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PHP - Php Dropdown List Populates Form Field
So I've been attempting to create a dropdown list that will post to mySQL.
The problem I'm running into is the way I populate the dropdown I can't figure out how to return the selected option to mySQL. <form name="pentry" method="post" action="" action="pentry.php"> <input name="username" type="hidden" value="<?php echo "$username"?>"> Date of Practice:<input type="text" id="dates" name="practdate" datepicker="true" datepicker_format="YYYY-MM-DD" maxlength="100"><br> <?php $query="SELECT eoptions,id FROM sm_options"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=eoptions value=''>Event Options</option>"; // printing the list box select command while($nt=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option value=$nt[id]>$nt[eoptions]</option>"; /* Option values are added by looping through the array */ } echo "</select>";// Closing of list box mysql_close(); ?><br> Event:<input type="text" name="event" maxlength="100" value="<?php echo "test $output" ?>"><br> Session Time:<input type="text" name="practtime" id="practtime">min. <input type="button" value=" + " onClick="addmin(practtime);"> <input type="button" value=" - " onClick="submin(practtime);"><br> Practice Content:<input type="text" name="practnotes"><br> <input type="submit" name="submit" value="Submit"><br> </form> I've been moving code around trying to solve the problem so its a bit messy. Similar TutorialsYesterday, I created a topic about how I could update records and I managed to achieve that successfully. Now I have another dilemma. When I have a specific record I want to update, I want to change a category ID of an product (e.g. change it from 1 to 2) but how do I go about doing this? Here is my code thus far: Code: [Select] <?php require_once ('./includes/config.inc.php'); require_once (MYSQL); $id=$_GET['prodID']; $results = mysqli_query($dbc, "SELECT * FROM product WHERE prodID=".$_GET['prodID'].""); $row = mysqli_fetch_assoc($results); ?> <form action="" method='POST'> Product ID: <input type="text" value="<?php echo $row['prodID'];?>" name="prodID" /> <br /> Product: <input type="text" value="<?php echo $row['product'] ;?>" name="product" /> <br /> Product Description: <input type="text" value="<?php echo $row['prod_descr'] ;?>" name="prod_descr" /> <br /> Category: <select name="category"> <option value="<?php echo $row['catID'];?>"></option> </select> Price: <input type="text" value="<?php echo $row['price'] ;?>" name="price" /> <br /> In Stock: <input type="text" value="<?php echo $row['stock'] ;?>" name="stock" /> <br /> <br /><input type="submit" value="save" name="save"> </form> <?php if(isset($_POST['save'])) { $id = $_POST['prodID']; $product = $_POST['product']; $descr = $_POST['prod_descr']; $price = $_POST['price']; $stock = $_POST['stock']; // Update data $update = mysqli_query($dbc, "UPDATE product SET product='$product', prod_descr='$descr', price='$price', stock='$stock' WHERE prodID=".$_GET['prodID'].""); header( 'Location: update_save.php' ) ; } ?> hi all,i have a php form that i would like to populate from a drop down list,my problem is that i cant place the drop down list at the correct place.what i want is that where the franchisee is the input should be the drop down,please help....this is the code that i am using Code: [Select] <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post"> <p>Add a Bus in the Bus table.</p> <p> Fleet Number:<br /> <input type="text" name="fleet_number" size="6" maxlength="10" value="" /> </p> <p> Registration Number:<br /> <input type="text" name="registration_number" size="10" maxlength="20" value="" /> </p> <p> Model:<br /> <input type="text" name="model" size="10" max length="15" value="" /> </p> <p> Franchisee :<br /> <input type="select" name="franchisee_id" size="10" max length="15" value="" /> <select name=dropdown_list> <?php while($row = oci_fetch_array($stid)) { echo "<option value='".$row['FRANCHISEE_NAME']."'>"; echo $row['FRANCHISEE_NAME']; echo "</option> "; } ?> </select> </p> <p> <input type="Submit" name="submit" value="Submit !" /> </p> </form> the line : echo $form['catcher_id'] gives me a dropdown list when i choose another item from the dropdown i want to do a few things but my code not working: $selected_catcher = $form['catcher_id']; foreach($selected_catcher as $val) { $catcher_name = $val->getName(); echo $catcher_name." ".$val->getId(); if ($catcher_name = "zed-catcher") { echo $form['service_code']->renderLabel(); echo $form['service_code']->renderError(); echo $form['service_code']; } } please help? thanks Can someone help here. I can't get the drop down to populate before the post. It does populate after post and the selected option doesn't post after it get populated.
