I am trying to pass variables from delivery.php to delivery-insert.php using HTTP POST. Upon submitting, no variables are passed in.
Below is the form in delivery.php
<form action="delivery-insert.php" method="post">
<div class="form-group">
<label>Delivery No.</label>
<input type="text" class="form-control" id="deliverynoinput" placeholder="D0" required> <br>
<label>Price</label>
<input type="number" class="form-control" id="deliverypriceinput" placeholder="0" required> <br>
<label>Quantity</label>
<input type="number" class="form-control" id="deliveryquantityinput" placeholder="0" required>
</div>
<div class="form-group">
<label for="selectTruckInput">Select Delivery Truck</label>
<select class="form-control" id="selectTruckInput">
<?php
foreach ($trucks as $i => $value) {
print "<option>" . $value . "</option>";
}
?>
</select>
</div>
<div class="form-group">
<label for="selectCustomerInput">Select Delivery Customer</label>
<select class="form-control" id="selectCustomerInput">
<?php
foreach ($customers as $i => $value) {
print "<option>" . $value . "</option>";
}
?>
</select>
</div>
<div class="form-group">
<label for="selectDriverInput">Select Delivery Driver</label>
<select class="form-control" id="selectDriverInput">
<?php
foreach ($drivers as $i => $value) {
print "<option>" . $value . "</option>";
}
?>
</select>
</div>
<input type="submit"/>
</form>
And below is delivery-insert.php where I attempt to gather the POST variables.
<html>
<body>
<?php
$host = "localhost";
$user = "root";
$pw = "root";
$db = "project";
$con = new mysqli($host, $user, $pw, $db);
if (! $con) {
die("Connection failed: " . mysqli_connect_error());
}
$dno = $_POST['deliverynoinput'];
$dprice = intval($_POST['deliverypriceinput']);
$dquant = intval($_POST['deliveryquantityinput']);
$dtruck = $_POST['selectTruckInput'];
$dcust = $_POST['selectCustomerInput'];
$ddriver = $_POST['selectDriverInput'];
print '<h1>Here</h1>';
print '<h1>'.$dno.'</h1>';
print $_POST['selectTruckInput'];
$sql = "INSERT INTO delivery (deliveryno, price, truckid, custphoneno, driverid) VALUES ('$dno','$dprice','$dtruck','$dcust','$ddriver')";
if (mysqli_query($con, $sql)) {
print "added to delivery successfully";
} else {
print "Error: " . mysqli_error($con);
}
$sql = "INSERT INTO deliveryquantity (deliveryno, quantity) VALUES ('$dno', '$dquant')";
if (mysqli_query($con, $sql)) {
print "added to deliveryquantity successfully";
} else {
print "Error: " . mysqli_error($con);
}
$con->close();
?>
</body>
</html>
Does anyone know why all my variables are empty?
You will need the attribute name on your inputs. With this the browser will send the corresponding names as a key for $_POST. This even works for arrays!
e. g.
<!-- Single Value -->
<input name="someValue" />
<!-- Array -->
<input name="someArray[]" />
<input name="someArray[]" />
can be accessed like this:
// Single value
$_POST['someValue']
// Array
$_POST['someArray'][$numericIndex]
Related
I have a php file named "add_report" with a form inside it. All my inputs are running, i can input data into my database, but everytime I use the select-option. my database accepts it as null. Why is that?
