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PHP - Dynamic Select Menus
I want to create 2 select menus - one static (Menu 1) and the other (Menu 2) populated according to the selection made in Menu 1.
Data will will be pulled from a CSV file. Menu 2 will not have any values in it until an option from Menu 1 has been selected. Then, the options available will be limited based on the Item 1 selection. PHP must be used to read the CSV file and to capture the data (using arrays) I met this problem before and my approach then was to call a function using the onChange event for Menu1 ... <script type = "text/javascript">function update_select(obj){ window.location.href="./call_stats_main.php?choice=" + obj.value;} </script> onchange="update_select(this) This would reload the page passing a parameter to the URL (?choice=) I could then use $_GET['choice'] to capture the selected value and dynamically populate Menu 2 according to this value. It is the onchange event that triggers the population of menu 2. I would like to know IF THERE IS A BETTER WAY, other than using ajax ? Steven M Similar TutorialsSo I have the following code, which is supposed to get the value selected by the user, then on submitting, will use that value in the URL. Code: [Select] $durations = array("1", "2", "3", "4", "5", "7", "10", "10000000"); echo "Choose Duration: ";?><form method="post" action=""><select name="duration_chosen"> <? foreach($durations as $duration){ ?><option name="duration_len" value="<?php echo $duration;?>"> <? if ($duration == "10000000") { echo "Forever"; } else { echo "$duration day(s)"; } ?></option><? } ?> </select></form><? $hrs = $_GET['duration_len'] * 24; $duration_chosen = $_POST['duration_chosen']; ?> <form method="get" action="index.php"> <input type="hidden" name="duration" value="<? echo $duration_chosen; ?>"> <input type="submit" value="Accept"></form> For some reason, though, it doesn't work. Can anyone help me? Thanks, Mark This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=347324.0 I'm having trouble getting the dynamic data from my <select> menus to write into my MYSQL database. Can anyone see what I'm doing wrong here? First post btw The output looks like this, which is obviously wrong: Code: [Select] <html> <head> </head> <link rel="stylesheet" type="text/css" href="./css/newuser.css" /> <body> <?php session_start(); require 'default.inc.php'; ?> <?php if (isset($_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, sorry for being a newbie, but I need help setting up a dynamic form. The form will be a reservation type form. I have a single table, which I will post below. What I am looking to do is choose a month from a select box and then have it populate another select box with the available dates. Here is my table info from mysql... Code: [Select] CREATE TABLE `daterange` ( `RID` int(5) NOT NULL auto_increment, `DEND` date NOT NULL, `MONTH` varchar(50) NOT NULL, `DATE` varchar(50) NOT NULL, `SITE` varchar(50) NOT NULL, `PRICE` varchar(10) NOT NULL, `STATUS` varchar(1) NOT NULL default 'A', `FNAME` varchar(50) NOT NULL, `LNAME` varchar(50) NOT NULL, `ADDR1` varchar(50) NOT NULL, `ADDR2` varchar(50) NOT NULL, `CITY` varchar(50) NOT NULL, `STATE` varchar(2) NOT NULL, `ZIP` varchar(5) NOT NULL, `PHONE1` varchar(3) NOT NULL, `PHONE2` varchar(3) NOT NULL, `PHONE3` varchar(4) NOT NULL, `EMAIL` varchar(50) NOT NULL, PRIMARY KEY (`RID`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=83 ; Here is a sample mysql dump... Code: [Select] INSERT INTO `daterange` VALUES(1, '2011-05-15', 'May 2011', '5/15-5/22/11', 'Carterville Pond/Adirondack', '$2625.00', 'N', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', ''); INSERT INTO `daterange` VALUES(2, '2011-05-22', 'May 2011', '5/22-5/29/11', 'Carterville Pond/Adirondack', '$2625.00', 'A', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', ''); INSERT INTO `daterange` VALUES(3, '2011-05-29', 'May 2011', '5/29-6/5/11', 'Carterville Pond/Adirondack', '$2625.00', 'A', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', ''); Finally, here is a php form that I