Here is the code I have
Once I get everything working I will be moving to mysqli or pod
<?PHP
require_once('../../lib/connections/db.php');
include('../../lib/functions/functions.php');
checkLogin('2');
$getuser = getUserRecords($_SESSION['user_id']);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Water Analysis Data(WAD)</title>
<style type="text/css">
</style>
</head>
<body>
<div align="right"><a href="../wad/index.php">Home</a> | <? if (!empty($getuser[0]['thumb_path'])){echo "<a href='.//manage_photo.php'>Manage My Photo</a> | ";} else {echo "<a href='../upload_photo.php'>Upload Photo</a> | ";} ?><a href="../change_pass.php">Change password</a> | <a href="../edit_profile.php">Edit Profile</a> | <a href="../log_off.php?action=logoff">Sign Out</a></div></td>
<p>Welcome <?php if(empty($getuser[0]['first_name']) || empty($getuser[0]['last_name'])){echo $getuser[0]['username'];} else {echo $getuser[0]['first_name']." ".$getuser[0]['last_name'];} ?></p>
<br><br><br>
<?php
// this is processed when the form is submitted
// back on to this page (POST METHOD)
if ($_SERVER['REQUEST_METHOD'] == "POST")
{ $usernow = $getuser[0]['username'];
$userid = $usernow;
# escape data and set variables
$tank = addslashes($_POST["tank"]);
$date = addslashes($_POST["date"]);
$temperature = addslashes($_POST["temperature"]);
$ph = addslashes($_POST["ph"]);
$ammonia = addslashes($_POST["ammonia"]);
$nitrite = addslashes($_POST["nitrite"]);
$nitrate = addslashes($_POST["nitrate"]);
$phosphate = addslashes($_POST["phosphate"]);
$gh = addslashes($_POST["gh"]);
$kh = addslashes($_POST["kh"]);
$iron = addslashes($_POST["iron"]);
$potassium = addslashes($_POST["potassium"]);
$notes = addslashes($_POST["notes"]);
// build query
// # setup SQL statement
$sql = " INSERT INTO water_parameters ";
$sql .= " (id, userid, tank, date, temperature, ph, ammonia, nitrite, nitrate, phosphate, gh, kh, iron, potassium, notes) VALUES ";
$sql .= " ('', '$userid', '$tank', '$date', '$temperature', '$ph', '$ammonia', '$nitrite', '$nitrate', '$phosphate', '$gh', '$kh', '$iron', '$potassium', '$notes') ";
// #execute SQL statement
$result = mysql_query($sql);
// # check for error
if (mysql_error()) { print "Database ERROR: " . mysql_error(); }
print "<h3><font color=red>New Water Parameters Were Added</font></h3>";
}
?>
<table width="810" border="2" align="center">
<tr>
<td>
<table width="800" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td align="center" bgcolor="#FFFFFF" scope="col"> <h2><b>Water Analysis Data(WAD)</b></h2></td>
</tr>
<tr>
<td bgcolor="#FFFFFF"> <form name="water_parameters" action="water_parameters.php" method="POST">
<table border="2" align="center" cellpadding="0">
<tr><td><div align="left"><b>Tank Name: </b> </div></td><td><div align="left">
<?php
echo "<select>";
$results = mysql_query("SELECT tank FROM tank WHERE userid = '$userid'");
while($row = mysql_fetch_array($results))
{
echo "<option value=". $row["tank"] .">". $row["tank"] . "</option>";
}
echo "</select>";
?>
</div></td></tr>
<tr><td><div align="left"><b>Test Date: </b> </div></td><td><div align="left">
<input type="text" name="date" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Temperatu </b> </div></td><td><div align="left">
<input type="text" name="temperature" size=25>
</div></td></tr>
<tr><td><div align="left"><b>pH: </b> </div></td><td><div align="left">
<input type="text" name="ph" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Ammonia: </b> </div></td><td><div align="left">
<input type="text" name="ammonia" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Nitrite: </b> </div></td><td><div align="left">
<input type="text" name="nitrite" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Nitrate: </b> </div></td><td><div align="left">
<input type="text" name="nitrate" size=25>
</div></td></tr>
<tr><td><div align="left"><b>phosphate: </b> </div></td><td><div align="left">
<input type="text" name="phosphate" size=25>
</div></td></tr>