This is my form "add_report.php"
<div class="wrapper">
<form action="add_report_backend.php" method="post">
<input type="hidden" name="id">
<label>Agency: </label> <input class="input1" type="text" name="agency" value="CAAP" required readonly><br>
<label>File Name: </label> <input class="input2" type="text" name="filename" placeholder="file.pdf/xlsx/xls/docx" required autofocus><br>
<label>File Type: </label> <select name="myselectbox">
<option name="myoption1" value="myoption1">pdf</option>
<option name="myoption2" value="myoption2">excel</option>
<option name="myoption3" value="myoption3">word</option>
</select><br>
<label>Date: </label> <input class="input4" type="Date" name="date" required><br>
<input class="submit-btn" type="submit" name="insert" value="Save">
</form>
</div>
And this another php file "add_report_backend.php"
<?php
if(isset($_POST['insert']))
{
try {
$pdoConnect = new PDO("mysql:host=localhost;dbname=annualdb","root","");
$pdoConnect->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $exc) {
echo $exc->getMessage();
exit();
}
$id = $_POST['id'];
$Agency = $_POST['agency'];
$FName = $_POST['filename'];
$FType = $_POST['filetype'];
$Date = $_POST['date'];
$pdoQuery = "INSERT INTO `company_report`(`agency`, `filename`, `filetype`, `date`) VALUES (:Agency,:FName,:FType,:Date)";
$pdoResult = $pdoConnect->prepare($pdoQuery);
$pdoExec = $pdoResult->execute(array(":Agency"=>$Agency,":FName"=>$FName,":FType"=>$FType, ":Date"=>$Date));
if($pdoExec)
{
$pdoQuery = 'SELECT * FROM company_report';
$pdoResult = $pdoConnect->prepare($pdoQuery);
$pdoResult->execute();
while ($row = $pdoResult->fetch()){
echo $row['id'] . " | " .$row['agency'] . " | " . $row['filename'] . " | " . $row['filetype'] . " | " . $row['date'];
}
header("Location: ../agencies/company.php");
exit;
} else {
echo 'Data Not Inserted';
}
}
$pdoConnect = null;
?>
The HTML name attribute and the $_POST name should be the same.
You need to change
$FType = $_POST['filetype'];
by
$FType = $_POST['myselectbox'];
Change $FType = $_POST['filetype']; to $FType = $_POST['myselectbox'];
I made a table in php and wanted to show the Id's in the dropdown select menu by making a separate file for php. So the code in main file is:
<?php include "functions.php";?>
<form action="login_update.php" method="post">
<div class="form-group">
<label for="username">username</label>
<input type="text" name="username" class="form-control">
</div>
<div class="form-group">
<label for="password">password</label>
<input type="password" name="password" class="form-control">
</div>
<div class="form-group">
<select name="id" id="">
<?php
showAllData();
echo "<br>"."askfkldfjl;adfafladfdf";
?>
</select>
</div>
<input class="btn btn-primary" type="submit" name="submit" value="update">
</form>
The code of functions.php is :
<?php
function showAllData(){
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
echo "We are connected. a=".$a."<br>";
}else{
die("Database connection failed");
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
echo("<br>"." <b><h6>We are successful</h6></b>");
}
else {
die("Query FAILED" . mysqli_error());
}
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
echo"<option value='$id'>$id</option>";
}
}
?>
The expected output was :
But the output is:
So the top two lines in the above screenshot are not printing.
These lines are shown in the INSPECT ELEMENT in chrome.
I forgot to mention the echo command:
echo "<br>"."askfkldfjl;adfafladfdf";
below show all data is also not working.
You made a mistake.
Actually you wrote a code in selectbox and you dont add option so thats why it is not show in html
So write a code like below code so its show in select box as option.
<div class="form-group">
<select name="id" id="">
<option> <?php
showAllData();
echo "<br>"."askfkldfjl;adfafladfdf";
?></option>
</select>
</div>
And If you want to show option from showAllData(); function, the you have return the html.
For this update your showAllData(); function with below code:
function showAllData(){
$options="";
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
echo "We are connected. a=".$a."<br>";
}else{
die("Database connection failed");
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
echo("<br>"." <b><h6>We are successful</h6></b>");
}
else {
die("Query FAILED" . mysqli_error());
}
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
$options.="<option value='$id'>$id</option>";
}
return $options;
}
Move the PHP function showAllData() before the HTML <select> element.
Because the <select> element awaits for an <option> element, but all another text will not be visible on page.
E.g.:
<div class="form-group">
<?php showAllData(); ?>
</div>
<?php
function showAllData(){
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
echo "We are connected. a=".$a."<br>";
}else{
die("Database connection failed");
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
echo("<br>"." <b><h6>We are successful</h6></b>");
}
else {
die("Query FAILED" . mysqli_error());
}
echo '<select name="id" id="">';
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
echo"<option value='$id'>$id</option>";
}
echo "</select>";
}
?>
Your code is mixed up.
