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PHP - Drop Down Box From Mysql
hey guys just a quick one
i want a drop down box which is populated from a mysql query which is eay enough but also at the same time i also want the dropdown box to look at the current record and have the value in the drop down box e.g. the drop down has values 1, 2, 3, 4, 5 in it which are pulled from a table called values the current record is from a table called staff and it has an id of 45 and a user level of 3 i want the drop down to have all the values selectable to update the record but on load it will have 3 as the current, i want to be able to chage this 3 to a 5 and update the record how can i do this.? just a brief code example would be awsome and ill adopt it to fit Similar TutorialsHey all - I (think) that I'm following the code exactly, but no luck... Essentially I want to populate a drop down menu from text stored in mysql (varchar). Here is my code (and where I'm pulling the code found below). Any ideas what's going wrong. (My error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in....). $sql = "SELECT * FROM table"; $results = mysql_query($sql); //or die("Query failed with error: ".mysql_error()); // execute the query $options=""; while ($row=mysql_fetch_array($results)){ $id=$row["id"]; $thing=$row["thing"]; $options.="<OPTION VALUE=\"$id\">".$thing; } <select name="id" style="font-size:20px;font-family:Arial;width:275px"> <option value=0>Choose<? echo $options?></option> </select> Here is where I'm getting my code sample: http://forums.devarticles.com/mysql-development-50/drop-down-menu-populated-from-a-mysql-database-1811.html Thank for any advice!! Hi, At present the PHP page reads the MySQL data and prints on screen and if the result if 'FREE' it shows a clickable image for the user to book. I am looking for something that allows the user to drag the result they fund to another location with in the table. I.e. user has box 1 filled with their name, they can drag their name to box 9 and it saves it, so they can go back on the it will be their. If anyone knows where I can firn maybe a tutorial, or guide to get started with this, as I have looked and cannot see anything. Cheers. Stu I got some help here about 6 months ago with drop down menus & I need a little more. I'm working on a form that has about 20 drop down menus, each populated from a mysql database table with about 50 entries each. I need to keep the selected menu option on the form after the form is submitted but each time the form is submitted the selected menu option reverts back to the first item in the database table. Is there any practical way to fix this problem other than using javascript to finish up? I'm a newbie on php. I'm really a system administrator and I was just task to do this simple task. For me its hard but I'm sure for a programmer this is very simple. My agenda is to pull out data on one of my column in mysql, select it and dump it on mysql. Here is the php for retrieving mysql data Code: [Select] <?php function database_connect($users) { $resource_link = mysql_connect("localhost", "root", "root"); if (mysql_select_db($users, $resource_link)) { return $resource_link; } else { echo "Cannot connect to DB"; return false; } } function print_dropdown($query, $link){ $queried = mysql_query($query, $link); $menu = '<select username="username">'; while ($result = mysql_fetch_array($queried)) { $menu .= ' <option value="' . $result['id'] . '">' . $result['username'] . '</option>'; } $menu .= '</select>'; return $menu; } //Some other form elements, or just start a form. echo '<form method="post" action="create2.php">'; //The important bit echo print_dropdown("SELECT username FROM mailbox;", database_connect("users")); //Some other form elements, or just end the form. echo '<input type="submit" name="submit" value="submit"/></form>'; Here is the content of my create2.php. This is the php page who do the insert on my mysql. Code: [Select] <?php // open the connection $conn = mysql_connect("localhost", "root", "root"); // pick the database to use mysql_select_db("users",$conn); // create the SQL statement $sql2 = "INSERT INTO mailbox values ('','locked','','$_POST[username]','',NOW(),'','locked','')"; // for troubleshooting $result = mysql_query($sql2, $conn) or die(mysql_error()); // execute the SQL statement //if (mysql_query($sql2, $conn)) { // echo "Success"; //} else { // echo "Fail"; //} } ?