='') && ($_POST['item']!=''))
{
$sql="SELECT supplier.id AS sid, supplier.name AS SNAME, supplier.category, supplier.website, supplier.email, supplier.phone, supplier.vat, supplier.pan, supplier_location.id, supplier_location.supplier_id, supplier_location.location, supplier_products.id, supplier_products.supplier_id, supplier_products.product_id, location.loc_id, location.name AS locname, products.product_id, products.name AS pname FROM supplier INNER JOIN supplier_location ON supplier.id = supplier_location.supplier_id INNER JOIN supplier_products ON supplier.id=supplier_products.supplier_id INNER JOIN location ON supplier_location.location = location.loc_id INNER JOIN products ON supplier_products.product_id=products.product_id WHERE supplier.id=".$sup." AND supplier_products.product_id=".$product;
$sql1 = mysql_query($sql) or die(mysql_error());
<table>
<thead><tr>
<th>Vendor ID</th>
<th>Vendor</th>
<th>Category</th>
<th>Website</th>
<th>Email</th>
<th>Phone</th>
<th>Products</th>
<th>Locations</th>
<th>VAT</th>
<th>PAN</th>
</tr>
</thead>
<tbody>
<tr>
<?php
while($row = mysql_fetch_array($sql1))
{
?>
<td><?php echo $row['sid'] ?></td>
<td><?php echo $row['SNAME'] ?></td>
<td><?php echo $row['category'] ?></td>
<td><?php echo $row['website'] ?></td>
<td><?php echo $row['email'] ?></td>
<td><?php echo $row['phone'] ?></td>
<td><?php echo $row['iname']; ?></td>
<td><?php echo $row['locname']; ?></td>
<td><?php echo $row['vat'] ?></td>
<td><?php echo $row['pan'] ?></td>
</tr>
<?php
}
?>
</tbody></table>
}
Similar TutorialsHi.
I have a product database w/ tables each for product series, model, sub-model and optionals. One item in each of these makes up one product. So Series K, Model 7, sub-model 2, optional 3 will be a product called K 7/2-3 .The permutations and combinations of these are all possible products.
In the same db, I have a quotations table. Each quotation has a unique id / record. A quotation can include any number of products : it can have K 7/2-3, KA 561/1-2 , FPN 311/2-1 or it can only have KA 561/1-2 etc.
Now I am stuck. Don't know how to
a. structure the quotations table
b. create a MySQL select so that one statement can make multiple selections . I thought of using a checkbox but that doesn't work either since the series/model/sub model/optionals must chosen sequentially.
Any ideas I can follow thru ?
Many thanks
Swati
For instance if i set the following 2x variables; "$varOne = $_GET[first_name]" "$varTwo = $_GET[last_name]" "$varThree = $_GET[userid]" how do i insert them both together into a DB field for example: $full = $varOne,$varTwo,$varThree (without the , ) this would then be echo as : joebloggs66985 i have the INSERT in there atm but i wanted to insert first name and last name together with there userid to generate a sort of username possibility. sorry if i didnt explain correcly Hello!
I have five websites on a single server, and each website uses the same geographic data, for instance US zip codes and city/state latitudes and longitudes. Each website uses it's own copy of these "large" tables.
I would like to avoid having to maintain the same data in 5 databases, but I am more concerned with database performance, e.g., the speed at which the data is retrieved for each website.
Question: Should I set up a new separate database that contains the shared tables for all websites and allow each website to connect to this new database for shared data? Or is it better for each website have it's own copy of the duplicate tables?
Thanks for any info!
