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PHP - Dynamic Lists With Dynamic Selected
Hi all
I need to combine these two scripts: Firstly, the following decides which out of the following list is selected based on its value in the mySQL table: <select name="pack_choice"> <option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option> <option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option> <option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option> </select> Although this works OK, I need it also to show dynamic values like this: select name="category"> <?php $listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC"); while($categoryReturned=mysql_fetch_array($listCategories)) { echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>"; } ?> </select> I'm not sure if this is possible? Many thanks for your help. Pete Similar TutorialsHi everyone, I've read through the FAQs for Dynamic Dropdown Menus on this site, as well as others, and I can't figure out why my code won't work. I have two dropdown boxes that need to be populated with data from a mysql table; one menu for route types, and one for route numbers. When I choose the route from the 'Route' menu, the 'Number' menu automatically populates with all of the possible numbers, rather than only those that correspond with the route type. I can tell that the problem has something to do with the value of 'Route' not being recognized, but I don't know why. I'm a beginner when it comes to PHP, so any suggestions or help would be much appreciated! Thanks! The code is as follows: Code: [Select] <html> <body> <basefont face='calibri' color='#7E2217'> <?php // set variables $mileTable = $_GET['mileTable']; $routeType = isset($_POST['Roadtype'])? $_POST['Roadtype']: 0; include 'opendbMile.php'; include 'error.php'; // Connect to the MySQL DBMS if (!($connection = @ mysql_connect($hostName, $username, $password))) die("Could not connect"); if (!mysql_select_db($databaseName, $connection)) showerror( ); // Start a query... $query = "SELECT ID, Roadtype FROM Alabama GROUP BY Roadtype"; // execute the SQL statement $result = mysql_query($query, $connection) or die(mysql_error()); echo '<form name="mileform" method="post" action="MileQuery.php">'; echo '<p>Route: <select name="routeType" id="routeType" onchange="this.form.submit();"> <option value="0"'.($routeType == 0? ' SELECTED': '').'>Route</option>'; while($row = mysql_fetch_array($result)){ echo ' <option value="'.$row[0].'"'.(($routeType == $row[0])? ' SELECTED': '').'>'.$row[1].'</option>'; } echo ' </select> </p>'; // create the SQL statement $query2 = "SELECT ID, Roadnumber FROM Alabama GROUP BY Roadnumber"; if($mnucategory != 0){ // Filter road numbers $query2 .= " WHERE Roadtype='".$routeType."'"; } // execute the SQL statement $result2 = mysql_query($query2, $connection) or die(mysql_error()); echo '<p>Number: <select name="routeNumber" id="routeNumber" onchange="this.form.submit();"> <option value="0"'.($routeNumber == 0? ' SELECTED': '').'>Number</option>'; while($row2 = mysql_fetch_array($result2)){ echo ' <option value="', $row2[0].'"'.(($routeNumber == $row2[0])? ' SELECTED': '').'>'. $row2[1], '</option>'; } echo ' </select> </p>'; // Close the DBMS connection mysql_close($connection); ?> </form> </body> </html> The rest of my coding works however this part does not and I'm trying to figure out why. I'm sure my syntax isn't right so I hope someone can correct my mistake. $contentpageID = $_GET['id']; $query = "SELECT contentpages.contentpage, contentpages.shortname, contentpages.contentcode, contentpages.linebreaks, contentpages.backlink, contentpages.showheading, contentpages.visible, contentpages.template_id FROM contentpages WHERE contentpages.id = '" . $contentpageID . "'"; <label for="template">Template</label> <select class="dropdown" name="template" id="template" title="Template"> <option value="0">- Select -</option> <?php $query = 'SELECT * FROM templates'; $result = mysql_query ( $query ); while ( $template_row = mysql_fetch_assoc ( $result ) ) { print "<option value=\"".$template_row['id']."\" "; if($template_row['id'] == $row['template_id']) { print " SELECTED"; } print ">".$template_row['templatename']."</option>\r"; } ?> </select> Hi, I wanted to make a dynamic php page. I want the user to choose an option you could select in an option box (html) and after that I wish that down the page some text would change. If I take option 1 -> text 1, option 2 -> text 2, ... But this doesn't work, do I need a button or something to made my code running, and how do I have to use it on the same page? Or is it possible to run it automaticly after a new choice? Thanx, Kevin My code: <?php /* Template Name: template voor Ieder1Trainer # */ get_header(); $Tacktiek = ""; $Tacktiek1 = "3-5-2"; $Tacktiek2 = "3-4-3"; $Tacktiek3 = "4-5-1"; $Tacktiek4 = "4-4-2"; $Tacktiek5 = "4-3-3"; $Tacktiek6 = "5-4-1"; $Tacktiek7 = "5-3-2"; $aantalDef; $aantalMid; $aantalAtt; ?