> When I click the submit button, I don't see any record being inserted on my table. I'm using the create2.php on my other page though it is only an insert/fill up form not like this one that I need to pull up the date, select and insert to mysql. hey all so I have this bit down: Code: [Select] $query="SELECT `2010 Region Code` AS codes FROM locations"; $results = mysql_query($query); $options=""; $options = "<select location='codes'>"; while($nt=mysql_fetch_assoc($results)) { $thing=$nt["codes"]; $options.="\r\n<option value ='{$nt['codes']}'> {$nt['codes']}</option>"; } $options .="\r\n</select>"; echo $options; what I'm trying to do is grab the selection from the drop down and display it as a table (the sql query would be extended should we manage to figure this one out I've tried Code: [Select] echo"<form name='LOCATIONS' action='".$_SERVER['PHP_SELF']."' target='iframe' method='post'>"; any ideas? I've got a HTML drop down box as similar to this: <select name="dropdown"> <option>Option 1</option> <option>Option 2</option> <option>Option 3</option> </select> If I run a mysql query and get a result of "Option 3", is there anyway using PHP to give Option 3 the selected value? Hello Friends!.... here is the big idea. I am trying to make a form in which there will be 2 drop down lists which will be populating directly from MySQL DB .... Actually i am developing a student management system as my first PHP project... here i want to have 2 drop down lists first is roll number and second is student name. i want that when some one select the roll number in the first drop down the student name against it is automatically populated in the next drop down. please help me friends i am new to php and dont know soo much.! any help PLZZZZZZZZZZ I am trying to update a mysql table called AvItems with the value 'Torso' in the Equip "section?" I have been through the forums and cannot see anything to match. I dont mind if the page looses the onsubmit() and has a button instead. Though I would like to update the database and link back to the same page: There is a display that shows the item that is currently equiped, I have put this in to show it works, or doesn't as the case may be. Hope I got the code /code right this time. many thanks in advance Andy Curtis Code: [Select] create table Items( ItemID integer unsigned auto_increment primary key, ItemName varchar(20) not null, Type varchar(10), UsedOn varchar(10), ); create table AvItems( AvItemID integer unsigned auto_increment primary key, AvID integer unsigned, ItemID integer unsigned, Equip varchar(8)); <?php $username="root"; $password="MyPassword"; $database="MyDataBase"; $AvName = "AndyJCurtis"; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $AvAccR = mysql_query( " SELECT AvID FROM AvAcc WHERE AvName = '$AvName' " ); $AvID = mysql_result($AvAccR, 0, 'AvID'); /////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// $Torso = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' AND UsedON = 'Body' "); $TorsoE = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' And UsedON = 'Body' AND Equip = 'Body' "); if(mysql_num_rows($TorsoE) != 0) { $TorsoItem = mysql_result($TorsoE ,0,"ItemName"); //mysql_close(); ?> <title></title> <head></head> <body> <form action="http://localhost/CI/Equip2.php" method="post"> <table border=1> <tbody> <tr> <td>Torso<BR> <?PHP echo "$TorsoItem <BR>"; ?> <select name="Torso" onchange="submit();" value =" Update"> <?PHP while($TorsoRow = mysql_fetch_array($Torso)) { echo "<option value=\"".$TorsoRow['ItemName']."\">".$TorsoRow['ItemName']."\n </option>"; } ?> </select> </td> </tr> //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// <?php if($_POST['Torso'] == 'Update') { mysql_query("update AvItems set Equip = '' where Equip='Torso'") or die("cant update unequip"); mysql_query("update AvItems set Equip = 'Torso' where ItemID='{$_POST['ItemName']}'") or die("cant update equip"); } ?