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PHP - Dynamic Dropdown Lists
Hi 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> Similar TutorialsHi 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']."\">".$categoryReturned['name']."</option>"; } ?> </select> I'm not sure if this is possible? Many thanks for your help. Pete hello. i have an issue where the data stored with an image is not saving to a mysql table. the image data is ok, just not the selections from the dropdown lists. here is the code <?php include ('connect.php'); // Insert any new image into database if(isset($_POST['xsubmit']) && $_FILES['imagefile']['name'] != "") { $fileName = $_FILES['imagefile']['name']; $fileSize = $_FILES['imagefile']['size']; $fileType = $_FILES['imagefile']['type']; $content = addslashes (file_get_contents($_FILES['imagefile']['tmp_name'])); $jeweltype = $_POST['jeweltype']; $jewelsize = $_POST['jewelsize_in']; $jewelcolour = $_POST['jewelcolour_in']; $jewelmaterial = $_POST['jewelmaterial_in']; $jewelgender = $_POST['jewelgender_in']; if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } // Checking file size if ($fileSize < 150000) { mysql_query ("INSERT into jewel_images (name,size,type,content,jeweltype,jewelsize,jewelcolour,jewelmaterial,jewelgender) " . "values ('$fileName','$fileSize','$fileType','$content','$jeweltype','$jewelsize','$jewelcolour','$jewelmaterial','$jewelgender')"); } else { $err = "The Image file to too large!"; } } // Find out about latest image $gotten = mysql_query("select * from jewel_images order by row_id desc"); $row = mysql_fetch_assoc($gotten); $bytes = $row['content']; // If this is the image request, send out the image if ($_REQUEST['pic'] == 1) { header("Content-type: $row[type];"); print $bytes; } ?> <html> <head> <title>Upload an image to a database</title> </head> <body> <font color="#FF3333"><?php echo $err ?></font> <table> <form name="Upload" enctype="multipart/form-data" method="post"> <tr> <td>Upload <input type="file" name="imagefile"><br /> Jewelery Type: <select> <?php $sql="SELECT jeweltype FROM jeweltypes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jeweltype" ><?php echo $data['jeweltype'] ?></option> <?php } ?> </select> <br /> Jewelery Size: <select> <?php $sql="SELECT jewelsize FROM jewelsizes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelsize" ><?php echo $data['jewelsize'] ?></option> <?php } ?> </select> <br /> Jewelery Colour: <select> <?php $sql="SELECT jewelcolour FROM jewelcolours"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelcolour_in" ><?php echo $data['jewelcolour'] ?></option> <?php } ?> </select> <br /> Jewelery Material: <select> <?php $sql="SELECT jewelmaterial FROM jewelmaterials"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelmaterial_in" ><?php echo $data['jewelmaterial'] ?></option> <?php } ?> </select> <br /> Jewelery Gender: <select> <?php $sql="SELECT jewelgender FROM jewelgenders"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelgender_in" ><?php echo $data['jewelgender'] ?></option> <?php } ?> </select> <br /> <input type="submit" name="xsubmit" value="Upload"> </td> </tr> <tr> <td>Latest Image</td> </tr> <tr> <td><img src="?pic=1"></td> </tr> </form> </table> </body> </html> ============================================== here is the query =============================================== <html> <head><title>Your Page Title</title></head> <body> <?php $database="josh_jewel"; mysql_connect ("localhost", "xxxxxxxxx", "yyyyyyyyyyyy"); @mysql_select_db($database) or die( "Unable to select database"); $result = mysql_query( "SELECT jewelcolour FROM jewel_images" ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); print "There are $num_rows records.<P>"; print "<table width=400 border=1>\n"; while ($get_info = mysql_fetch_row($result)){ print "<tr>\n"; foreach ($get_info as $field) print "\t<td><font face=arial size=1/>$field</font></td>\n"; print "</tr>\n"; } print "</table>\n"; ?