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PHP - Drop Down Menu Select?
Hi, I'm no pro at PHP but I am trying to get a drop down menu to a authenticate before moving to the next part of the form.
What I want is once a selection has been made, ONLY THEN can the user move on, OTHERWISE a message echo appears. This is the html menu box <select size="1" name="title"> <option>Please Select</option> <option value="Mr">Mr</option> <option value="Mrs">Mrs</option> <option value="Miss">Miss</option> <option value="Ms">Ms</option> <option value="Dr">Dr</option> </select> Then this is what I have in the form PHP: $visitortitle = $_POST['visitortitle']; if ( HOW DO I GET THIS PART TO AUTHENTICATE AN OPTION HAS BEEN SELECTED? ) { echo "<p>Please enter a title correctly<br />before you try submitting the form again.</p>\n"; die ( '<a href="pef.html">click here go back and try again</a>' ); echo $id;} If anyone can help me sort out this part of the form I can move on as the rest is working fine? Thanks Gary Similar TutorialsI am using jquery .change function to perform an operation when a month is selected from a drop down menu. The change works but I am unable to update the value of the drop down menu with the updated month. My drop down shows the starting value as default even on change. Can anyone help. Following is the code snippet that does change and then the drop down menu form. Code: [Select] $("#monthName").change(function() { alert($("#monthName").val()); if ($("#post").val() == 1) { $("#monthselect").submit(); } }); Code: [Select] <form id="monthselect" action="<?=$_SERVER['PHP_SELF']?>" method="get"> <input id="post" type="hidden" name="post" value="1"> <label>SELECT MONTH</label> <select id="monthName" name="monthName"> <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> </select> </form> Even after the change, January shows up by default even if I select say June or July. I tried something like following but did not work. Code: [Select] $("#monthName option[value=" + $("#monthName").val() +"]").attr("selected","selected") ; the same page? Hi there, i am relatively new to php, mysql, css etc but learning fast. My problem is such; i have a php file which is doing a SELECT mysql_query, WHILE results to strings, then ECHO the resulting rows to produce a list formatted using <table> and finally this <table> is inside a <form> which will POST the changes back to the specific database.tble.row. I wish to have a drop down menu within the <form><table> which will be populated from a separate database.table. I have accomplished the drop down menu outside the <?php ?> tags inside <form><table> which POSTS to a php file but my problem is to add the populating drop down menu inside <?php ?> an already ECHOing resulting rows from the sql query. i.e <?php blurb and stuff ?> <form><table><tr><td> <select name etc> <?php $result = mysql_query("SELECT * FROM tbl WHERE string = tble.rw ORDER BY column"); while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['column'].">".$row['column']."</OPTION>"; } ?> </select> WORKS!!!! but placing this inside <?php $x =mysql_query[select] while {strings = conditions; echo ("<form><table><tr><td> insert populated drop menu here </td> etc "); echo"";}?> doesnt work and just leaves the select drop menu blank Hoep you understand my problem. I do not think i can attached the population WHILE loop to a string and just insert the string to the form but maybe i am wrong. thanks in advance and if you go tthis far reading you must be on lots and lots of coffee zark I am currently creating a form and I want to populate a drop down selection menu with data from two fields in a form. For example, I want it to pull the first and last name fields from a database to populate names in a drop down menu in a form. I want the form to submit to the email address of the person selected in the drop down. Is this possible to do? The email is already a field in the record of the person in the database. Can anyone give me some pointers or advice on how I should go about setting up the "Select" box drop down? I am not sure how to code it to do what I am wanting. Any links to relevant help would be appreciated too. Thanks in advance! Hi everyone - this code is doing my head in! You might want to take a look at the page as it will help me explain it better (www.bradleystokejudoclub.co.uk/inttest.php Basically what I have is a database with the results from international competitions, and I am trying to build a kind of search for it. I have 3 drop down boxes, one with player names, one with competiton and one with year. The only one that works is the competiton. I can do competition + year, but not year by itself, and player doesnt work at all .... this is the relevant code: (before head) <?php include("includes/dbconnect120-gem.php"); include("includes/db_auth_bits.php"); include("includes/db_stp.php"); if($_POST) { if($name == 'select') { $sql1 = ""; } else { if(($award == 'select') && ($year == 'select')) { $sql1 = "r.pname_id = '$pname'"; } else { $sql1 = "r.pname_id = '$pname' AND "; } } if($award == 'select') { $sql2 = ""; } else { if($year == 'select') { $sql2 = "r.comp_id = '$comp'"; } else { $sql2 = "r.comp_id = '$comp' AND "; } } if($year == 'select') { $sql3 = ""; } else { $sql3 = "r.year_id = '$year'"; } if(($sql1 == "") && ($sql2 == "") && ($sql3 == "")) { $where = ""; } else { $where = " WHERE "; } $sql = "select p.pname, i.comp, m.place_name, yr.year_full from intcomp_result r left join playername p on r.pname_id=p.name_id left join intcomp i on i.comp_id = r.comp_id left join place m on m.place_id = r.place_id left join yearname yr on r.year_id = yr.year_id $where $sql1 $sql2 $sql3 order by r.year_id desc, r.comp_id, r.place_id"; $search_result = mysql_query($sql); } ?