> /////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// </tbody> </table> </.form> </body> </html> your MySQL server version: 5.1.36 Code: [Select] SELECT * FROM game_weapons Table: Picture Attached the EXPLAIN output for your query, if applicable: I wish to connect a drop down menu's selection of weapons with the correlated table information. What do I want to happen: Click on a drop down Menu and have a list of weapons, these weapons are associated with a set number in the database and passed onto the next screen. (The larger number wins, this part I have figured out). <p> <select name="weapon2" style="font-size:20px;font-family:Arial;width:275px"> <option value="power">Power</option> <option value="intelligence">Intelligence</option> <option value="speed">Speed</option> <option value="reserve">Reserve</option> </select> </p> I do have the battle code figured out! (This should be the last step). Hi all, I am currently learning PHP and have the homework to produce a function that can delete a row in a MySQL database table by clicking on an item in a drop-down menu in a web page. The code I have produced up until now is this:
<!DOCTYPE HTML>
<html lang="de">
<head>
<meta charset="utf-8" />
<title>E3_Artikel_Löschen</title>
</head>
<body>
<form method = "GET">
<?php
$anr="";
try {
$pdo = new PDO ('mysql:dbname=bestelldatenbank;host=localhost;charset=utf8', 'root', '');
} catch (PDOException $error){
die ($error->getMessage());
}
?>
<div>
<p>
<label for="artikel">Artikel: </label>
<select id="artikel" name="artikel">
<?php
$sqlSelect = "SELECT anr, name FROM artikel ORDER BY anr ASC";
foreach ($pdo->query($sqlSelect) as $row) {
echo "<option value=$row[0]>$row[0] | $row[1]</option>\n";
$anr = $row[0];
}
?>
</select>
<input type = "submit" value = "Delete row" />
</p>
</div>
<?php
function artLoeschen($anr) {
echo "Function called $anr";
if(isset($_GET[$anr])) {
$anr = $_GET[$anr];
$sqlDelete = $pdo->query("DELETE FROM artikel WHERE anr = :anr");
if ($stmt = $pdo->prepare($sqlDelete)) {
$stmt->bindParam(':anr', $anr);
$stmt->execute();
}
echo "<h2><b>Artikel gelöscht!</b></h2>";
}
}
?>
</form>
</body>
</html>
So, I have observed the following when I run the script in a browser: 1. The HTML works as expected and I get a drop-down list with the article number and description of each item in the affected table. 2. I can click on an item in the list and it populates the top item in the drop-down list. 3. When I click delete row, the selected item is not deleted. 4. There are no error messages returned but the function is not executed (at least not as I would like to expect).
I have obviously missed something or made a mistake in my code. I would be very grateful for any help...this is driving me mad! :) Regards, Kevin Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike. In each table their are 2 fields, one INDEX_ID and MODELS. Under that for rows I have every model bike named. A quick question before I get onto what I want to do: Can data be added underneath the rows? For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit. To make it easier to understand I'll post the SQL code for the brand Honda: CREATE TABLE `Honda` ( `INDEX_ID` int(3) NOT NULL auto_increment, `MODELS` varchar(20) collate latin1_general_ci NOT NULL, PRIMARY KEY (`INDEX_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ; -- -- Dumping data for table `Honda` -- INSERT INTO `Honda` VALUES(1, 'CR85'); INSERT INTO `Honda` VALUES(2, 'CR125'); INSERT INTO `Honda` VALUES(3, 'CR250'); INSERT INTO `Honda` VALUES(4, 'CRF100'); INSERT INTO `Honda` VALUES(5, 'CRF150'); INSERT INTO `Honda` VALUES(6, 'CRF230'); INSERT INTO `Honda` VALUES(7, 'CRF250X'); INSERT INTO `Honda` VALUES(8, 'CRF250R'); INSERT INTO `Honda` VALUES(9, 'CRF450X'); INSERT INTO `Honda` VALUES(10, 'CRF450R'); INSERT INTO `Honda` VALUES(11, 'CRF50'); INSERT INTO `Honda` VALUES(12, 'CRF70'); INSERT INTO `Honda` VALUES(13, 'CRF80'); INSERT INTO `Honda` VALUES(14, 'XR650'); INSERT INTO `Honda` VALUES(15, 'CR500'); INSERT INTO `Honda` VALUES(16, 'XR100'); INSERT INTO `Honda` VALUES(17, 'XR200'); INSERT INTO `Honda` VALUES(18, 'XR250'); INSERT INTO `Honda` VALUES(19, 'XR400'); INSERT INTO `Honda` VALUES(20, 'XR50'); INSERT INTO `Honda` VALUES(21, 'XR70'); INSERT INTO `Honda` VALUES(22, 'XR80'); Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?: <? $connection = mysql_connect("localhost","user","pass"); $fields = mysql_list_fields("dbname", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; ?