started to develop but it only has a single select box, which works but I wanted another select box that will filter the list down a little... Code: [Select] <? // script to display all the Available Reservations in the daterange table // connection information $hostName = "my_host"; $userName = "my_user"; $password = "my_pass"; $dbName = "my_db"; // make connection to database mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName"); mysql_select_db($dbName) or die("Unable to select database $dbName"); // Select all the fields in all the records of the daterange table $query = "SELECT * FROM daterange WHERE DEND > DATE(NOW()) AND STATUS='A' ORDER BY RID, DATE, SITE"; $result = mysql_query($query); // Determine the number of reservation dates $number = mysql_numrows($result); // Create drop-down menu of reservation dates print "<font size=\"3\" face=\"Arial\"><b>Select Reservation Date:</b> <form action=\"test.php\" method=\"post\"> <select name=\"RID\"> <option value=\"\">Choose One</option>"; for ($i=0; $i<$number; $i++) { $RID = mysql_result($result,$i,"RID"); $DATE = mysql_result($result,$i,"DATE"); $SITE = mysql_result($result,$i, "SITE"); $PRICE = mysql_result($result,$i, "PRICE"); print "<option value=\"$RID\">$DATE, $SITE, $PRICE</option>"; } print "</select><p align=left><label><font size=\"3\" face=\"Arial\">First Name: <input type=\"text\" name=\"FNAME\" size=\"50\" maxlength=\"50\" tabindex=\"1\"<br>"; print "<p align=left><label>Last Name: <input type=\"text\" name=\"LNAME\" size=\"50\" maxlength=\"50\" tabindex=\"2\"<br>"; print "<p align=left><label>Address Line 1: <input type=\"text\" name=\"ADDR1\" size=\"50\" maxlength=\"50\" tabindex=\"3\"<br>"; print "<p align=left><label>Address Line 2: <input type=\"text\" name=\"ADDR2\" size=\"50\" maxlength=\"50\" tabindex=\"4\"<br>"; print "<p align=left><label>City: <input type=\"text\" name=\"CITY\" size=\"50\" maxlength=\"50\" tabindex=\"5\"<br>"; print "<p align=left><label>State (abbrev.): <input type=\"text\" name=\"STATE\" size=\"2\" maxlength=\"2\" tabindex=\"6\"<br>"; print "<p align=left><label>Zip Code: <input type=\"text\" name=\"ZIP\" size=\"5\" maxlength=\"5\" tabindex=\"7\"<br>"; print "<p align=left><label>Contact Phone Number: (<input type=\"text\" name=\"PHONE1\" size=\"3\" maxlength=\"3\" tabindex=\"8\""; print "<label>)<input type=\"text\" name=\"PHONE2\" size=\"3\" maxlength=\"3\" tabindex=\"9\""; print "<label>-<input type=\"text\" name=\"PHONE3\" size=\"4\" maxlength=\"4\" tabindex=\"10\"<br>"; print "<p align=left><label>Email: <input type=\"text\" name=\"EMAIL\" size=\"50\" maxlength=\"50\" tabindex=\"11\"<br>"; print "<p align=left><label>Payment Method: <select name=\"PM\"> <option value=\"\">Choose One</option> <option value=\"Visa\">Visa</option> <option value=\"Mastercard\">Mastercard</option> <option value=\"Discover\">Discover</option>"; print "</select><p align=left><label>Credit Card Number: <input type=\"text\" name=\"C1\" size=\"4\" maxlength=\"4\" tabindex=\"12\""; print "<label> <input type=\"text\" name=\"C2\" size=\"4\" maxlength=\"4\" tabindex=\"13\""; print "<label> <input type=\"text\" name=\"C3\" size=\"4\" maxlength=\"4\" tabindex=\"14\""; print "<label> <input type=\"text\" name=\"C4\" size=\"4\" maxlength=\"4\" tabindex=\"15\"<br>"; print "<p align=left><label>Expiration Month: <select name=\"EXM\"> <option value=\"\">Choose One</option> <option value=\"January\">January</option> <option value=\"February\">February</option> <option value=\"March\">March</option> <option value=\"April\">April</option> <option value=\"May\">May</option> <option value=\"June\">June</option> <option value=\"July\">July</option> <option value=\"August\">August</option> <option value=\"September\">September</option> <option value=\"October\">October</option> <option value=\"November\">November</option> <option value=\"December\">December</option>"; print "</select><p align=left><label> Expiration Year: <select name=\"EXY\"> <option value=\"\">Choose One</option> <option value=\"2011\">2011</option> <option value=\"2012\">2012</option> <option value=\"2013\">2013</option> <option value=\"2014\">2014</option> <option value=\"2015\">2015</option> <option value=\"2016\">2016</option> <option value=\"2017\">2017</option> <option value=\"2018\">2018</option> <option value=\"2019\">2019</option> <option value=\"2020\">2020</option>"; print "</select><p align=left><label>Security Code (3 or 4 digits): <input type=\"text\" name=\"CSC\" size=\"4\" maxlength=\"4\" tabindex=\"16\"<br>"; print "<p align=left><input type=\"submit\" value=\"Book Now!