> </body> </html> Hello to all. I'm new here and i am very glad to be a part of this community. I am new in PHP/mysql but i try to learn as much as i can in php & mysql. And here is the problem: Into a php or html page i want to display 2 dropdown list to act as a filter for displaying a table. I managed to display a table based on option select from first dropdown but i don'y know how to make or activate the second dropdown list to act as a filter, and when i press "result" to display on the same page the results. Thanks a lot. Hi guys, I've got this php script which display the users of my database in a dynamic dropdown: <?php include "leadscript/connect_to_mysql.php"; $canvass_name=""; $sql = mysql_query("SELECT * FROM csj_canvasser"); $appointmentCount6 = mysql_num_rows($sql); // count the output amount if ($appointmentCount6 > 0) { while($row = mysql_fetch_array($sql)){ $c_employee = $row["c_employee"]; $canvass_name .='<option value="' . $c_employee . '">' . $c_employee . ' </option>'; } } ?> <form> <select name="c_employee"> <option value="">Select a person:</option> <?php echo $canvass_name; ?> </select> </form> I was wondering if there's a way I can write a code to GET value I select from the dynamic dropdown and use it to write a select query. Thank Hi guys. I am having a hard time finding a solution for this, is it possible to get not the value of a dropdown (oh what's it called??? ) but what is in between of the <option> tag?like, Code: [Select] <select name="catID"> <option value=$row['c_id']>$row['c_name']</option> and save it to the database??cuz I'm using a dynamic dropdown which bases the content of another dropdown by the id of the previous. And so, if i save it to the database, instead of for example "BSA" is saved, the id of "BSA" which is "1" is saved..any ideas guys? 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> I want make the following, (I have already a database with three tables (Countries, Timeline and Category)). 1: list of countries (drop down menu 1), Timeline of the countries history (drop down 2) and Category (drop down 3). 2: The selected values of the drop down menus must show take the information from the database. Can any one help me with the coding? Hi all, here's my code: Code: [Select] <?php foreach ($_SESSION['topping'] as $value) { echo "<tr><td width='30%'>Topping</td><td width='50%'>$value</td><td width='20%'><select name='notopping'>"; foreach ($_SESSION['cupcake'] as $number) { '<option name="notoppings[]" value="'.$number.'">".$number."</option>'; } echo "</select></td></tr>"; } ?> $_SESSION['cupcake'] is a value from either 6, 12, 24 or 36. What I want to do is put them into a drop down box (second foreach) as the value and the displayed value - counting up from 1 (so 1,2,3,4,5,6 or up to 12,24 etc). Also by creating this as an array, does this mean than for each topping (say Vanilla and Chocolate) the value dynamically created can be used on the next page by using $_POST['notoppings'] to display each type (two different numbers - one for Vanilla and one for Chocolate). Does that make sense? Thanks! Jason Hi, I am hoping someone can help me out with a slight issue I have with php and mySQL. I have an ajax-powered form with a select (dropdown) field populated through a php function. Based on the user-selected values in this field, data is displayed on the webpage; i.e. selected value 1 returns values x and y on the page. I am now trying to call additional data (value z) from a different table in the same database, and as before, use the selected values from the dropdown to display the data. For some reason, value z is not changing according to the user-selected value. This is my code: [The function to populate the select field] Code: [Select] function kfl_get_funds_names() { $result = array(); $result['CDF'] = 'Crosby Dragon Fund'; $result['CPF'] = 'Crosby Phoenix Fund'; $result['AMZPIF'] = 'AMZ Plus Income Fund'; $result['KASBIIF'] = 'KASB Islamic Income Opportunity'; $result['KASBCPGF'] = 'KASB Capital Protected Gold Fund'; $result['KASBLF'] = 'KASB Income Opportunity Fund'; $result['KASBCF'] = 'KASB Cash Fund'; $result['KASBBF'] = 'KASB Asset Allocation Fund'; $result['KASBSMF'] = 'KASB Stock Market Fund'; return $result; } [the code calling and using the function to interact with the database] Code: [Select] $funds_to_display = kfl_get_funds_names(); $current_symbol = key( $funds_to_display ); $current_nav_rates = kfl_get_latest_rates( $current_symbol ); [the code calling additional data, value z, from the database, and using the info in the select field to filter it] Code: [Select] $cutoff = kfl_cutoff( $current_symbol ); The display of each of these items is as follows: Code: [Select] <?php echo $current_nav_rates['nav_date']; ?