> <div id="newsfeed"> <div class="banderol"> <div class="banderol_left"></div> <h1>Dynamic page</h1> </div> <?php if(is_user_logged_in()) { get_currentuserinfo(); //echo 'Username: ' . $current_user->user_login . "\n"; //echo 'User email: ' . $current_user->user_email . "\n"; ?> <!-- Here you choose a tactiek; 1,2,3, .. after that has been choosen I whise the next php-code would be run (the if-then-elseif-...) --> <tr> <td><p class="formulier">Kies je Tacktiek:</p></td> <td> <select size="1" name="$Tacktiek"> <option selected><?php echo $Tacktiek1;?></option> <option><?php echo $Tacktiek2;?></option> <option><?php echo $Tacktiek3;?></option> <option><?php echo $Tacktiek4;?></option> <option><?php echo $Tacktiek5;?></option> <option><?php echo $Tacktiek6;?></option> <option><?php echo $Tacktiek7;?></option> </select> </td> </tr> <!-- This must be run after choosen a tactiek. --> <?php if($Tacktiek == '3-5-2') { $aantalDef = 3; $aantalMid = 5; $aantalAtt = 2; } elseif($Tacktiek == '3-4-3') { $aantalDef = 3; $aantalMid = 4; $aantalAtt = 3; } elseif($Tacktiek == '4-5-1') { $aantalDef = 4; $aantalMid = 5; $aantalAtt = 1; } elseif($Tacktiek == '4-4-2') { $aantalDef = 4; $aantalMid = 4; $aantalAtt = 2; } elseif($Tacktiek == '4-3-3') { $aantalDef = 4; $aantalMid = 3; $aantalAtt = 3; } elseif($Tacktiek == '5-4-1') { $aantalDef = 5; $aantalMid = 4; $aantalAtt = 1; } elseif($Tacktiek == '5-3-2') { $aantalDef = 5; $aantalMid = 3; $aantalAtt = 2; } ?> <!-- After this run, the next code must be run and made a choose of text --> <tr> <td><p class="formulier"> <?php if($aantalDef == '3') { echo "3 defs"; //Choosen text if aantalDef == 3 } elseif($aantalDef == '4') { echo "4 defs"; //Choosen text if aantalDef == 4 } elseif($aantalDef == '5') { echo "5 defs"; //Choosen text if aantalDef == 5 } ?> </p></td> <!-- Other choose of options, but always the same options. --> <td> <select size="1" name="$Speler4"> <option selected><?php echo $NaamDef1;?></option> <option><?php echo $NaamDef2;?></option> <option><?php echo $NaamDefMid1;?></option> <option><?php echo $NaamDefMid2;?></option> <option><?php echo $NaamDefMid3;?></option> <option><?php echo $NaamDefMid4;?></option> <option><?php echo $NaamDefMid5;?></option> <option><?php echo $NaamDefMid6;?></option> <option><?php echo $NaamDefMid7;?></option> <option><?php echo $NaamDefMid8;?></option> <option><?php echo $NaamDefMidAtt1;?></option> <option><?php echo $NaamDefMidAtt2;?></option> <option><?php echo $NaamDefMidAtt3;?></option> </select> </td> </tr> <?php } else{ echo 'U dient in te loggen.'; } ?> </div> <?php get_footer(); ?> I'm trying to create a dynamic option menu with one alert selected based on the first query to the db. Any help would be greatly appreciated. Code: [Select] //function to get alerts and create select menu with current alerts pre-selected function getALERTS1($id){ require('db.php'); $alert = mysqli_query($conn, "SELECT alert1 FROM visit_data WEHERE patientid = $id AND discharged IS NULL"); $row = mysqli_fetch_array($alert); $selects=null; $query = mysqli_query($conn, "SELECT alertid, name FROM alerts"); while($row1 = mysqli_fetch_array($query)) { $selects .= "<option value=\"" . $row1['alertid'] . "\"> if($row1['alertid']==$row['alert1']) { echo ' selected'; } ".$row1['name']."</option>"; } return $selects; } Folks, I need help (Php code ) to generate a Dynamic Text on a Base Image. What i want to do is, to make this Image as header on my Site and to make this Header Specific to a Site, i want to Add the Domain Name on the Lower Left of the Image. Got the Idea? Here is the Image link: Quote http://img27.imageshack.us/i/shoppingheader1.jpg/ PHP Variable that holds the Domain name is: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hi, A quick question please , I am trying to assign a selected index to a drop down list based on data retrieved from the database. I tried using JavaScript : document.getElementById('Country').selectedIndex ='".