\" name=\"submit\"></form>"; print " <input type=\"reset\" value=\"reset\" name=\"reset\"></form>"; // Close the database connection mysql_close(); ?> Sorry this is so long, any help would be appreciated...... -Bob Brand new to the forum, and have only been working with HTML, PHP, & MySQL for about 3 months. I've run into an issue I can't seem to figure out how to work around. I want to have two select boxes, that are dependent on one another. For Example: First box lists types of games: RPG LARP Board Game Miniatures CCG The second box displays game names of the type selected above. I have a MySQL table setup for the second box. To be honest I kinda already have it working... with one small but MAJOR glitch. I select the first box and the page submits using onchange='this.form.submit()'. When the page reloads the first select box reverts back to it's default setting, but the second box does correctly filter & show the content based on the selection that was made in the first box. My Question: How do I set this up so that the first box displays the user selected setting while also passing the $_POST information to the second box? My Code: <table> <tr> <td width=50%> Game Type: </td> <td width=50%> <form name='game_type' action='event_submit.php' method='post'> <select name='game_type' align=center onchange='this.form.submit()' style="width:150px;margin:5px 0 5px 0;"> <option value=''>Chose a Game Type</option> <option value='RPG'>RPG</option> <option value='LARP'>LARP</option> <option value='Board Game'>Board Game</option> <option value='CCG'>Collectable Cards & Games</option> <option value='Miniatures'>Miniatures</option> </select> </form> </td> </tr> <tr> <td width=50%> Game System: </td> <td width=50%> <?php $game_type = $_POST['game_type']; echo "$game_type"; if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("DrowCon", $con); $result = mysql_query("SELECT * FROM Games WHERE Type='$game_type' ORDER BY System ASC"); echo "<form name='game_system' action='event_submit_form.php' method='post'>"; echo "<select name='game_system' align=center style='width:150px;margin:5px 0 5px 0;'>"; echo "<option value=''>Chose a Game System</option>"; while($row_game = mysql_fetch_array($result)) { echo "<option value=''>".$row_game['System']."</option>"; } echo "</form>"; mysql_close($con); ?> </td> </tr> </table> I have a table in a database that I need to filter upon using dependent select boxes. The database was poorly designed and would be difficult to break into separate tables without affecting code written in other places. I have found some good code and understanding of how to implement the dependent select boxes if you have two tables with one column functioning as a lookup value between the two. I have provided a short bit of sql of what the table looks like with some column names changed. I would need to first sort by lastname and have firstname dependent on the selection of lastname. Any help would be greatly appreciated. Code: [Select] create table `user` ( `id` int(3) Not Null Auto_increment, `firstname` varchar (60), `lastname` varchar (24), `age` double , `hometown` varchar (75), `job` varchar (75), `birthdate` date ); insert into `user` (`id`, `firstname`, `lastname`, `age`, `hometown`, `job`, `birthdate`) values ('1','Peter','Griffin','41','Quahog','Brewery','1960-01-01'), ('2','Lois','Griffin','40','Newport','Piano Teacher','1961-08-11'), ('3','Joseph','Swanson','39','Quahog','Police Officer','1962-07-23'), ('4','Glenn','Quagmire','41','Quahog','Pilot','1960-02-28'), ('5','Megan','Griffin','16','Quahog','Student','1984-04-24'), ('6','Stewie','Griffin','2','Quahog','Dictator','2008-03-03'); Dear Support, Here is my code: <?php if ( strtolower($_SERVER['REQUEST_METHOD']) === 'post' ) { $name = isset($_POST['name]) ? htmlspecialchars($_POST['name']) : ""; $area = isset($_POST['area']) ? htmlspecialchars($_POST['area']) : ""; $choice = isset($_POST['choice ']) ? htmlspecialchars($_POST['choice ']) : ""; //Do Others } ?