Cheers
Ok so I am pretty good with mysql and performing queries to get the data that I need. However, I have a question for any of you gurus out there that may be able to help me with an issue that I always run into with certain types of queries. Say I have 2 tables like this: Galleries id name 1 My Photos Gallery Photos id gallery_id photo 1 1 photo1.jpg 2 1 photo2.jpg 3 1 photo3.jpg Now usually when I pull this info from the database I have to do 2 separate queries in order to get the data and then link them so I would do something like. <?php //make the gallery query $query = "SELECT * FROM galleries"; $results = mysql_query($query); //setup an array for the galleries data $gallery_data = array(); //loop through the galleries for($i=0;$i<mysql_num_rows($results);$i++){ //add the gallery info to the gallery data array $gallery_data[$i] = mysql_fetch_array($results); //make the photos query with the gallery id $photo_query = "SELECT * FROM gallery_photos WHERE gallery_id='".mysql_real_escape_string($gallery_data[$i]['id'])."'"; $photo_results = mysql_query($photo_query); //setup the photos array $photo_data = array(); //put the photos in the array for($n=0;$n<mysql_num_rows($photo_results);$n++){ $photo_data[$n] = mysql_fetch_array($photo_results); } //add the photos to the gallery array $gallery_data[$i]['photos'] = $photo_data; } //now to display it is like this if(is_array($gallery_data)){ foreach($gallery_data as $gallery){ echo $gallery['name']; //show the photos if(is_array($gallery['photos'])){ foreach($gallery['photos'] as $photo){ echo $photo['photo']; } } } } ?> So my question is there a way to get all of this data at one time. I know how to do multiple queries in one and to do joins but they can only return one row as far as I know of. The only other way that I know how to do this is by ordering them by name and selecting them directly from the photos table and then just getting the gallery name like this: SELECT *,(SELECT name FROM galleries WHERE id=gallery_photos.id) AS gallery_name FROM gallery_photos ORDER BY gallery_id Then I could just do one loop and see if the name has changed and if so to output the new name. But I would like to know if there is a way to get a second set of results as an array from a query so I could just select the galleries and photos all in one query. Any help is appreciated. this is the line in my script that I have to show the image: Code: [Select] $output .= "<img>{$row['disp_pic']}</img></br>\n"; As you can see I added the image tag, but it wont show the actual image. IE shows it as a small square with another small square picture icon in the middle of it (i'm sure you guys know what i mean). Often times I need to convert between PHP Arrays, MySQL, XML, JSON sqlite etc. Lately I've been just building PHP arrays with all the data. Its a big hassle though when theres a lot of data. I'm thinking I should have one central location that I store data in, then export to other formats from there, but I dont know what that central location should be. I like MySQL because I'm used to it. Sometimes its not an option though, like if I need to make a really light script, I dont want the hassle of connecting to MySQL. Would XML be the best place to store data, then from the XML file I can convert it to whatever format I want?
hey guys i have this code FILE NAME "index.php" Code: [Select] <?php // make a connection to the database mysql_connect ("localhost", "root", "vertrigo") or die ('Error: I Failed To Connect To The Database ' . mysql_error()); mysql_select_db ("test"); // Get Data $query = mysql_query("SELECT * FROM TestTable"); // display the data and loop while ($row = mysql_fetch_array($query)) { echo "<br /> ID: ".$row['ID']."<br /> First Name: ".$row['FName']."<br /> Last Name: ".$row['LName']."<br /> Contact Number: ".$row['CNumber']."<br />";} ?