$array['Country']."'; but it does work. Any suggestion? Hello! If you search Google, you'll notice the URL: http://www.google.com/webhp?hl=en#hl=en&source=hp&q=php+freaks&aq=f&aqi=g10&aql=&oq=&gs_rfai=CtdaTdtdRTNnPD5HuzASZtMWiCgAAAKoEBU_Q0ZT5&pbx=1&fp=19d754eee0b4f223 You can copy that URL anywhere you like... and the user will still see the same results. So basically the URL dynamically does an action, and accesses the database. How on do I make a URL like this? Meaning when you change the URL parameter values, it request a slightly different database query? Does that make sense? Similarly with NexTag.com: http://www.nextag.com/serv/main/buyer/ProductCompare.jsp?search=camera&page=0&node=500001&psort=%2FDigital-Cameras--zzcameraz500001zB6z5---html&zipcode=&cptitle=657166355&cptitle=656751324&cptitle=620051906&cptitle=705150048 That was a Dynamically made URL (I selected from check boxes which products to compare) and it makes that URL so anyone can see those products. Any ideas on how to do this? More examples on the URL: http://www.cars.com/go/compare/modelCompare.jsp?myids=9721,11439 (i select the cars, it generates that URL. Notice the IDs 9721,11439) So we just started php after learning a month of HTML, and we need to create a website with a page that allows the admin to change the colors and font size, header, etc on the webpage itself via forms. E.g. I'm on the site and I can use a drop down menu to change the background to "blue" for example. No need to go edit the CSS file or anything. How can be this done? My second and final question is how would I write code that allows my website to have users "log in" to their account so they can edit their page? Thanks a lot, I'm not good at PHP yet I've just started so please keep that in mind:) - Kranti Hi, I'm learning php and I'm trying to get to grips with dynamic xml and php coming from a database. The trouble I am having is I can't get my code to give any output. I don't get an error I message, I just can't get the xml to display. If anyone can give me any pointers as to what I'm doing wrong that would be great... Here's the code I have. Code: [Select] <?php DEFINE ('DB_USER', 'root'); DEFINE ('DB_PASSWORD', 'password'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'flashphpbible'); $link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $query = "SELECT * FROM dynamicxml WHERE id=1"; $result = mysqli_query($link, $query); $xmlData .= " "; $xmlData .= "<store>\n"; while($row = mysqli_fetch_array($result)); { $xmlData .= " <item>\n"; $xmlData .= " <name>".$row['name']."</name>\n"; $xmlData .= " <section>".$row['section']."</section>\n"; $xmlData .= " <price>".$row['price']."</price>\n"; $xmlData .= " <inStock>".$row['instock']."</inStock>\n"; $xmlData .= " </item>\n"; } $xmlData .= "</store>\n"; echo "response=".$xmlData; ?> Thanks for any help offered. Ben. Hi all This probarly looks really bad but im trying to replase tags that change. So far it only replaces the last tag and not well either. I will try to show you Code: [Select] function testfunc($value) { if ($value=='gallery'){$return .= "this gallery has worked = --$value--";} if ($value=='email'){$return .= "this email has worked = --$value--";} return $return; } $test = "this is a test {%email%} message {%gallery%} rtfgdsfgdsfg"; $test = preg_replace("#(.*){%(.*?)%}(.*)#is",'$1' . testfunc('$2') . '$3', $test); echo $test; this outputs this is a test {%email%} message rtfgdsfgdsfg but im trying to get this this is a test this email has worked = --email-- message this gallery has worked = --gallery-- rtfgdsfgdsfg Im sure im way off but I hope you can understand. Thank you for taking the time to read this post. Is there a way to create dynamic content without the use of a Content Management System? I simply want to click on a link in my navigation menu and have that display the content in a div section and keep the content in a separate folder on the server. I didnt want to have to use a CMS to accomplish that because every other feature of it will be unused as I just simply want to display the content in a div when I click on the link. I use a simple test server on my Home PC that I use to format the articles in html from documents I have wrote in LibreOffice, so I dont really need all the features and sophistication of full blown CMS. But I also dont want a bunch of different files in my root folder if I can avoid it. I would rather keep the content in a separate folder and just pull it from there and display it on the main page when the link is clicked. Anyway, I dont know if its even possible or if its the best way, so I am open to suggestions as well. Some of the documents are quite long (like several chapters) and some are much shorter (just a few pages). Any help and ideas are much appreciated. Hey guys does anyone know of a good dynamic directory solution? What i mean by this is when i am creating an application and i use require_once 'includes/config.php'; and then i create a class that needs the config file so i include it into the class like so: require_once '../include/config.php'; then i make a new folder named member then i have to require this class (created above) and it fails to find the directory. What would be a good solution for this? Thanks guys!! Hello: I am trying to