> (body) <form method="post" action="<?php echo $PHP_SELF;?>"> Name:<select name="name"> <option value="select" selected="selected" >-SHOW ALL-</option> <?php $sql="select distinct r.pname_id, p.pname from intcomp_result r left join playername p on r.pname_id=p.name_id order by p.pname"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $nameid=$row['pname_id']; $pname=$row['pname']; ?> <option value="<?php echo $nameid; ?>"><?php echo $pname;?></option> <?php } ?> </select><br /> Competition:<select name="comp"> <option value="select" selected="selected" >-SHOW ALL-</option> <?php $sql="select * from intcomp order by comp_id asc"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $compid=$row['comp_id']; $comp=$row['comp']; ?> <option value="<?php echo $compid; ?>"><?php echo $comp;?></option> <?php } ?> </select><br /> Year:<select name="year"> <option value="select" selected="selected" >-SHOW ALL-</option> <?php $sql="select distinct r.year_id, y.year_full from intcomp_result r left join yearname y on r.year_id=y.year_id order by y.year_id"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $yearid=$row['year_id']; $year=$row['year_full']; ?> <option value="<?php echo $yearid; ?>"><?php echo $year;?></option> <?php } ?> </select><br /> <input name="Submit" type="submit" class="button" tabindex="14" value="Submit" /> </form> <?php if(isset($search_result)) { while($row = mysql_fetch_array($search_result)) { echo $row['pname'].' - '.$row['comp'].' - '.$row['year_full'].' - '.$row['place_name'].'<br />'; } } ?> Hope you can help! Thanks Gem Hey guys i am new around here and i need some help with php. So basically i want to make a submit form and i found a problem for myself here's some images to understand what i am talking about. so here's what i've made at first. This is meant for new data. <p><label>Status :</label> <select name="status" > <option value=""></option> <option value="Ongoing">Ongoing</option> <option value="Completed">Completed</option> </select> basically what I want is when I edit the data, that selection returns the value that was stored in the database rather than just null value. See image below to understand what i am trying to do. i dont really get how to do it with select. Please help me and thanks before. Sorry if my question was already asked before. Tried to search but didnt really found what i am looking for. Hey there, i have a cookie which echos ok but would like to know how to automatically select the corresponding option from a drop down list when the page loads. Normal List <form id="switchform"> <select name="switchcontrol" size="1" class="topusernav" onChange="chooseStyle(this.options[this.selectedIndex].value, 60)"> <option value="none">-----</option> <option value="noir">Noir</option> <option value="crimson">Crimson</option> <option value="forrest">Forrest</option> <option value="ocean">Ocean</option> <option value="petal">Petal</option> </select> </form> works fine but <form id="switchform"> <select name="switchcontrol" size="1" class="topusernav" onChange="chooseStyle(this.options[this.selectedIndex].value, 60)"> <option value="none" <?php if (!(strcmp("none", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>-----</option> <option value="noir" <?php if (!(strcmp("noir", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Noir</option> <option value="crimson" <?php if (!(strcmp("crimson", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Crimson</option> <option value="forrest" <?php if (!(strcmp("forrest", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Forrest</option> <option value="ocean" <?php if (!(strcmp("ocean", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Ocean</option> <option value="petal" <?php if (!(strcmp("petal", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Petal</option> </select> </form> gives me a whitespace error any thoughts?? thanks Hey Guys. I am trying to create a time dropdown and increment eat by an an hour until it is lless than or equal to the end time. I am using the DateTime class to accomplish this.
Now when I am trying to use a for loop to accompish this it does not work. Can anyone please help me out with this issue?
Below is the code that I have
$start_hour = new DateTime("now",new DateTimeZone("America/New_York"));
$start_hour->setTime(6,00);
$formatted_start_time = $start_hour->format("H:i:s");
$end_hour = mktime(11,45);
for($start_hour; $start_hour <= $end_hour; $start_hour->modify("+60 minutes"));{
echo "<select name='cat_display_timeslot'>";
echo "<option>{$formatted_start_time}</option>";
echo "</select>";
}
Thanks!