> <form method="post" action="update.php"> <table border="1" align="center"> <tr> <td align="right" width="220">ID: </td> <td align="left" width="220"> <input type="text" name="ID" size="30" /></td> </tr> <tr> <td align="right" width="220">First Name: </td> <td align="left" width="220"> <input type="text" name="FName" size="30" /></td> </tr> <tr> <td align="right" width="220">Last Name: </td> <td align="left" width="220"> <input type="text" name="LName" size="30" /></td> </tr> <tr> <td align="right" width="220">Contact Number: </td> <td align="left" width="220"> <input type="text" name="CNumber" size="30" /></td> </tr> <tr> <td align="right" width="220"><input type="reset" value="Reset" /> </td> <td align="left" width="220"> <input type="submit" value="Update Database" /></td> </tr> </table> </form> and i also have this code FILE NAME "update.php" Code: [Select] <?php $ID = $_POST['ID']; $FName = $_POST['FName']; $LName = $_POST['LName']; $CNumber = $_POST['CNumber']; mysql_connect ("localhost", "root", "vertrigo") or die ('Error: I Failed To Connect To The Database ' . mysql_error()); mysql_select_db ("test"); $query="INSERT INTO testtable (ID, FName, LName, CNumber)VALUES ('".$ID."','".$FName."', '".$LName."', '".$CNumber."')"; mysql_query($query) or die ('Error Updating Database'); echo "Database Updated Sucsessfully With: ".$ID." ".$FName." ".$LName." ".$CNumber ; ?> ok so the script is working like a charm its sending the data to the database as i want it to. the problem i have is that i want to be able to update the info that is already on the database lets say i want to change a phone/contact number i have typed the ID number into the ID text field and the same first and last name into there correct boxes and then typed in the new phone number i then click submit and i get the error ""Error Updating Database"" i have looked all over the forum and net to see what i have done wrong to not allow this code to update can anyone help me out here please im quite new to the php language and could really do with some pointers thanks Steve Hey Guys, I have got 3 chained select boxes working. Basically what you do is check for a region with the country, then once you pick a region, you select a club within that region. Once you have selected the club, you can then choose a team from within that club. Screenshot: Attchment As part of this chained select boxes's, a user can register. Although, the user has to FIRST check if the TEAM has already been registered in the database. I have taken a screenshot of my database, which shows the regions, clubs, and teams from the chained select boxes. What i want to do now is be able to check if a TEAM is already registered in the database. So as you can see in the database, the team 'NPOB, Senior As' has already been taken. Database Screenshot: Attachment *** How do i check if a user has been registered to one of the teams? And if there is not a registered user to that team, they can then register it. Here is my code: <?php include"database.php"; ?> <script type="text/javascript"> /* Triple Combo Script Credit By Philip M: http://www.codingforums.com/member.php?u=186 Visit http://javascriptkit.com for this and over 400+ other scripts */ var categories = []; categories["startList"] = ["Taranaki","Auckland"] // Regions + Clubs categories["Taranaki"] = ["NPOB","Tukapa"]; categories["Auckland"] = ["Marist","Takapuna"]; // Clubs + Teams within that Club categories["NPOB"] = ["Senior As","Senior Bs","Colts U21s"]; categories["Tukapa"] = ["Senior As","Senior Bs","Colts U21s"]; categories["Marist"] = ["Senior As","Senior Bs","Colts U21s"]; categories["Takapuna"] = ["Senior As","Senior Bs","Colts U21s"]; var nLists = 3; // number of select lists in the set function fillSelect(currCat,currList){ var step = Number(currList.name.replace(/\D/g,"")); for (i=step; i<nLists+1; i++) { document.forms['tripleplay']['List'+i].length = 1; document.forms['tripleplay']['List'+i].selectedIndex = 0; } var nCat = categories[currCat]; for (each in nCat) { var nOption = document.createElement('option'); var nData = document.createTextNode(nCat[each]); nOption.setAttribute('value',nCat[each]); nOption.appendChild(nData); currList.appendChild(nOption); } } // function getValue(L3, L2, L1) { // alert("Your selection was:- \n" + L1 + "\n" + L2 + "\n" + L3); // } function init() { fillSelect('startList',document.forms['tripleplay']['List1']) } navigator.appName == "Microsoft Internet Explorer" ? attachEvent('onload', init, false) : addEventListener('load', init, false); </script> <form name="tripleplay" action="testingdropdown.php" method="post"> <select