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PHP - Select Drop Menu Unable To Populate
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 Similar TutorialsI 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, 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 I 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? I am having a helluva time trying to populate my state option box with a state array. I have the array set but now how does this get called into the foreach block where the select box is? I can populate it i guess with options manually but I want to get the array to display in the box instead. I tried everything with the minimal experience I have so I need a little help, I got the select box up and displaying where I want it but I cant populate it with an array call. See foreach block near the end of the code: Thanks a bunch! here is my include file code so far: <?php /* Program name: form_test_state.inc * Description: Defines a form that collects a user's * name and mailing address. */ $rows = array( "first_name" => "First Name", "last_name" => "Last Name", "phone" => "Phone Number", "city" => "City", "state" => "State", "address" => "Address", "zip_code" => "ZIP Code", "e_mail" => "E-Mail"); $states_list = array('AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",'AR'=>"Arkansas",'CA'=>"California",'CO'=>"Colorado",'CT'=>"Connecticut",'DE'=>"Delaware",'DC'=>"District Of Columbia",'FL'=>"Florida",'GA'=>"Georgia",'HI'=>"Hawaii",'ID'=>"Idaho",'IL'=>"Illinois", 'IN'=>"Indiana", 'IA'=>"Iowa", 'KS'=>"Kansas",'KY'=>"Kentucky",'LA'=>"Louisiana",'ME'=>"Maine",'MD'=>"Maryland", 'MA'=>"Massachusetts",'MI'=>"Michigan",'MN'=>"Minnesota",'MS'=>"Mississippi",'MO'=>"Missouri",'MT'=>"Montana",'NE'=>"Nebraska",'NV'=>"Nevada",'NH'=>"New Hampshire",'NJ'=>"New Jersey",'NM'=>"New Mexico",'NY'=>"New York",'NC'=>"North Carolina",'ND'=>"North Dakota",'OH'=>"Ohio",'OK'=>"Oklahoma", 'OR'=>"Oregon",'PA'=>"Pennsylvania",'RI'=>"Rhode Island",'SC'=>"South Carolina",'SD'=>"South Dakota",'TN'=>"Tennessee",'TX'=>"Texas",'UT'=>"Utah",'VT'=>"Vermont",'VA'=>"Virginia",'WA'=>"Washington",'WV'=>"West Virginia",'WI'=>"Wisconsin",'WY'=>"Wyoming"); $submit = "Submit mailing information"; ?> <html> <head> <style type='text/css'> <!-- body { background: #42413C; color: #000; font-family: Tahoma, Geneva, sans-serif; font-size: 90%; } form { margin: 1em 2 2 2; padding: 1; width: 1100px; margin-left: 35%; } .field {padding-bottom: 1em; } label { font-weight: bold; float: left; width: 10%; margin-right: 1em; text-align: right; background-color: #9F6; padding-right: 5px; } select {margin-bottom: 1em;} #submit { margin-left: 15%; } h3 { width: 500px; margin-left: 35%; } --> </style> </head> <body> <h3>Please enter your mailing information below:</h3> <?php /* loop that displays the form */ echo "<form action='checkBlankOnly_latest.php' method='post'>"; foreach($rows as $field => $label) if($field == "state") { echo "<label for='$field'>$label</label> <select name='State'> <<<< Insert $states_list array here I am trying to figure out >>>> <option value='state'>Alabama </option> </select> "; } else { echo "<div class='field'><label for='$field'>$label</label> <input id='$field' name='$field' type='text' value='".@$$field."' size='25%' maxlength='65' /></div>\n"; } echo "<div id='submit'> <input type='submit' value='$submit'></div>"; ?> </form> </body> </html> hi, I was going through this tutorial: http://www.electrictoolbox.com/json-data-jquery-php-mysql/ here is my HTML Code: [Select] <!