Edited by eldan88, 24 August 2014 - 07:21 PM. 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> Hi, I am trying to create a drop down menu that will select a value that is stored in the database - right now the code creats a dropdown (with nothing selected) - hope someone can help. in the database, the values are stored as --null- Option1 Option2 Option3 my code is Code: [Select] $instruction = $_GET['instruction']; <?php <select id="instruction" name="instruction"> <option value="" <?php if (!empty($instruction) && $instruction == '' ) echo 'selected = "selected"'; ?>></option> <option value="Option1" <?php if (!empty($instruction) && $instruction == 'Option1') echo 'selected = "selected"'; ?>>Option1</option> <option value="Option2" <?php if (!empty($instruction) && $instruction == 'Option2') echo 'selected = "selected"'; ?>>Option2</option> <option value="Option3" <?php if (!empty($instruction) && $instruction == 'Option3') echo 'selected = "selected"'; ?>>Option3</option> </select> <? Hello guys I've hit a problem whicle trying to validate my form. I have 3 drop down boxes where the user chooses from three options. But I can't seem to figure out how to set the validation so the user does not select the same option in each drop down. Can anyone help me solve this please. Thank you. <p><b>Course Choice 1</b> <select name="course1"> <option value="0"></option> <option value="Business computer Systems">Business computer Systems</option> <option value="Business computer Science">Business computer Science</option> <option value="Business computer Science (Games)">Business computer Science (Games)</option> <option value="Business Information Systems">Business Information Systems</option> <option value="Digital Media Development">Digital Media Development</option> <option value="Digital Media">Digital Media</option> </select></p> <p><b>Course Choice 2</b> <select name="course2"> <option value="Leave Blank">Leave Blank</option> <option value="Business computer Systems">Business computer Systems</option> <option value="Business computer Science">Business computer Science</option> <option value="Business computer Science (Games)">Business computer Science (Games)</option> <option value="Business Information Systems">Business Information Systems</option> <option value="Digital Media Development">Digital Media Development</option> <option value="Digital Media">Digital Media</option> </select></p> <p><b>Course Choice 3</b> <select name="course3"> <option value="Leave Blank">Leave Blank</option> <option value="Business computer Systems">Business computer Systems</option> <option value="Business computer Science">Business computer Science</option> <option value="Business computer Science (Games)">Business computer Science (Games)</option> <option value="Business Information Systems">Business Information Systems</option> <option value="Digital Media Development">Digital Media Development</option> <option value="Digital Media">Digital Media</option> </select></p> <div align="centre"><input type="submit" name="submit" value="send request" /></div> //Validate course choice 1 if (!empty($_REQUEST['course1'])) { $course1 = $_REQUEST['course1']; } else { $course1 = NULL; echo '<p><font color="red">Please enter your first choice</font></p>'; } //Validate course choice 2 if (!empty($_REQUEST['course2'])) { $course2 = $_REQUEST['course2']; } else { $course2 = NULL; echo '<p><font color="red">Please enter your second choice</font></p>'; } //Validate course choice 3 if (!empty($_REQUEST['course3'])) { $course3 = $_REQUEST['course3']; } else { $course3 = NULL; echo '<p><font color="red">Please enter your third choice</font></p>'; } //If everything is ok, print the message if ($name && $email && $course1 && $course2 && $course3) { echo "<p>Thank you, <b>$name</b>, You have chosen the following courses for information:<br /><br /> <b>$course1</b><br /> <b>$course2</b><br /> <b>$course3</b></p> <p>We will reply to you at <i>$email</i>.</p>\n"; } else { // One form element was not filled out properly echo '<p><font color="red">Please go back and fill out the form again.</font></p>'; } 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]')"; Hello, I have the following function function make_agent_drop($dropname,$parent=''){ $agents = mysql_query("SELECT * FROM ad_category WHERE cat_status='1' AND parent_id='".(is_numeric($parent)?$parent:"0")."'") or die(mysql_error()); $anum = mysql_num_rows($agents); if($anum>0){ $agentdrop='<select style="width:150px; height:20px; margin-left:100px; font-size:11px;" name="'.$dropname.'" id="'.$dropname.'" class="text" '.(is_numeric($parent)?'':'onchange="update_subcatdrop($(this).val());').'"> <option value="0">Select a Category</option>'; while($row= mysql_fetch_array($agents)){ $agentdrop.='<option value="'.$row['cat_id'].'">'.$row['cat_name'].'</option>'; } $agentdrop.='</select>'; }else{ $agentdrop= 'No '.