You can use below code. Create an array which gives you values which you needs to show in select.
<?php
function showAllData(){
$idArr = array('msg'=>'','data'=>'','status'=>0);
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
$idArr['msg'] = "We are connected";
$idArr['status'] = 1;
}else{
$idArr['msg'] = "Database connection failed";
$idArr['status'] = 0;
}
if($idArr['status'] == 1){
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
$idArr['msg'] = "We are successful";
$idArr['status'] = 1;
}else {
$idArr['msg'] = "Query FAILED" . mysqli_error();
$idArr['status'] = 0;
}
if($idArr['status'] == 1){
while($row = mysqli_fetch_assoc($result)) {
$idArr['data'][] = $row["section_id"];
}
}
}
return $idArr;
}
$idArr = showAllData();
?>
<?php
if(!empty($idArr['data'])){
echo "We are connected<br>";
echo("<br>"." <b><h6>We are successful</h6></b>");
?>
<form action="login_update.php" method="post">
<div class="form-group">
<label for="username">username</label>
<input type="text" name="username" class="form-control">
</div>
<div class="form-group">
<label for="password">password</label>
<input type="password" name="password" class="form-control">
</div>
<div class="form-group">
<select name="id" id="">
<option value="0">--Select--</option>
<?php
foreach ($idArr['data'] as $key => $value) {
echo"<option value='$value'>$value</option>";
}
?>
</select>
</div>
<input class="btn btn-primary" type="submit" name="submit" value="update">
</form>
<?php }else{
echo $idArr['msg'];
}
?>
Why you are putting <br> inside the select tag? select tag only accept the options tag under it, so please remove br tag and all extra strings inside the select tag. and you should echo the message above the select tag if you want to show your users.
Thanks
I am working on a project and would like to give the user per-determined values when updating a record.
Here is my code so far.
<?php
// if there are any errors, display them
if ($error != '')
{
echo '<div style="padding:4px; border:1px solid red; color:red;">'.$error.'</div>';
}
?>
<form action="" method="post">
<input type="hidden" name="id" value="<?php echo $id; ?>"/>
<div>
<p><strong>ID:</strong> <?php echo $id; ?></p>
<strong>School Name:</strong> <input type="text" name="Name" value="<?php echo $Name; ?>"/><br/><br>
<strong>Status:</strong> <input type="text" name="Status" value="<?php echo $Status; ?>"/><br/><br>
<strong>Comments:</strong> <input type="text" name="Comments" value="<?php echo $Comments; ?>"/><br/><br>
<strong>Type:</strong> <input type="text" name="Type" value="<?php echo $Type; ?>"/><br/><br>
<input type="submit" name="submit" value="Submit">
</div>
</form>
</body>
</html>
<?php
}
// connect to the database
include('connect-db.php');
// check if the form has been submitted. If it has, process the form and save it to the database
if (isset($_POST['submit']))
{
// confirm that the 'id' value is a valid integer before getting the form data
if (is_numeric($_POST['id']))
{
// get form data, making sure it is valid
$id = $_POST['id'];
$Name = mysql_real_escape_string(htmlspecialchars($_POST['Name']));
$Status = mysql_real_escape_string(htmlspecialchars($_POST['Status']));
$Comments = mysql_real_escape_string(htmlspecialchars($_POST['Comments']));
$Type = mysql_real_escape_string(htmlspecialchars($_POST['Type']));
// check that firstname/lastname fields are both filled in
if ($Name == '' || $Type == '')
{
// generate error message
$error = 'ERROR: Please fill in all required fields!';
//error, display form
renderForm($id, $Name, $Status, $Comments, $Type, $error);
}
else
{
// save the data to the database
mysql_query("UPDATE Schools SET Name='$Name', Status='$Status', Comments='$Comments', Type='$Type' WHERE id='$id'")
or die(mysql_error());
// once saved, redirect back to the view page
header("Location: view.php");
}
}
else
{
// if the 'id' isn't valid, display an error
echo 'Error!';
}
}
else
// if the form hasn't been submitted, get the data from the db and display the form
{
// get the 'id' value from the URL (if it exists), making sure that it is valid (checing that it is numeric/larger than 0)
if (isset($_GET['id']) && is_numeric($_GET['id']) && $_GET['id'] > 0)
{
// query db
$id = $_GET['id'];
$result = mysql_query("SELECT * FROM Schools WHERE id=$id")
or die(mysql_error());
$row = mysql_fetch_array($result);
// check that the 'id' matches up with a row in the databse
if($row)
{
// get data from db
$Name = $row['Name'];
$Status = $row['Status'];
$Comments = $row['Comments'];
$Type = $row['Type'];
// show form
renderForm($id, $Name, $Status, $Comments, $Type, '');
}
else
// if no match, display result
{
echo "No results!";
}
}
else
// if the 'id' in the URL isn't valid, or if there is no 'id' value, display an error
{
echo 'Error!';
}
}
?>
I am wanting to replace the status text filed with a drop down list of options.