> Thanks. I am trying to insert a date into a mySQL table from html drop downs with php. Right now when I enter the date it goes into the database as 0000-00-00. Can anybody see why it might be doing this? My html: Code: [Select] <label for='birthdate' >Birthdate (Optional):</label><br/> <select name='month' id='month' value='<?php echo $fgmembersite->SafeDisplay('month') ?>'> <option value="01" selected="January">January</option> <option value="02">February</option> <option value="03">March</option> etc... </select> <select name='day' id='day' value='<?php echo $fgmembersite->SafeDisplay('day') ?>'> <option value="01" selected="1">1</option> <option value="02">2</option> <option value="03">3</option> etc... </select> <select name='year' id='year' value='<?php echo $fgmembersite->SafeDisplay('year') ?>'> <option value="2010" selected="2010">2010</option> <option value="2009">2009</option> <option value="2008">2008</option> etc... </select> And my php to collect info: Code: [Select] $formvars['birthdate'] = $this->Sanitize($_POST['year'], $_POST['month'], $_POST['day']); php where I make table: Code: [Select] "birthdate DATE NOT NULL ,". php to insert into mysql: Code: [Select] $insert_query = 'insert into '.$this->tablename.'( name, address, birthdate, sex, program, guide, email, username, password, confirmcode ) values ( "' . $this->SanitizeForSQL($formvars['name']) . '", "' . $this->SanitizeForSQL($formvars['address']) . '", "' . $this->SanitizeForSQL($formvars['birthdate']) . '", "' . $this->SanitizeForSQL($formvars['sex']) . '", "' . $this->SanitizeForSQL($formvars['program']) . '", "' . $this->SanitizeForSQL($formvars['guide']) . '", "' . $this->SanitizeForSQL($formvars['email']) . '", "' . $this->SanitizeForSQL($formvars['username']) . '", "' . md5($formvars['password']) . '", "' . $confirmcode . '" )'; Thank you! I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information. How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> 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 I am working on a project that uses a drop down list to chose the category when inserting new data into the database. What I want to do now is make the drop down list default to the chosen category on the list records page and the update page. I have read several tutorials, but they all say that I have to list the options and then select the default. But since it is possible to add and remove categories, this approch won't work. I need the code to chose the correct category on the fly. There are two tables, one that has the category ID and category name. The second table has the data and the catid which is referenced to the category id in the first table. Code: [Select] -- -- Table structure for table `categories` -- DROP TABLE IF EXISTS `categories`; CREATE TABLE IF NOT EXISTS `categories` ( `id` int(11) NOT NULL AUTO_INCREMENT, `categories` varchar(37) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ; -- -------------------------------------------------------- -- -- Table structure for table `links` -- DROP TABLE IF EXISTS `links`; CREATE TABLE IF NOT EXISTS `links` ( `id` int(4) NOT NULL AUTO_INCREMENT, `catid` int(11) DEFAULT NULL, `name` varchar(255) NOT NULL DEFAULT '', `url` varchar(255) NOT NULL DEFAULT '', `content` varchar(255) NOT NULL DEFAULT '', PRIMARY KEY (`id`), KEY `catid` (`catid`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ; Then is the list records file, I have Code: [Select] <?php include ("db.php"); include ("menu.php"); $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; $id = $_GET['id']; $query="SELECT * FROM links ORDER BY catid ASC"; $result=mysql_query($query); ?