<tr><td><div align="left"><b>GH: </b> </div></td><td><div align="left">
<input type="text" name="gh" size=25>
</div></td></tr>
<tr><td><div align="left"><b>KH: </b> </div></td><td><div align="left">
<input type="text" name="kh" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Iron: </b> </div></td><td><div align="left">
<input type="text" name="iron" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Potassium: </b> </div></td><td><div align="left">
<input type="text" name="potassium" size=25>
</div></td></tr>
<tr><td><div align="left"><b>Notes: </b> </div></td><td><div align="left">
<p><textarea name="notes" cols="50" rows="10"></textarea></p>
</div></td></tr>
<tr><th colspan=2><p><input type="submit" value="Add New Test"></p></th></tr>
</table>
</form> </td>
</tr>
<tr>
<td align="center" valign="top" bgcolor="#FFFFFF"><div align="center"><font size=2>
© 2014 </font> </div></td>
</tr>
</table>
</td>
</tr>
</table>
</body>
</html>
Hi, I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code: mysql: Code: [Select] create table years ( yearID integer auto_increment, year varchar(30), primary key (yearID) ); insert into years (yearID, year) values ('1', '2007-2008'); insert into years (yearID, year) values ('2', '2008-2009'); insert into years (yearID, year) values ('3', '2009-2010'); insert into years (yearID, year) values ('4', '2010-2011'); insert into years (yearID, year) values ('5', '2011-2012'); insert into years (yearID, year) values ('6', '2012-2013'); PHP: Code: [Select] <?php require_once('../Connections/connection.php'); ?> <?php $result = @mysql_query( "select yearID, year, from sss.years"); print "<p>Select a year:\n"; print "<select name=\"yearID\">\n"; while ($row = mysql_fetch_assoc($result)){ $yearID = $row[ 'yearID' ]; $year = $row[ 'year' ]; print "<option value=$yearID>$year\n"; } print "</select>\n"; print "</p>\n"; ?> Thank you! i have been trying to get this code to get a list of usernames from a database and i have now got that to work but when i try and save it it saves all the usernames from the drop down list and not just the one i have selected how can i get it to just use the one i have selected Code: [Select] <?php include "connect.php"; //connection string include("include/session.php"); print "<link rel='stylesheet' href='style.css' type='text/css'>"; print "<table class='maintables'>"; print "<tr class='headline'><td>Post a message</td></tr>"; print "<tr class='maintables'><td>"; // Write out our query. $query = "SELECT username FROM users"; // Execute it, or return the error message if there's a problem. $result = mysql_query($query) or die(mysql_error()); $dropdown = "<select name='username'>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>"; } $dropdown .= "\r\n</select>"; if(isset($_POST['submit'])) { $name=$session->username; $yourpost=$_POST['yourpost']; $subject=$_POST['subject']; $to=$dropdown; if(strlen($name)<1) { print "You did not type in a name."; //no name entered } else if(strlen($yourpost)<1) { print "You did not type in a post."; //no post entered } else if(strlen($subject)<1) { print "You did not enter a subject."; //no subject entered } else { $thedate=date("U"); //get unix timestamp $displaytime=date("F j, Y, g:i a"); //we now strip HTML injections $subject=strip_tags($subject); $name=strip_tags($name); $yourpost=strip_tags($yourpost); $to=strip_tags($to); $insertpost="INSERT INTO forumtutorial_posts(author,title,post,showtime,realtime,lastposter,name) values('$name','$subject','$yourpost','$displaytime','$thedate','$name','$to')"; mysql_query($insertpost) or die("Could not insert post"); //insert post print "Message posted, go back to <A href='forum.php'>Forum</a>."; } } else { print "<form action='newtopic.php' method='post'>"; print "Your name:<br>"; print "$session->username<br>"; print "User to send to:<br>"; print "$dropdown"; print "Subject:<br>"; print "<input type='text' name='subject' size='20'><br>"; print "Your message:<br>"; print "<textarea name='yourpost' rows='5' cols='40'></textarea><br>"; print "<input type='submit' name='submit' value='submit'></form>"; } print "</td></tr></table>"; ?