name='List1' onchange="fillSelect(this.value,this.form['List2'])"> <option selected>Choose Region</option> </select><br /><br /> <select name='List2' onchange="fillSelect(this.value,this.form['List3'])"> <option selected>Choose Club </option> </select><br /><br /> <select name='List3' onchange="getValue(this.value, this.form['List2'].value, this.form['List1'].value)"> <option selected >Choose Team </option> </select> <input type="submit" name="tripleplay" value="Register"> </form> <?php if (isset($_POST['tripleplay'])) { $region = addslashes(strip_tags($_POST['List1'])); $club = addslashes(strip_tags($_POST['List2'])); $team = addslashes(strip_tags($_POST['List3'])); $email = 'email'; $check = mysql_query("SELECT * FROM managers WHERE email='$email'"); if ($email == '') { echo "You can register that club"; } else { echo "Sorry that team has already been registered"; } } ?> i want the name of a picture stored in my db after i upload it the data is not stored in the db after i run this script, but i dont get errors either i print the two vars before sending them, and they get printed fine any help on this would be greatly appreciated thanks ! <?php error_reporting(E_ALL); ini_set("display_errors", 1); // INCLUDE THE CLASS FILE include('ImageLib.Class.php'); include("./includes/egl_inc.php"); $displayMessage = ''; if($_POST){ if(isset($_FILES['image_file'])){ // SEE THE MAGIC HAPPEN $destination_path = 'uploads/'; $post_file_name = 'image_file'; $width = 600; $height = 400; $scale = false; $trim = true; $uniqueName = true; $img = ImageLib::getInstance()->upload($post_file_name, $destination_path, $uniqueName)->resize($width, $height, $scale, $trim)->save(); $imgstr = mysql_real_escape_string ($img); $fileName = $_FILES['image_file']['name']; $displayMessage = '<div class="image"><img src="'.$destination_path.$fileName.'" /><br />Uploaded And Resized...With new file name : "'.$img.'"</div><br /><br />'; $playerid=$_SESSION['tid']; $matchdetails = mysql_fetch_array(mysql_query("SELECT id FROM ffa_matches WHERE status=2 and admin=$playerid")); $id = $matchdetails[id]; print $img;print $imgstr; print $id; mysql_query(" INSERT INTO ffa_screens (imgname,match) VALUES( '" . mysql_real_escape_string($imgstr) . "', '" . mysql_real_escape_string($id) . "' )"); }} ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>ImageLib Samples By Rahul Kate</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <style> body{font-family: arial; font-size:12px; color:#444444; padding:20px;} li{margin-top:10px;} .image{color:green;} .image img{margin-bottom:5px;} </style> </head> <body> <h1>ImageLib | Upload Image, move it to Uploads folder and Resize it and Trim</h1> <?php echo $displayMessage; ?> <form method="post" enctype="multipart/form-data"> Select Image<br /> <input type="file" name="image_file" id="image_file" /> <br /> <br /> <input type="submit" name="submit" value="Submit" /> <br /> <br /> <a href="index.html">Back TO Home</a> </form> </body> </html> sorry if i have posted this in the wrong place, wasnt sure wether to post here or mysql. I have made a php calendar, and i am now wanting it to show if there is an event on that day and if so show it in a tool tip. the tool tip is populated by what is in the title="" of the link so i need my events to be shown in there. I have figured out how to show the event but now i am stuck on how to show all events if there is more than one one that day, how i have set it seems to only the second event, probably as the second variable overwrote the first variable. this is the code i have at the minute .............. $result = mysql_query("SELECT * FROM events WHERE day='$day_num' AND month='$fullmonth' AND year='$year'") or die(mysql_error()); $rows= mysql_num_rows($result); if ($rows !="false"){ while($rowout = mysql_fetch_array($result)) { $todayis = $rowout['day'] ."-". $rowout['month'] ."-". $rowout['year'] ."<br>"; $title= $rowout['event'] ."<br><br>";} $firstl = "<a href='' title='". $todayis . $title ."'>"; $lastl = "</a>";} else {$firstl = ""; $lastl = "";} then to display the day and links ................. Code: [Select] <td><? echo $firstl; ?><? print $day_num; ?><? echo $lastl; ?