> <form class="customersForm" method="post"> <div class=""> <p>Name<br /> <select style="width:120px;" name="name"> <?php $sql = "select * from `name_info` order by `name`"; $query = mysql_query($sql); $dd = '<option>Name</option>'; while($row = mysql_fetch_object($query)) { $dd .= '<option>'.$row->name.'</option>'; } echo $dd; ?> </select> </p> </div> <div class=""> <p>Area<br /> <select style="width:120px;" area="area"> <?php $sql = "select * from `area_info` order by `area`"; $query = mysql_query($sql); $dd = '<option>Area</option>'; while($row = mysql_fetch_object($query)) { $dd .= '<option>'.$row->area.'</option>'; } echo $dd; ?> </select> </p> </div> <div class=""> <p>Choice<br /> <select style="width:120px;" choice="choice"> <?php $sql = "select * from `choice_info` where `name`='".$name."' and `area`='".$area."' order by `choice`"; $query = mysql_query($sql); $dd = '<option>Choice</option>'; while($row = mysql_fetch_object($query)) { $dd .= '<option>'.$row->choice.'</option>'; } echo $dd; ?> </select> </p> </div> ......//Others input fields ...some text fields ... </ form> I have the above code, where I would like to show the choice Drop-down box, depends on $name and $area match of the choice_info table. $name field and $area field needs to have any dependency and straight from the MySQL table. But, choice needs to depend on $name & $area. How can I do it without much using Javacript? I do not mind to use JavaScript, but prefer to minimal. By the way, I will have some other input fields too. It's an input form but, one drop-down box will be depends on two others choice. Thanks for your help. Hi, I had a form which contains, 8 select boxes. In those 8 select boxes, 3 of the boxes are interlinked, like, I am able to generate data in other 2 select boxes, if one box is been selected. i.e. based upon selection of one select box, and i am able to generate data in other select box. So, 3 select boxes are interlinked. And remaining 5 selects are optional, if the user selects, then we need to generate data based upon selection of the user. Here the main issue is how to grab data in the post back form, like, how to capture the number of selects boxes selected and what are those selected, and based upon that we need to generate data. Here one more important thing is that, in the report generation form(postback form), I need to show the data in a table. So , in table I had 9 columns and these would be common to whatever the selection made, but, only the requeired data would be changed based upon the selection of the remaining 5 optional boxes. Hope you got me!!!! 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: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hi all I need to combine these two scripts: Firstly, the following decides which out of the following list is selected based on its value in the mySQL table: <select name="pack_choice"> <option value="Meters / Pack"<?php echo (($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 hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks I have 2 queries that I want to join together to make one row