make a StyleSheet updateable from my admin panel, but I'm not sure if what I want to do is possible. So ... I have a TEXTAREA in my admin panel that writes to the DB just fine. it has styles like: Code: [Select] body { margin: 5px 0; padding: 0; background-color: #ebe9e6; background-image: url('../images/Site-BG2.jpg'); background-position: center bottom; background-repeat: no-repeat; background-attachment: scroll; font-family: arial, sans-serif; font-size: 100%; line-height: 1.4em; } ... Etc... On the frontend, it writes into the StyleSheet just fine like: Code: [Select] <?php include('myConn.php'); //contains your mysql connection and table selection $query=mysql_query("SELECT * FROM myStyleSheet") or die("Could not get data from db: ".mysql_error()); while($result=mysql_fetch_array($query)) { $myPageData=$result['myPageData']; } ?> <?php echo $myPageData; ?> So, what I want to see is if I can somehow pull the styles onto the page like: Code: [Select] <link rel="stylesheet" type="text/css" href="include/StyleSheet.css" /> Now, if I do it as listed above it and look at the code it displays the Code: [Select] <?php include('myConn.php'); //contains your mysql connection and table selection Etc... If I do it like Code: [Select] <link rel="stylesheet" type="text/css" href="include/StyleSheet.php" /> It displays the styles fine, but the browser doesn't read it as a StyleSheet and therefore the page doesn't get formatted. So, is there any way to make this work? Anyone have any ideas about this? The idea is so I can manage the Styles via an admin panel remotely without having to login with web editing software. Thanks. Hi, I hope someone can help. I currently have a page that includes the main page after login, however I am looking into changing this so that when a user logs in they get 3/4 linked images that, when clicked dynamically load/include the page based on the select i.e. user selects the image laptop, that has a hyperlink, it loads the laptop page. If anyone has any ideas or suggestion,i'd appreciate it. Let's see if I can explain this ok.. Basically I have a site and the page product categories use dynamically created URL's. I want to use these URL's to put on my products so I can list all the categories each product is in. But, as these URL's are not stored anywhere I'm not sure how to go about calling them. Is the best way going to be to modify the site to store the page URL's in MySQL? Or could they be stored within variables in PHP? A little guidance would be great! Thanks in advance I want to get the ibase_blob_echo function and I was told I need to uncomment the extension=php_interbase.dll in my php.ini. So I did and now I get the message: Unable to Load Dynamic Library in a pop-up window. I'm lost and don't know how to solve this I'm having trouble getting the dynamic data from my <select> menus to write into my MYSQL database. Can anyone see what I'm doing wrong here? First post btw The output looks like this, which is obviously wrong: Code: [Select] <html> <head> </head> <link rel="stylesheet" type="text/css" href="./css/newuser.css" /> <body> <?php session_start(); require 'default.inc.php'; ?> <?php if (isset($_POST['amount'])): $host = 'localhost'; $user = 'user'; $pass = 'password'; $conn = mysql_connect($host, $user, $pass); if (!$conn) { exit('<p>Unable to connect to the database server</p>'); } if (!@mysql_select_db('spikesusers')) { exit('<p>Unable to locate the database</p>'); } $locationname = $_POST['donor']; $donorid = mysql_query("SELECT id FROM donors WHERE locationname='$locationname'"); $amount = $_POST['amount']; $year = $_POST['year']; $type = $_POST['type']; $typeid = mysql_query("SELECT id FROM donationtype WHERE type='$type'"); $player = $_POST['player']; //$playerid = mysql_query("SELECT id FROM players WHERE $sql = "INSERT INTO donations SET donorid='$donorid', amount='$amount', yearofdonation='$year', typeid='$typeid'"; mysql_query($sql); ?> <div class='standard'> <h1>Donation Management</h1> <?php if ($sql) { echo "New donation added "; echo "<p></p>"; echo "<a href=managedonations.php>Back to donation management</a>"; exit(); } else { echo "Error adding new donation"; echo "<a href=adddonation.php>Try again</a>"; exit(); } ?></div> <?php else: $host = 'localhost'; $user = 'user'; $pass = 'pass'; $conn = mysql_connect($host, $user, $pass); if (!$conn) { exit('<p>Unable to connect to the database server</p>'); } if (!@mysql_select_db('spikesusers')) { exit('<p>Unable to locate the database</p>'); } $donor=@mysql_query('SELECT id, locationname FROM donors'); ?> <div class='standard'> <h1>Donation Management</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <label>Donor: <select class="text" name="donor"> <option value=new>Add a new donor...