></td> could some one be so kind and help me write it so that it displays all events? here is a link to the calendar, incase its needed. http://www.scripttesting.htmlstig.com/calendar/index.php Many thanks Carl I can use the first while loop, but there is no data in the second while loop. Is there a better way as I have never done this before... Code: [Select] <?php $featured_results = mysql_query("SELECT * FROM products LEFT JOIN product_images ON products.product_id=product_images.product_id WHERE products.product_featured='1' AND products.product_active='1' AND thumb='1'"); $fa=0; while($featured_row = mysql_fetch_assoc($featured_results)) { $fthumb_result = mysql_query("SELECT image_name FROM product_images WHERE product_id='".$featured_row['product_id']."' AND thumb='1'"); $fthumb = mysql_fetch_row($fthumb_result); if ($fa==0) { echo "\n<img id=\"home-slider-photo-".$fa."\" class=\"home-slider-photo preload\" src=\"/includes/getimage.php?img=".$fthumb[0]."&w=370&h=370\" alt=\"\" />"; } else { echo "\n<img id=\"home-slider-photo-".$fa."\" class=\"home-slider-photo preload home-slider-photo-unsel\" src=\"/includes/getimage.php?img=".$fthumb[0]."&w=370&h=370\" alt=\"\" />"; } $fa++; } echo "<div id=\"home-slider-photo-price\">"; $fb=0; while($featured_row2 = mysql_fetch_assoc($featured_results)) { if ($fb==0) { echo "\n<div id=\"home-slider-photo-price-".$fb."\" class=\"home-slider-photo-price\">\n<span>only</span>$".$featured_row2['product_price']."\n</div>"; } else { echo "\n<div id=\"home-slider-photo-price-".$fb."\" class=\"home-slider-photo-price home-slider-photo-price-unsel\">\n<span>only</span>$".$featured_row2['product_price']."\n</div>"; } $fb++; } echo "</div>"; ?> Hi all, slightly confusing title! What I am trying to do is make a simple stores system whereby I can add and remove stores. It is purely php and no mySQL. My idea was to create a text file with the individual number and then include the file on the table on the page. I have worked out how to increment one value by one, but the others I am struggling with. Example I wish to have two links one which I click to increase and another to decrease the value by one. It's the multiple issue that is confusing me, I create a txt file and a php file using a variable from a submitted form (part number). The php file holds info and the text file holds the number which is default at '0' (Zero). My idea was to write some php to the file, perhaps when I click the plus it will open and write to the file + 1 .. Any guidance or points in the right direction would be greatly appreciated. Thanks. Shugs Hey guys, I'll be the first to say that I'm a real php n00b, and only understand the basics. I have a website already in place that someone else coded and I just need to add something to. There's a form that posts to another page and then back to a sql database. What I want to do is have the form also email me the contents of the form when it is submitted. Is there a way to have a form do multiple actions? I'm not sure the best way to go about this, so if someone has some pointers, I'd really appreciate it. Thanks! Hi there im trying to set a single variable multiple rows of data that are echoed using a single variable. The trouble is i just cant seem to make it work by trying to add a while or do loop.. The variable is $alderaanfleetalt and consists of: Code: [Select] $fleetname = "FleetName"; $shipname = "Ship Name"; Which is just text stored in two other variables. The select query row of data is added the $alderaanfleetalt variable. Code: [Select] $alderaanfleetalt = $fleetname." ".$row_Alderaanfleet['FleetName']."</br>".$shipname." ".$row_Alderaanfleet['ShipName']; At the moment only a single row appears. ive tried to add a while/do loop so that multiple rows are outputted but its not working. Code: [Select] do{ $alderaanfleetalt = $fleetname." ".$row_Alderaanfleet['FleetName']."</br>".$shipname." ".$row_Alderaanfleet['ShipName']; } while ($row_Alderaanfleet = mysql_fetch_assoc($Alderaanfleet)); Im a bit lost here and not even sure it can be done this way... Any help would be ace. Thank you I have the form:
Hi, I have a table of content in my page. For each row it has a check box. How do i achieve the function of when i click on submit button, those rows which are checked will be downloaded as a single pdf file. For example, for each checked row is a PDF file here. Suppose if i checked for 5 rows , 5 PDFs will be downloaded. Hi is it possible to update multiple records of the same table using a single hidden field with a while loop??? For example: Code: [Select] <?php if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1")) { $updateSQL = sprintf("UPDATE ships SET HealthA=%s WHERE ShipID=%s", GetSQLValueString($_POST['healtha'], "int"), GetSQLValueString($_POST['shipid'], "text")); ?> <input name="healtha" type="hidden" id="healtha" value="<?php echo $row_Ships['HealthA'] + 1; ?>" /> <input name="shipid" type="hidden" id="shipid" value="<?php while($row = mysql_fetch_assoc($Ships)){ echo $row_Ships['ShipID']; }?>" /> If its not possible what else could i do to achieve this please? Thank you Hi, I believe this is my first post on here. First off let me say thank you for silently helping out in the past while I keep on learning PHP. I have decided to build an image gallery, and I am having problems with the page numbering links and the breadcrumbs. Reson I decided to build my own is that I wanted a challange to learn more. I could not find anything that would intregrate with the site as I wanted it to (pulled as a function so offers some customising with options). The problem that I am having is that the <a href= tag is starting to give multipul "/"s when I only need one, so that the url at the top is coming up as http://localhost/mmv4/gallery/pics///Ax/Ax_Jo/ when I am wanting http://localhost/mmv4/gallery/pics/Ax/Ax_Jo/IMG_4244.JPG. I have used str_replace to catch //s and more, but this then adds odd slashes on the end so the page numbeing is getting odd results. some dirs have the / on the end and some do not, so the links are sometimes "visited" and others not. I beleve that the problem is generated in Code: [Select] <?php if(isset($_GET['album'])) { //load the available dirs into an array so that the sneeky users cannot get further back. $files = scandir($gal); $dirs = array(); foreach($files as $file){ if($file != '.' && $file != '..' && $file != 'thumbs') { if(is_dir($gal."/".$file)) { $dirs[] = $file; //echo $file ."<br />\n"; } } } //if(in_array($_GET['album'],$dirs)) { $album = "/".$_GET['album']."/"; $album = str_replace("..", "", $album); //} //else { // $output .= "<p>There is no album called \"$_GET[album]\", please select an album from below.</p>"; //} } or the breadcrumbs bit Code: [Select] <?php if($crumbs == 1) { //bread crums $path = $gal.$album; $path = explode("/", $path); $crumbs = array(); foreach($path as $key => $value) { if($value != ""){ $crumbs[] = $value; } } // create Breadcrumbs div $output .= "<div id=\"breadcrumbs\">"; $number = count($crumbs); $real = ""; foreach($crumbs as $num => $link) { if ($link == $gal) { $link = "Gallery Home"; } if ($num == $number -1) { $output .= $link; } else{ if ($num >= 1){ $real =""; for($x=1; $x<=$num; $x++) { $real .= "/".$crumbs[$x]; } $output .= "<a href=\"?album=".str_replace("/", "",$real)."\">".$link."</a>"." <- "; } else{ $output .= "<a href=\"?album=".str_replace("/", "",$real)."\">".$link."</a>"." <- "; } } $num +1; } //close breadcrumbs div $output .= "</div>"; } Can any one please help me shed some light on this one, I am confused. I can send / attach the full source if needed. MOD EDIT: tags fixed for syntax highlighting . . . Hey Im always trying to remove code and cut corners to reduce work in the long run, soIim wondering how I could link my menu bar from say a template to ALL my php pages for my site so I don't have to write/change links on every page when I need to. Thanks Hey, I was wondering if there is a way to pull multiple rows at once using a list of unique identifiers. For example, I want to pull the rows with the IDs of 4,13,91 and 252 I know the WHERE part of this query is incorrect, but I'm putting it to hopefully help you guys understand what I'm looking for. $result = mysql_query("SELECT * FROM $table WHERE id='4' OR '13' OR '91' OR '252'"); while($row = mysql_fetch_array($result)) { echo($row['name']); } Or is the best way simply to do it one query at a time without a while statement? There could be as many as a few dozen records being pulled per page. |
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