> MOD EDIT: Changed PHP manual link [m] . . . [/m] tags to [code] . . . [/code] tags. I have the following code currently: Code: [Select] <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php print $item['view'] ?> <?php } ?> I would like to make the list a drop down with a link so that when a user selects, he goes to a new page. I tried the following: Code: [Select] <select name="select"> <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php $url = $node->field_buy_at[0]['url']; $store = $item['view']; ?> <? echo "<option value='$url'>$store</option>";?> <?php } ?> </select> I'm pretty sure it's this "$url = $node->field_buy_at[0]['url'];" that I don't have correct. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=316599.0 Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. Hey, I have the following coding: Quote <? $dbuser="*******"; $dbpass="*******"; $dbname="virtuda_db"; //the name of the database $chandle = mysql_connect("localhost", $dbuser, $dbpass) or die("Connection Failure to Database"); mysql_select_db($dbname, $chandle) or die ($dbname . " Database not found. " . $dbuser); $mainsection="license"; $query1="select name from license"; $result = mysql_db_query($dbname, $query1) or die("Failed Query of " . $query1); //do the query while($thisrow=mysql_fetch_row($result)) { $i=0; while ($i < mysql_num_fields($result)) { $field_name=mysql_fetch_field($result, $i); echo $thisrow[$i] . " "; //Display all the fields on one line $i++; } echo "<br>"; //put a break after each database entry } ?> How would I set up this so that instead of just "listing" them out on new lines, it would list the results into a drop down list? Thanks! I'm trying to sort this dropdown box. It reads from a directory, and lists the file name in the dropdown box. Here's the tricky part... the filename is listed differently in the dropdown than in the directory by using explode(). I want to sort it though since it's still being sorted by the directory listings... For example: Filename starts out as: 123_abc_567.pdf then gets listed as abc_123_567.pdf in the dropdown, but it's still getting sorted as if it were 123_abc_567.pdf How can I do that? Here's my code: // Define the full path to folder from root $path = "C:/Work_Orders/"; // Open the folder $dir_handle = @opendir($path) or die("Unable to open $path"); echo "<form method=\"POST\" action='".$_SERVER['PHP_SELF']."' name='selectworkorder'><select name='ordernumber2'>"; // Loop through the files while ($file = readdir($dir_handle)) { //Remove file extension $ext = strrchr($file, '.'); if($ext !== false) { $file = substr($file, 0, -strlen($ext)); } if($file == "." || $file == ".." || $file == "index.php" ) continue; //explode file name $changedordernumber = explode("_",$file); //put in new order $changedordernumber = $changedordernumber[1]."_".$changedordernumber[0]."_".$changedordernumber[2]; $changedordernumber=trim($changedordernumber,"_"); //list options echo "<option name='$file' value='$file'>$changedordernumber</option>\n"; } echo "</select><input type='submit' value='Change' name='submit'/></form></div>"; // Close closedir($dir_handle); Hey Guys, I know it may seem pretty simple, but im having trouble populating a drop down list. Here is my code at the moment, but what it's doing is displaying the names all in one value, where it should be in separate select values. *Note that i have only done it to the first one. See attachment. 'AntonMatt' are next to each other, they should be separate select values. Code: [Select] <? $id = $_GET['id']; $selectplayers="SELECT * FROM players WHERE club='$club' AND team='$team'"; $player=mysql_query($selectplayers); ?