> <?php echo $funds_to_display[$current_symbol]; ?> <?php echo $cutoff['cutoff']; ?> I can't get the $cutoff code to display the correct values. It picks up the first symbol to display and doesn't change with user selection. The code for the selection box, by the way: Code: [Select] <select id="dailynav-funds" autocomplete="off" name="dnf"> <?php foreach ($funds_to_display as $fund_symbol => $fund_name) { echo '<option'; if( $fund_symbol == $current_symbol ) { echo ' selected="selected"'; } echo ' value="' . $fund_symbol . '">'; echo $fund_name; echo '</option>'; } ?> </select> I've tried to get data using $_GET['dnf'] into the cutoff code, but that throws up parse errors. What am I doing wrong, and how can I resolve this issue? Thanks in advance! I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID. Index.php
<!doctype html>
<?PHP
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass", "Database" =>
"database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
?>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<section id="formaT2" class="formaT2 formContent">
<div class="row">
<div class="col-md-2 col-3 row-color remove-mob"></div>
<div class="col-md-5 col-9 bg-img" style="padding-left: 0;
padding-right: 0;">
<h1>Form</h1>
<div class="rest-text">
<div class="contactFrm">
<p class="statusMsg <?php echo
!empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
<form action="connection.php" method="post">
<div>machinery</div>
<select id="machinery">
<option value="0">--Please Select Machinery--</option>
<?php
// Fetch Department
$sql = "SELECT Machinery FROM T013";
$machanery_data = sqlsrv_query($conn2,$sql);
while($row = sqlsrv_fetch_array($machanery_data) ){
$id = $row['Id'];
$machinery = $row['Machinery'];
// Option
echo "<option value='".$id."' >".$machinery."</option>";
}
?>
</select>
<div class="clear"></div>
<div>Sub Machinery</div>
<select id="sub_machinery">
<option value="0">- Select -</option>
</select>
<input type="submit" name="submit"
id="submit" class="strelka-send" value="Insert">
<div class="clear"> </div>
</form>
</div>
</div>
</div>
</div>
</section>
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#machinery").change(function(){
var machinery_id = $(this).val();
$.ajax({
url:'ajaxfile.php',
type: 'post',
data: {machinery:machinery_id},
dataType: 'json',
success:function(response){
var len = response.length;
$("#sub_machinery").empty();
for( var i = 0; i<len; i++){
var machinery_id = response[i]['machinery_id'];
var machinery = response[i]['machinery'];
$("#sub_machinery").append("<option
value='"+machinery_id+"'>"+machinery+"</option>");
}
}
});
});
});
</script>
</body>
</html>
Ajaxfile.php
<?php
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass",
"Database" => "database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
$machineryID = $_POST['machinery']; // department id
$sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
$result = sqlsrv_query($conn2,$sql);
$machinery_arr = array();
while( $row = sqlsrv_fetch_array($result) ){
$machinery_id = $row['ID'];
$machinery = $row['MachineID'];
$machinery_arr[] = array("ID" => $machinery_id, "MachineID" =>
$machinery);
}
// encoding array to json format
echo json_encode($machinery_arr);
?>
Edited May 6, 2019 by davidd