> <table width="65%" align="center" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="100%" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="7"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Name</strong></td> <td align="center"><strong>URL</strong></td> <td align="center"><strong>Content</strong></td> <td align="center"><strong>Update</strong></td> <td align="center"><strong>Delete</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td> <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> </td> <td><? echo $rows['catid']; ?></td> <td><? echo $rows['name']; ?></td> <td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td> <td><? echo $rows['content']; ?></td> <td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td> <td align="center"><a href="delete.php?id=<? echo $rows['id']; ?>">delete</a></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> So, how do I get this code Code: [Select] $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> to give me an output that will be something like if catid exactly matches categories.id echo categories.categorie ??? so far everything I have done produces either a default category of the last category, the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing. How do I get just the category name? I will keep reading and try to figure this out, but any help would be greatly appreciated. Thanks in advance I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue... Here is my select box code which is working perfect: <select name="city"> <?php $sql = "SELECT id, city_name FROM cities ". "ORDER BY city_name"; $results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!"); while($row = mysqli_fetch_array($results_set)) { echo "<option value=$row[id]>$row[city_name]</option>"; } ?> </select> Are any variables defined in a while loop global to the while loop only? Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form. Any ideas for a workaround to get this value passing as normal? if (isset($_POST['addPosting'])) { $query = "INSERT INTO Postings (id, city_id, title, description) VALUES ('','$row[id]','$_POST[title]','$_POST[description]')"; Hi I have tried the mysql forum but have had no joy with an answer to my problem so wondered if php would be better. I want my users to be able to select from 5 different drop down lists where they can chose any combination from 1 up to all 5, I have attached the front end. These lists are being populated from mysql tables. Code for the drop down lists is as follows Code: [Select] <form action="horse-events-devon.php?url_countyid=<?php echo urlencode ($url_countyid ['url_countyid']) ; ?>&go" method="POST"> <table id="searchtable"> <tr> <th>Find By Discipline</th> <th>Find By Venue</th> <th>Find By Championship</th> <th>Find By Organiser</th> <th>Equine Association</th> <th>Submit Your Selections</th> </tr> <tr> <td><select name="dis_id"> <?php $upcomingdis = upcomingdis($url_countyid); $upcoming_dis_bycounty = mysql_fetch_assoc ($upcomingdis); ?> <?php do { ?> <option value="<?php echo $upcoming_dis_bycounty ['dis_id']; ?>" > <?php echo $upcoming_dis_bycounty ['dis_description']; ?></option> <?php } while ($upcoming_dis_bycounty = mysql_fetch_assoc ($upcomingdis)); ?></select></td> <td><select name="ven_id"> <?php $upvenbycounty_set = upcoming_venevents_bycounty($url_countyid); $upcoming_ven_bycounty = mysql_fetch_assoc ($upvenbycounty_set); ?> <?php do { ?> <option value="<?php echo $upcoming_ven_bycounty ['ven_id']; ?>" > <?php echo $upcoming_ven_bycounty ['ven_name']; ?></option> <?php } while ($upcoming_ven_bycounty = mysql_fetch_assoc ($upvenbycounty_set)); ?></select></td> <td><select name="champ_id"> <?php $championship_set = findchampionships(); $champlist = mysql_fetch_assoc ($championship_set); ?> <?php do { ?> <option value="<?php echo $champlist ['champ_id']; ?>" > <?php echo $champlist ['champ_description']; ?></option> <?php } while ($champlist = mysql_fetch_assoc ($championship_set)); ?></select></td> <td> <select name="org_id"> <?php $uporgbycounty_set = upcoming_organevents_bycounty($url_countyid); $upcoming_org_bycounty = mysql_fetch_assoc ($uporgbycounty_set); ?> <?php do { ?> <option value="<?php echo $upcoming_org_bycounty ['org_id']; ?