</option> <?php while ($donors=mysql_fetch_array($donor)) { $donorname=$donors['locationname']; echo "<option value='<?php echo $donorname;?>'>$donorname</option>"; //echo "<option value='hi'>hi</option>"; } ?> </option> </select> </label><br /> <label>Amount: <input class="text" type="text" name="amount" class="text" /></label><br /> <label>Year of donation: <select class="text" name="year"> <option value='2011'>2011</option> <option value='2010'>2010</option> <option value='2009'>2009</option> </select> </label><br /> <?php $player=@mysql_query('SELECT id, firstname, lastname FROM players'); ?> <label>Player: <select class="text" name="player"> <option value="player" selected="selected"></option> <?php while ($players=mysql_fetch_array($player)) { $playerfirstname=$players['firstname']; $playerlastname=$players['lastname']; echo "<option value=player>$playerfirstname $playerlastname</option>"; } ?> </select> </label><br /> <?php $type=@mysql_query('SELECT id, type FROM donationtype'); ?> <label>Donation type: <select class="text" name="type"> <?php while ($types=mysql_fetch_array($type)) { $donationtype=$types['type']; echo "<option value=type>$donationtype</option>"; } ?> </select> </label><br /> <input type="submit" value="SUBMIT" class="buttons"/> <input type="button" name="Cancel" value="CANCEL" onclick="window.location = 'managedonations.php'" class="buttons"/> </form> </div> <?php endif; ?> </body> </html> Hi, I am using php dynamic listing from database. it is working but I wanna integrate to a horizontal css menu how can i integrate my codes to any css horizantal dropdown menu. thans. here my dynamic list. <?php function sinirsiz_kategori($parent) { $sql = mysql_query("SELECT * FROM kategoriler ORDER BY id DESC"); while($row = mysql_fetch_array($sql)) { $diziler[$row['id']] = array('baslik' => $row['baslik'],'parent' => $row['parent']); } $has_childs = false; foreach($diziler as $key => $value) { if ($value['parent'] == $parent) { if ($has_childs === false){ $has_childs = true; echo "\t<ul>"; } echo "<li><a href=\"".$value['baslik'].".php\">".$value['baslik']."</a>"; sinirsiz_kategori($key); echo "</li>\n"; } } if ($has_childs === true) echo "</ul>"; } ?> <?=sinirsiz_kategori(0)?> trying to be able to store client information in mysql db for reference later in an admin area. The code I have so far allows me to list the client names, but once I "select" the name I want it to show the rest of the database information stored for that user (email, height, weight, phone number, etc...) which it currently is not doing. Any and all help will be greatly appreciated, tired of struggling through this blindly. Code: [Select] <? // Connect database mysql_connect("localhost","",""); mysql_select_db("mydb"); if(isset($select)&&$select!=""){ $select=$_GET['select']; } ?> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> </head> <body> <form id="form1" name="form1" method="get" action="<? echo $PHP_SELF; ?> "> Client Name : <select name="select"> <option value="">--- Select ---</option> <? // Get records from database (table "users"). $list=mysql_query("select * from users order by id asc"); while($row_list=mysql_fetch_assoc($list)){ ?> <option value="<? echo $row_list['id']; ?>" <? if($row_list['id']==$select){ echo "selected"; } ?>><? echo $row_list['name']; ?></option> <? } ?> </select> <input type="submit" name="Submit" value="Select" /> </form> <hr> <p> <? if(isset($select)&&$select!=""){ // Get records from database (table "users"). $result=mysql_query("select * from users where id='$select'"); $row=mysql_fetch_assoc($result); ?> Information about <strong><? echo $row['name']; ?></strong> client...</p> <p><? echo $row['email']; ?><br> ........................................<br> ........................................ <? // End if statement. } // Close database connection. mysql_close(); ?> </p> </body> </html> Hey All, Im having trouble tring to figure out how to save dynamic radio buttons. I have a project where i have to use radiobuttons instead of checkboxes. Basically, there is a database table with "roles" in it ... i.e administrator, user, etc etc etc and these are updateable but the user. On a specific page, these get spit out with a radio button next to them, like so: <input name="wfa" type="radio" id="wfa" class="radio" value="1" /><label for="wfa">administrator</label> These go on down the page, for as many entries there are in the database. What i need to do is save the checked ones to the database, but being dynamic i cant save each value to a individual row ..... any help? Someone mentioned serialize, i had a look at the php documentation but it didnt make much sense to me. Cheers. |
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