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 freaks, I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database. The code I'm using to populate dropdown: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if(!$con) { die ('Can not connect to : '.mysql_error()); } mysql_select_db("ims",$con); $result=mysql_query("select cat_id,cat_name from category"); echo "<select name=cat>"; while($nt=mysql_fetch_array($result)) { echo "<option value=$nt[cat_id]> $nt[cat_name] </option>"; } echo "</select>"; mysql_close($con); ?> Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]"); Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow. Help on this will be highly appreiable . (Tell me if I'm not clear with my question) Hi I have to multi select tables, the table on the left is field with the contents of an array, I then have it so the user can make there selections from the left box and move them into the right. They then hit submit and it gets stored inmysql table as an array. The problem I am having is keeping the selections in that right box after any sort of page refresh. Any ideas I have pasted the php and html code below, the add/remove function is controlled by javascript, I won't bother pasting that as i don't think it has anything to do with it. PHP ******** Code: [Select] <?php / ------- PARSING PERSONAL TRAINER DETAILS VENUE --------- if ($_POST['parse_var'] == "fromPool"){ $venue = $_POST['venue']; if (is_array($_POST['venue'])) { $venue = explode("\n", $_POST['venue']); } else { $venue = $_POST['venue']; } // Update the database data now here for all fields posted in the form $sqlUpdate = mysql_query("UPDATE ptdata SET venue='$venue' WHERE id='$id' LIMIT 1"); if ($sqlUpdate){ $success_msg = '<img src="images/Form/round_success.png" width="20" height="20" alt="Success" />Your venue information has been updated successfully'; } else { $error_msg = '<img src="images/Form/round_error.png" width="20" height="20" alt="Failure" /> ERROR: Problems arose during the information exchange, please try again later</font>'; } } // ------- END PARSING PERSONAL TRAINER DETAILS VENUE --------- ?> <?php ////////// Venue Array PHP For Multi Selection Boxes //////////// $dbString = ''; // To be pulled from Database $toPool = explode(',', $dbString); $pool = Array('Your Home','My Home','Outside','Private Studio','Your Work','Gym'); if($_SERVER['REQUEST_METHOD'] == 'POST'){ $toPool = (count($_POST['venue']) > 0)? $_POST['venue'] : Array(); $newDbString = implode(',', $toPool); // Store in Database; } $fromPool = array_diff($pool, $toPool); ?> html *********** <form action="ptmemberaccount.php" method="post" enctype="multipart/form-data"onsubmit="multipleSelectOnSubmit()"> <table width="500" align="left"> <tr> <td width="500" height="300" align="center"> <select multiple="multiple" name="fromBox[]" id="fromBox"> <?php foreach($fromPool as $itm){ echo "\t" . '<option value="' . $itm . '">' . $itm . '</option>' . PHP_EOL; } ?> </select> <select multiple="multiple" name="venue[]" id="toBox"> <?php foreach($toPool as $itm){ echo "\t" . '<option value="' . $itm . '">' . $itm . '</option>' . PHP_EOL; } ?> </select> </td> </tr> <tr> <td align="center" height="35"><script type="text/javascript"> createMovableOptions("fromBox","toBox",400,200,'Training Venues','Selected Training Venues'); </script> <p>Use the buttons to Add/Remove selections</p></td></td> </tr> <tr> <td align="center" height="100"> <input type="submit" value="OK"> </tr> </table> </form> If you need anymore info let me know, also if anyone knows a better way of achieving this i am open to suggestions. Many Thanks MOD EDIT: code tags added Hi, for uni i've got to make a shopping cart application in php/mysql. I'm almost done with it, managed to teach myself some nifty jquery and have been able to learn loads more PHP, I'm really enjoying it. I've come to a bit of a stand still though, I showed my application as it is to my lecturer and he had one little point. My application, you have lists, each list has its own categories (users can add, edit, delete their own) and in each category, there are items (which like categorys can be added, edited or deleted). So basically, you have a browse_list.php, which browses all the lists, each list has it's own unique hyperlink to browse_list_category.php?list_id= using the get method, it selects all the categories associated with that list, which, like browse_list.php have their own hyperlinks, but this time they link to browse_category_item.php?category_id= and they list what ever items are associated with that category. Sounds a mouth full, but it's