>" ><?php echo $upcoming_org_bycounty ['org_name']; ?></option> <?php } while ($upcoming_org_bycounty = mysql_fetch_assoc ($uporgbycounty_set)); ?></select> </td> <td><select name="ass_id"> <?php $upassbycounty_set = upcoming_assevents_bycounty($url_countyid); $upcoming_assbycounty = mysql_fetch_assoc ($upassbycounty_set); ?> <?php do { ?> <option value="<?php echo $upcoming_assbycounty ['ass_id']; ?>" > <?php echo $upcoming_assbycounty ['ass_description']; ?></option> <?php } while ($upcoming_assbycounty = mysql_fetch_assoc ($upassbycounty_set)); ?></select></td> <td><input name="submit" type="submit" /><input name="countyid" type="hidden" value="<?php echo $url_countyid ['url_countyid']; ?>" /></td> </tr> </table> </form> My search processing is as follows Code: [Select] <?php if (isset($_POST['submit'])){ if (isset($_GET['go'])){ $countyid = $_POST['countyid']; $ven_id = $_POST['ven_id']; $dis_id = $_POST['dis_id']; $champ_id = $_POST['champ_id']; $org_id = $_POST['org_id']; $event_id = $row['event_id']; $sql = "SELECT DATE_FORMAT (events.startdate, '%a, %d, %b') AS stdate, events.event_id, events.title, events.ven_id, events.org_id, venue.county_id, venue.ven_id, eventdisciplines.event_id, eventdisciplines.dis_id, county.county_id, discipline.dis_id \n" . "FROM events \n" . "LEFT OUTER JOIN eventdisciplines \n" . "ON events.event_id = eventdisciplines.event_id \n" . "LEFT OUTER JOIN discipline \n" . "ON eventdisciplines.dis_id = discipline.dis_id \n" . "LEFT OUTER JOIN venue \n" . "ON events.ven_id = venue.ven_id \n" . "LEFT OUTER JOIN county \n" . "ON venue.county_id = county.county_id \n" . "WHERE events.ven_id = ({$ven_id} OR events.org_id = {$org_id})\n" . "AND events.startdate > NOW()\n" . "AND venue.county_id = {$countyid} \n" . "ORDER BY startdate ASC"; $result = mysql_query ($sql, $connection); ?> How am I best to do this please? my OR within the mysql does not work, should I not be doing this with php in the search processing? someones help would really be appreciated, just to point me in the right direction. [attachment deleted by admin] I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? Hi All, I have 2 tables: one CarMake - CarMakeID - CarMakeDesc two CarModel - CarModelID - CarModelMake - CarModelDesc Depending on what the user selects in the first dropdown (carmake) the possible selection in the second dropdown (model) needs to be limited to only the models from the selected carmake. in the second table (Carmodel : the 'CarModelMake' = CarMakeID, to identify the make) How do I limit the dropdown 'CarModel' based on the selected CarMake in the first dropdown. link : http://98.131.37.90/postCar.php code : -- -- -- Code: [Select] <label> <select name="carmake" id="CarMake" class="validate[required]" style="width: 200px;"> <option value="">Select CAR MAKE...</option> <?php while($obj_queryCarMake = mysql_fetch_object($result_queryCarMake)) { ?> <option value="<?php echo $obj_queryCarMake->CarMakeID;?>" <?php if($obj_queryCarMake->CarMakeID == $CarAdCarMake) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarMake->CarMakeDesc;?></option> <?php } ?> </select> </label> <label> <select name="carmodel" id="CarModel" class="validate[required]" style="width: 200px;"> <option value="">Select MODEL...</option> <?php while($obj_queryCarModel = mysql_fetch_object($result_queryCarModel)) { ?> <option value="<?php echo $obj_queryCarModel->CarModelID;?>" <?php if($obj_queryCarModel->CarmodelID == $CarAdCarModel) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarModel->CarModelDesc;?></option> <?php } ?> </select> </label> I'm a novice.. and appreciates all the help ! I am creating a form that will allow the user to select the make of vehicle "FORD" for example. If that make of vehicle is selected among different makes of vehicles, then another box will appear, with all the models for that particular model "Fiesta" for example. What type of code accomplishes this setup in my web page? I do not want to list 500 models in one drop down list, but just those for each make in the first drop down list. Thanks much! |
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