> <table class='lineups' width="560" cellpadding="5"> <tr> <td colspan="2">Starting Lineup</td> <td colspan="2">On the Bench</td> </tr> <tr> <td width="119"> </td> <td width="160"> </td> <td width="69"> </td> <td width="160"> </td> </tr> <tr> <td>Prop</td> <td><select name="secondary" style="width: 150px"> <option value='' selected="selected"><? while($rowplayer = mysql_fetch_array($player)) { echo $rowplayer['fname']; } ?></option> </select></td> <td>16.</td> <td><select name="secondary16" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> <tr> <td style="padding-top: 8px;">Hooker</td> <td style="padding-top: 8px;"><select name="secondary2" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> <td style="padding-top: 8px;">17.</td> <td style="padding-top: 8px;"><select name="secondary17" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> </table> </div> </div> If I can get this fixed, I will have completed all but the admin login for this project - my first php/mysql project. Here is what I need. I have a list_records.php that list all the records in the table 'links' and the category each entry is in from the table 'categories'. Here are my table structures. 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 ; On the update.php file, I have a form that lets me make changes to the record. Here is the codes for update.php Code: [Select] <? include "menu.php" ?> <? include "db.php" ?> <?php $id=$_GET['id']; $sql = "select * from links where id =$id"; $query = mysql_query($sql); while ($row = mysql_fetch_array($query)){ $id = $row['id']; $catid = $row['catid']; $name = $row['name']; $url = $row['url']; $content = $row['content']; //we will echo these into the proper fields } mysql_free_result($query); ?> <table width="65%" align="center"> <tr><td align="left"> <form action="updated.php" method="post"> <input type="hidden" value="<?php echo $id; ?>" name="id"/> <br> <b>Website Name:</B><br> Change the name of the website listing.<br> <input type="text" value="<?php echo $name; ?>" name="name"/> <br> <br> <b>URL:</b><br> Change the URL of the website listing.<br> <input type="text" value="<?php echo $url; ?>" name="url"/> <br> <br> <b>Description:</b><br> Change the description of the website listing.<br> Limit 255 characters.<br/> <textarea name="content" cols="45" rows="4" wrap="soft"><?php echo($content);?></textarea> <br> <?php $result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $id = $rows["id"]; $categories=$row["categories"]; $options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>'; }; ?> <SELECT NAME=catid> <OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION> </SELECT> <?php mysql_close(); ?> <div align="center"> <input type="submit" value="submit changes"/> </div> </form> <br> </td></tr></table> The part of he code I need help with is Code: [Select] <?php $result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $id = $rows["id"]; $categories=$row["categories"]; $options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>'; }; ?> <SELECT NAME=catid> <OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION> </SELECT> I want it to default to the category that the entry is in. If you look, you will see in the select portion that I I have Code: [Select] <$ echo $catid; ?> which echos the proper category ID, but if I use Code: [Select] <? echo $categories; ?> it echos Writing, which is the last category in the list. Yet, the $options echo the catid and it corresponding category. How can I get the default option to echo BOTH the catid and category name while also listing all the other categories so that the records can be moved to a new category is needed? Any help will be appreciated. Thank you in advance. Hi. I am using this script to populate a dropdown list box from sql, it works but does anyone know how to sort the list in alphabetical order? $sql="SELECT * FROM Fish WHERE ***** = '".$_GET['stocktype']."'"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $ID=$row["ID"]; $Stock=$row["Commonn"]; $Options.="<OPTION VALUE=\"$ID\">".