Replace your <input by <select :
<form action="" method="post">
<input type="hidden" name="id" value="<?php echo $id; ?>"/>
<div>
<p><strong>ID:</strong> <?php echo $id; ?></p>
<strong>School Name:</strong> <input type="text" name="Name" value="<?php echo $Name; ?>"/><br/><br>
<!-- <strong>Status:</strong> <input type="text" name="Status" value="<?php echo $Status; ?>"/><br/><br>-->
<strong>Status:</strong> <select name="Status">
<option value="1">Status 1</option>
<option value="2">Status 2</option>
</select>
<strong>Comments:</strong> <input type="text" name="Comments" value="<?php echo $Comments; ?>"/><br/><br>
<strong>Type:</strong> <input type="text" name="Type" value="<?php echo $Type; ?>"/><br/><br>
<input type="submit" name="submit" value="Submit">
</div>
</form>
If your statuses are in a table, fill the <select> with a query :
<form action="" method="post">
<input type="hidden" name="id" value="<?php echo $id; ?>"/>
<div>
<p><strong>ID:</strong> <?php echo $id; ?></p>
<strong>School Name:</strong> <input type="text" name="Name" value="<?php echo $Name; ?>"/><br/><br>
<!-- <strong>Status:</strong> <input type="text" name="Status" value="<?php echo $Status; ?>"/><br/><br>-->
<strong>Status:</strong> <select name="Status">
<?php
$result = mysql_query("SELECT * FROM tbl_status",$cnx);
while ( $row = mysql_fetch_array($result) )
echo "<option value='" . $row["id"] . "'>" . $row["text"] . "</option>";
?>
</select>
<strong>Comments:</strong> <input type="text" name="Comments" value="<?php echo $Comments; ?>"/><br/><br>
<strong>Type:</strong> <input type="text" name="Type" value="<?php echo $Type; ?>"/><br/><br>
<input type="submit" name="submit" value="Submit">
</div>
</form>
You could use the html <datalist> or the <select> tag.
I hope I could help.
First of all you need to switch from mysql_* to mysqli_* as it going to get removed in php 7.0 I'm using this function i created and it might help you
here is the php code
function GetOptions($request)
{
global $con;
$sql = "SELECT * FROM data GROUP BY $request ORDER BY $request";
$sql_result = mysqli_query($con, $sql) or die('request "Could not execute SQL query" ' . $sql);
while ($row = mysqli_fetch_assoc($sql_result)) {
echo "<option value='" . $row["$request"] . "'" . ($row["$request"] == $_REQUEST["$request"] ? " selected" : "") . ">" . $row["$request"] . "$x</option>";
}
}
and the html code goes like
<label>genre</label>
<select name="genre">
<option value="all">all</option>
<?php
GetOptions("genre");
?>
</select>
I can not understand why the application delete the las upload.I am new in php and I hope to help me. Thank you.