really quite simple. I showed my lecturer that and he wasn't happy with the whole category/item layout. He said it would be better if the category list had a nested item list underneath it. He was happy with the whole list to category part with the hyperlink, though. Kinda like: Quote -list -------category -------item -list1 ------category1 ------item1 ------category2 ------item2 ------category3 ------item3 My database layout is tbllist -main list table, list_id, list_name tblcategory - main category table, category_id, category_name tblcategory_list - link table for list and category (many-many) list_id, category_id tblitem - main item table, item_id, item_name tbl_category_item - link table for category and item (many-many) category_id, item_id So, now you've kinda got the gist of what I'm doing.. I just need help with displaying each item in a category, under the category.. like a nested list <ul> <li>category 1</li> <ul> <li> item 1</li> <li> item 2</li> <li> item 3</li> </ul> </ul> I did try and have a go earlier, kinda rushed and managed to come up with this: <ul id="sortme" class="shoppinglist"> <?php // Split records in $result by table rows and put them in $row. while($row = mysql_fetch_array($result)) { echo '<li class="list" id="' . $row['category_id'] . '"><div class="text"><a href="browse_item_category.php?category_id=' . $row['category_id'] . '">' . $row['category_name'] . "</a></div> <div class=\"actions\"><a href=\"#\" class=\"tick\">Tick</a> <a href=\"categoryupdater.php?category_id=" . $row['category_id'] . "\" class=\"edit\">Edit</a><a href=\"#\" class=\"delete\">Delete</a></div> </li>\n"; } ?><ul id="sortme" class="shoppinglist"> <?php while($row = mysql_fetch_array($resultitem)) { echo '<li class="list" id="' . $row['item_id'] . '"><div class="text">' . $row['item_name'] . "</div> <div class=\"actions\"><a href=\"#\" class=\"tick\">Tick</a> <a href=\"itemupdater.php?item_id=" . $row['item_id'] . "\" class=\"edit\">Edit</a><a href=\"#\" class=\"delete\">Delete</a></div> </li>\n"; } ?> </ul> </ul> but this i get this not the whole list nested'ness it's a bit long winded I know but could anyone possible give me a hand? i've done pretty well so far not asking for help, but I think this is going to be the end of me! Hi guys, I would like to display data from MySQL in the format of 5 lists (of one column each) one next to each other on a webpage I am quite good with SQL, but I lack knowledge about php and displaying data. SQL version: 5.1 Does anyone have any good website that would show how to do such a thing? I want to produce a pdf document using mysql and TCPDF. I can do this OK in some circumstances but I now have a problem. I have two tables - documents and parties - they are linked by a common field docid. I have for table documents, the fields: docid, doctitle, idcode and for the parties table: partyid, docid, party (the field party is the name of the party) For any document there could be any number of parties and for a single idcode there could be any number of documents. For example, for idcode = 12345 there might be 2 documents called doctitle 1 and doctitle 2. For doctitle 1 there are 3 parties to it party1, party 2 and party 3. For doctitle 2 there are two parties - party4 and party 5. For any idcode (which is passed through by the url) I want the pdf to give a list such as: Doctitle 1: party name 1 party name 2 party name 3 Doctitle 2: party name 4 party name 5 Part of my code is: mysql_select_db($database_process, $process); $result2 = mysql_query("SELECT parties.docid, GROUP_CONCAT(party SEPARATOR '<br>') AS partylist, documents.doctitle FROM parties INNER JOIN documents ON parties.docid=documents.docid WHERE parties.idcode = '$appidpassed' GROUP BY parties.docid LIMIT 1 ") or die(mysql_error()); while($row = mysql_fetch_array($result2)) { $doctitle = $row['doctitle']; $parties = $row['partylist']; } and then further down the page: $doctitle<br> $parties This only gives one document and the parties associated with it. I know the solution is probably easy but I can't see it. Does anyone have any ideas? |
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