(is_numeric($parent)?'Sub':'').'Categories Found.'; } return $agentdrop; ; } I creates a drop down from database cats and sub cats.. I am trying to figure out how to add a submit button to dynamically appear when it displays the sub category... Thanks! Dan Hi I want to add student from a dropdown list to database but I have some problem. This is my select dropdown menu code <form name ="student" method = "POST" action ="confirmation.php"> <select name="name"> <option selected>Select Student</option> <?php $arrStudent = executeSelectQuery("select * FROM user "); for ($i = 0; $i < count($arrStudent); $i++) { $student_result = $arrStudent[$i]['student_id']; $name_result = $arrStudent[$i]['student_name']; ?> <option value="<?php echo $id_result; ?>"><?php echo $id_result; ?>, <?php echo $name_result; ?></option> <?php } ?> </select> </form> The output in the dropdown menu look something like this: 1, Alvin 2, Benny 3, Charles 4, Daniel 5, Eva and so on... After submitting the form, it will proceed to confirmation.php page. At the confirmation page, I have the following variable: $student_result = $_REQUEST['student_id']; $name_result = $_REQUEST['student_name']; I want to insert to database with the following insert query $sql = "INSERT INTO student(student_id, student_name) VALUES ('". $student_result . "', '". $name_result ."')"; $insert = executeInsertQuery($sql); It can insert successfully but, it will not insert the student_name. May I know where I did wrongly? Thanks Ben Chew I have a search function in php where the text characters are matched to characters in a tables field--- works perfectly.... I need to make the input box have a droplist of words from database, this is also easy for me to do. the problem here is there is no definitive value! the options list always outputs a blank in the url--- its supposed to search a matching value and then output the matching value to url... Here is the droplist code: $output['RATESTITLE']='<input class="inputbox" type="text" size="24px" name="ratestitle" value="'.$sch->filter['ratestitle'].'" onfocus="if (this.value ==\''.$output['LANGUAGE_SEARCH_RATESTITLE'].'\') {this.value = \'\'}" />'; this outputs a input text box--- i want to have a droplist of options to populate this text box... If you must know this is the third day im at it... Hi I have tried the mysql forum but have had no joy with an answer to my problem so wondered if php would be better. I want my users to be able to select from 5 different drop down lists where they can chose any combination from 1 up to all 5, I have attached the front end. These lists are being populated from mysql tables. Code for the drop down lists is as follows Code: [Select] <form action="horse-events-devon.php?url_countyid=<?php echo urlencode ($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, $connection); ?> How am I best to do this please? my OR within the mysql does not work, should I not be doing this with php in the search processing? someones help would really be appreciated, just to point me in the right direction. [attachment deleted by admin] Hello, I'm using a dynamically created select menu for a user to make a choice which will then be put in my database with the following code: <select> <?php foreach ($course_number as $row) { echo "<option value = '{$row['course_id']}'"; if ($errors && $_POST["course_id"] == $row['course_id']) {echo 'selected = "selected"'; } echo ">{$row['course_number']}</option>"; } ?> Unfortunately, I'm having some problems figuring out how to pull off the selected value. Right now my database portion looks like this: $data = array('assignment_name' => $_POST['assignment_name'], 'due_date' => $_POST['due_date'], 'course_id' => $_POST["row['course_id']"]); $inserted = $dbWrite->insert('assignments_instructors',$data) While the assignment_name and due_dates work (they come from text fields), my course_id gives me an Undefined index: row['course_id'] error. Any help would be appreciated. Thank you. Could someone help me I really dont know how to go about coding this, so i would be happy if someone could point me in the right way Well what I am trying to do is use mysql_num_rows to call up how many rows in the table. The using how many rows, use a menu with the numbers of rows that are in the table ex below mysql_num_rows gets 5 rows so menu is <select name="order" > <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> I have a test database, I have two names in the database which get returned just fine in my html dropdown. I am trying to figure out how to figure out which item in the list was selected so that I can return information from another table based on the selected id. This is what I am trying now but I don't know how to proceed or if it is correct Code: [Select] $result = $mysql->query("SELECT * FROM names") or die($mysql->error); ?> <select> <?php if($result){ while($row = $result->fetch_object()){ $id = $row->nameID; $name = $row->firstName . " " . $row->lastName; ?> <option value"<?php $id ?>"><?php echo $name ?></option> <?php }?> </select> <?php } ?> Hi I have a temporary web page with a drop down menu. Problem is to get rid of the gap on the drop down menus. Any help please
www.des-otoole.co.uk/top_menu
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