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=UTF-8" /> <title>Untitled Document</title> <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js"></script> <script language="javascript" type="text/javascript"> function populateFruitVariety() { $.getJSON('search-by.php', {fruitName:$('#fruitName').val()}, function(data) { var select = $('#fruitVariety'); var options = select.attr('options'); $('option', select).remove(); $.each(data, function(index, array) { options[options.length] = new Option(array['variety']); }); }); } $(document).ready(function() { populateFruitVariety(); $('#fruitName').change(function() { populateFruitVariety(); }); }); </script> </head> <body> <form> Search by: <select name="name" id="fruitName"> <option>Please Select</option> <option id="Town" value="Town">Town</option> <option id="County" value="County">County</option> </select> Variety: <select name="variety" id="fruitVariety"> </select> </form> </body> </html> here is my PHP Code: [Select] $dbhost = "xxx"; $dbname = "xxx"; $dblogin = "xxx"; $dbpass = "xxx"; function dbConnect() { global $dbhost; global $dbname; global $dblogin; global $dbpass; $db = mysql_connect($dbhost, $dblogin, $dbpass) or die("could not connect to database: ".mysql_error()); mysql_select_db($dbname) or die("could not select database"); return $db; } function dbClose($db) { mysql_close($db); } // basis code ^^ $db = dbConnect(); $rows = array(); if(isset($_GET['Town'])) { $query = "SELECT DISTINCT rsCounties FROM pubs"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo json_encode($rows); } } if(isset($_GET['County'])) { $stmt = $pdo->prepare("SELECT rsCounty FROM pubs"); $stmt->execute(array($_GET['rsCounty'])); $rows = $stmt->fetchAll(PDO::FETCH_ASSOC); } echo json_encode($rows); Can someone please help? I have the JQuery working but not the PHP?! here is a link to what I have at the moment: http://www.mypubspace.com/dashnew/index.html thanks Hello, I am running into a problem. I am trying to convert : Code: [Select] $queryyp = "SELECT YEAR(date) as year, MONTHNAME(date) as month, title FROM post ORDER BY date DESC"; // query to get the rows you want in the order that you want them, with the year and monthname specifically selected as well $resultyp = mysql_query($queryyp); $last_heading = null; // remember the last heading (initialize to null) while($rowyp = mysql_fetch_assoc($resultyp)){ $new_heading = $rowyp['year']; // get the column in the data that represents the heading $new_subheading = $rowyp['month']; // get the column in the data that represents the subheading if($last_heading != $new_heading){ // heading changed or is the first one $last_heading = $new_heading; // remember the new heading $last_subheading = null; // (re)initialize the subheading // start a new section, output the heading here... echo "{$rowyp['year']}<br />"; } // subheading under each heading if($last_subheading != $new_subheading){ // subheading changed or is the first one $last_subheading = $new_subheading; // remember the new subheading // start a new section, output the subheading here... echo " {$rowyp['month']}<br />"; } // output each piece of data under a heading here... echo " {$rowyp['title']}<br />"; } And be able to put it in a javascript tree with very little luck: Code: [Select] <ul id="yeartree" class="tree"> <li>{$rowyp['year']} <ul> <li>{$rowyp['month']} <ul> <li>{$rowyp['title']}</li> </ul> </li> </ul> </li> Yes I know the phpcode in the tree is not correct. That is just an example of what I am attempting to do. I can get the tree to display without the js but I need to add it to js due to the amount of entries and need to beable to collapse it. Hi, I am making a login page which consists of 2 pages, index.php and config.php Following is a code of config.php Code: [Select] <?php define("HOST", "localhost"); define("USER", "aaaabbbccc"); define("PASS", "aaabbbccc"); define("DATABASE", "pricetagindia"); $connection = @mysql_connect(HOST, USER, PASS); $select_db = @mysql_select_db(DATABASE, $connection); if(!$connection) { echo "<h1>Cannot connect to database</h1>"; exit(0); } if(!$select_db) { echo "<h1>Cannot select database</h1>"; exit(1); } ?> Here is a coding of index.php Code: [Select] <?php include('config.php'); $_error = "<p></p>"; if (isset($_POST['username']) || isset($_POST['password'])) { $username = $_POST['username']; $password = $_POST['password']; if($username == NULL || $password == NULL) { $_error = "<p>Username and Password, both fields are required.</p>"; } else { // Check in DB $password = md5($password); $checkuser_query = "SELECT * FROM users WHERE username = '$username.' AND password = '$password'"; $checkuser_result = mysql_query($checkuser_query); $checkuser_count = mysql_num_rows($checkuser_result); if($checkuser_count == 1) { echo "match found"; } else { echo "nothing found error, username was ".