Right now I'm feeling like a little baby novice asking a ridiculiously simple question but I just can't find a solution to it! So here it goes.... please don't laugh!!! How do I create a menu which doesn't have to be changed on every existing page whenever I need to add a new page? See... novice stuff and I know it can be done in Dreamweaver. I have Dreamweaver software (old version) but how would I write the code to do so ? Thank you everyone. I have 3 dropdown menus which have the same name <select name="dropdown">, how to get the value from 3 of them by using php for loop and array? I did try to put array on them before, but seem it can't work. My codes : <?php for ( $counter = 1; $counter <= 3; $counter ++) { if (isset($_POST['submit'])) { //Form has been submitted. $dropvalue = $_POST['dropdown[$counter]']; echo $dropvalue; } } ?> <form action="test8.php" name="frmtest8" method='post'> <select name="dropdown[1]"> <option value="1st">first</option> <option value="2nd">second</option> <option value="3rd">third</option> </select> <select name="dropdown[2]"> <option value="1st">first</option> <option value="2nd">second</option> <option value="3rd">third</option> </select> <select name="dropdown[3]"> <option value="1st">first</option> <option value="2nd">second</option> <option value="3rd">third</option> </select> <input type="submit" name="submit" value="Send" /> </form> It gives error message Parse error: parse error in C:\wamp\www\plekz\test8.php on line 23 I guess the error is because of $_POST['dropdown[$counter]']; I cannot put the two symbols [] inside $_POST[' '] right? How to call the array if cannot put the two symbols? Hi Everyone,
Someone is asking me a bit of help on a website, I know a bit php and HTML5, I would like to know what would be the best of doing this please:
1)I need to create 2 big buttons, according to the choice of the buttons a different dropbox menu with towns will showup.
2)When the user chooses one of the town, in any drop down, I would like to show a different DIV with content in it.
Here is a practical example:
BUTTON A BUTTON B
if Button A is pressed, show dropdown menu A
if BUTTON B is pressed, show dropdown menu B
If town 1 from dropdown menu A is chosen show DIV 1
If town 2 from dropdown menu A is chosen show DIV 1
If town 3 from dropdown menu A is chosen show DIV 2
If town 4 from dropdown menu A is chosen show DIV 3
If town 5 from dropdown menu A is chosen show DIV 1
If town 1 from dropdown menu B is chosen show DIV 6
If town 2 from dropdown menu B is chosen show DIV 5
If town 3 from dropdown menu B is chosen show DIV 5
If town 4 from dropdown menu B is chosen show DIV 6
If town 5 from dropdown menu B is chosen show DIV 4
How to do this without reloading the page please?
Thank you,
Bambinou
Hi I was just wondering if someone could send me a good tutorial on pulling several fields from a table and displaying them all in one drop down menu for people to select one of them. I dont know what this is called to I dont know what to research online. 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? Hi all, I'm new to PHP and just looking for any advice anyone might have on the following problem I'm having. I have created a database and tables for registering and logging in users. Each record has three different User Types (or roles): 1, 2 and 3. Once the session has started, I would like to make certain menu items available to certain User Types. I suppose it is some sort of if...elseif statement such as: <?php if($UserType==1) echo {"<p>Menu1</p>"} elseif($UserType==2) echo {"<p>Menu2</p>"} elseif($UserType==3) echo {"<p>Menu3</p>"}; ?> Would anybody be able to point me in the right direction? I'd really appreciate any help at all. I'm simply trying to set up a form where, if when a user clicks 'Submit', and then 'Back', the values from the form are preserved. My question is, how do I preserve the values of drop down menus. The following is a snippet of my code: Code: [Select] <select name="dropdown_dept" id="dept_list"> <option value=0><?php echo "Please select one..."?></option> <?php $dropdown_dept = "select dept_name from departments"; $result_dept = $db_conn->query($dropdown_dept); if (!$result_dept) { echo '<p>Unable to get department data.</p>'; return false; } for($i=0; $i<$result_dept->num_rows; $i++) { $app_name_row = $result_dept -> fetch_array(); ?> <option><?php echo($app_name_row[0]); ?></option> <? } ?> </select> Above is where I have set up a drop down menu of departments. Given that code, how can I preserve the department name after a user clicks 'Submit'? 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! |
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