$Stock; } <SELECT NAME='stock1'> <OPTION VALUE='$Options'>$Options</option> </SELECT> Alright, so I have an xml file differences.xml that is being parsed in XML. This is what the xml looks like: <item code="lM" name="dog"> <cost>5000</cost> <Start>12/15/2010</Start> <End>01/13/2011</End> </item> <item code="lF" name="cat"> <cost>5000</cost> <Start>04/15/2010</Start> <End>04/23/2011</End> </item>[/ I want to have the item names (dog, cat) show in a dropdown menu so that I can select these items for editing before storing in my mysql database. This is the php code I have so far: <?PHP $xml = simplexml_load_file("differences.xml"); $object = $xml->xpath("//item"); echo '<SELECT name=object>'; foreach ($object['name'] as $key => $value) { echo '<OPTION value='.$value.'> '.$value; } echo '</select>'; ?> I do have a dropdown list but there are no values inside it (it is empty). Can anyone help me figure out why? I do have this code that does work which lists the items in plaintext (not in a dropdown) so hopefull this will help us out: <?PHP $xml = simplexml_load_file("differences.xml"); $object = $xml->xpath("//item"); $count = count($object); $i = 0; while($i < $count) { echo '<h1>'.$object[$i]['name'].'</h1>'; $i++; } ?> i skipped solving this the other day with an easier way but now i am stuck with this stumble again in another area and i have no way out.... so here it goes, i have an ajax dropdown box...i need to get the value that is selected by the user when it is clicked and then pass this value to a new pop window by appending to its url....any suggestions? how come the lines: PHP Code: echo $form['catcher_id']; $catcher_names = $form['catcher_id'] foreach($catcher_names as $val) { echo "testing loop"; } produce nothing?? its not doing the foreach. $form['catcher_id']; is a dropdown list containing catcher names please help? thanks How to create Dependent Dropdown List?
<div class="form-group">
<label class="control-label col-sm-2" for="i_have">I have :</label>
<div class="col-sm-10">
<select id="old" class="form-control" placeholder="Select I have" name="i_have">
<option value = "select_option">Select Option</option>
<option value = "three_compact">3 Compact</option>
<option value = "three_regular">3 Regular</option>
<option value = "three_triple">3 Triple</option>
<option value = "five_compact">5 Compact</option>
<option value = "five_regular">5 Regular</option>
<option value = "five_triple">5 Triple</option>
<option value = "seven_compact">7 Compact</option>
<option value = "seven_regular">7 Regular</option>
<option value = "seven_triple">7 Triple</option>
<option value = "nine_compact">9 Compact</option>
<option value = "nine_regular">9 Regular</option>
<option value = "nine_triple">9 Triple</option>
</select>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="i_want">I want :</label>
<div class="col-sm-10">
<select id="new" class="form-control" placeholder="Select I want" name="i_want">
<option value = "select_option">Select Option</option>
<option value = "three_compact">3 Compact</option>
<option value = "three_regular">3 Regular</option>
<option value = "three_triple">3 Triple</option>
<option value = "five_compact">5 Compact</option>
<option value = "five_regular">5 Regular</option>
<option value = "five_triple">5 Triple</option>
<option value = "seven_compact">7 Compact</option>
<option value = "seven_regular">7 Regular</option>
<option value = "seven_triple">7 Triple</option>
<option value = "nine_compact">9 Compact</option>
<option value = "nine_regular">9 Regular</option>
<option value = "nine_triple">9 Triple</option>
</select>
</div>
</div>
If the first dropdown "3 Compact " is selected second dropdown should not be able to select "3 Compact" & if first dropdown "5 Regular" is selected second dropdown should not be able to select below values of 5 Regular like "3 Compact, 3 Regular, 3 Triple, 5 Compact. The logic is, if first dropdown value selects like 5, second dropdown value must select above 5 not 4 or 3 or 2 or 1. |
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