$username." and pass is ".$password; echo "<p></p>"; echo $checkuser_result; } } } ?> <!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=utf-8" /> <title>Price Tag India</title> </head> <body> <center> <h1>Price Tag India</h1> <p> </p> </center> <center> <form action="index.php" method="post" target=""> <table width="389" height="168" border="0"> <tr> <td width="141">Username :</td> <td colspan="2"><label for="username"></label> <input type="text" name="username" id="username" /></td> </tr> <tr> <td>Password :</td> <td colspan="2"><label for="password"></label> <input type="password" name="password" id="password" /></td> </tr> <tr> <td> </td> <td width="56"><input type="submit" name="button" id="button" value="Submit" /></td> <td width="223"><input type="reset" name="button2" id="button2" value="Reset" /></td> </tr> <tr> <td> </td> <td colspan="2"> <?php echo $_error; ?> </td> </tr> </table> <p> </p> </form> </center> </body> </html> I have attached the screenshot of database. Problem is when I do valid login i.e. with user admin and its password, it is not working, I spent 1 hour for solving this issue but didnt find any solution. Its frustrating now.....I know its simple but still its not working. It is giving me error as Code: [Select] nothing found error, username was admin and pass is 62cc2d8b4bf2d8728120d052163a77df Resource id #5 Hi: I have the foll. code. The table "Reports" has multiple records for a given value of CID in the Field CID. I'd like to be able to select only 1 of them so that a list of customers appearing in the Reports table is available for selection in the dropdown alphabetically. The foll. code does it but it doesnt list the Customers alphabetically. And when I use Join, the query doesnt run. I get a blank list . The Field CID is common to both tables- Reports and Customers. Could someone help me with the Join ? Thanks. Swat Code: [Select] <?php $sqlco = "SELECT DISTINCT CID FROM `Reports` "; $resultco = mysql_query($sqlco) or die (mysql_error() ) ; if ($myrowco = mysql_fetch_array($resultco) ) { do { $cid = $myrowco["CID"]; $sqlrep = "SELECT * FROM `Customers` WHERE `CID` = '$cid' " ; $resultrep = mysql_query($sqlrep) or die (mysql_error() ) ; $myrowrep = mysql_fetch_array($resultrep); $company = $myrowrep["Company"]; printf("<option value=%d> %s , %s", $myrowco["CID"], $myrowrep["Company"], $myrowco["Mdate"]); } while ($myrowco = mysql_fetch_array($resultco)); } else { echo "No records found." ; } ?></select></a> What i tried was this : Code: [Select] <?php $sqlco = "SELECT DISTINCT CID FROM `Reports` r JOIN `Customers` c WHERE r.CID = c.CID ORDER BY c.Company asc "; $resultco = mysql_query($sqlco) or die (mysql_error() ); if ($myrowco = mysql_fetch_array($resultco) ) { do { printf("<option value=%d> %s ", $myrowco["CID"], $myrowco["Company"]); } while ($myrowco = mysql_fetch_array($resultco)); } else { echo "No records found." ; } ?> Hi everyone - this code is doing my head in! You might want to take a look at the page as it will help me explain it better (www.bradleystokejudoclub.co.uk/inttest.php Basically what I have is a database with the results from international competitions, and I am trying to build a kind of search for it. I have 3 drop down boxes, one with player names, one with competiton and one with year. The only one that works is the competiton. I can do competition + year, but not year by itself, and player doesnt work at all .... this is the relevant code: (before head) <?php include("includes/dbconnect120-gem.php"); include("includes/db_auth_bits.php"); include("includes/db_stp.php"); if($_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 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 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 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. 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 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> 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... |
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