Code: HTML
<div class="row">
<div class="col-md-6 col-centered">
<div class="newboxes" id="newboxes3">
<form class="form" method="POST">
<input type="text" id="nmPic" name="nmPic" placeholder="име на снимката" onfocus="this.placeholder = ''" onblur="this.placeholder = 'име на снимката'"></br>
<input type="text" id="price" class="priceFrom" name="priceFrom" placeholder="цена от" onfocus="this.placeholder = ''" onblur="this.placeholder = 'цена от'"></br>
<input type="text" id="price" class="priceTo" name="priceTo" placeholder="цена до" onfocus="this.placeholder = ''" onblur="this.placeholder = 'цена до'"></br>
<select name="picCat" id="picCat">
<option value="" selected disabled>Изберете категория</option>
<option value="Детски">Детски</option>
<option value="Сватби">Сватби</option>
<option value="Рожден ден">Рожден ден</option>
<option value="18+">18+</option>
<option value="Други">Други</option>
</select></br>
<input type="text" id="numPic" name="numPic" placeholder="номер на снимката" onfocus="this.placeholder = ''" onblur="this.placeholder = 'номер на снимката'"></br>
<input type="submit" name="showFilter" value="покажи" />
</form>
</div>
</div>
</div>
Code: php
<div class="row">
<div class="col-md-12 col-centered pic">
<form class='form' method='POST'>
<?php
if (isset($_POST["showFilter"]))
{
$picName = $_POST['nmPic'];
$priceFrom = $_POST['priceFrom'];
$priceTo = $_POST['priceTo'];
$picCat = isset($_POST['picCat']) ? $_POST['picCat'] : '';
$numPic = $_POST['numPic'];
$filter = " SELECT * FROM images WHERE status = '1'";
if ($numPic && !empty($numPic)) {
$filter .= " AND id='$numPic'";
}
if ($picName && !empty($picName)) {
$filter .= " AND img_content='$picName'";
}
if ($picCat && !empty($picCat)) {
$filter .= " AND category='$picCat'";
}
if ($priceTo && !empty($priceTo)) {
$filter .= " AND price < '$priceTo'+1";
}
if ($priceFrom && !empty($priceFrom)) {
$filter .= " AND price > '$priceFrom'";
}
$resFilter = $connect->query($filter);
if ($resFilter->num_rows > 0) {
while($row = mysqli_fetch_array($resFilter))
{
echo "<div class='col-md-3 picture'>
<img class='child-img' src='".$row["picture"]." '/></br>
<div class='number'>
<span class='id'>№ ".$row['id']."</br>
име: ".$row['img_content']."</br> категория: ".$row['category']."</br>
цена: ".$row['price']."лв.</br>
дата: ".$row['time']."ч.</br>
</span>
<input type='hidden' name='del' value=" .$row['id'].">
<input class='btn btn-danger' name='delete' type='submit' value='истрии'/>
</div>
</div>";
}
}
}
?>
</form>
</div>
</div>
<form method="POST">
<?php
if (isset($_POST['delete']))
{
$sql = "SELECT * FROM images WHERE status = '1'";
$res = $connect->query($sql);
while($row = mysqli_fetch_array($res))
{
$id = $_POST['del'];
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'paspartu';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$sql = 'DELETE FROM images
WHERE id='.$id;
mysql_select_db('login');
mysql_query($sql);
mysql_close($conn);
}
}
?>
</form>
If pressed the delete button <input class="btn btn-danger" name="delete" type="submit" value="истрии"> , this delete the last upload image???
And can you tell me how to delete the image from upload folder "uploads/"? Thank you verry much.
It's because you don't have an input in your form that's named "showFilter". You need to either rewrite or remove:
if (isset($_POST["showFilter"]))
Try modified query delete
$sql = "DELETE FROM `your_database`.`images` WHERE `your_database`.`id` = $del_id";
As I can see your submit button is in separate form tag. That is why the button is responsible only for the first parent form and if it is pressed the page is only reloaded. It should be placed in the general form you use, so that it will be related with the general form action.
You need to take your submit button inside the form. You have used two different forms. It is the problem.
Try to write your code as below:-
<?php
// Take div and form tag outside the loop
echo "<div class='col-md-3 picture'>
<form class='form' method='POST'>";
// Loop start
while($row = mysqli_fetch_array($resFilter))
{
echo "<img class='child-img' src='".$row["picture"]." '/></br>
<div class='number'>
<span class='id'>№ ".$row['id']."
<input class='check' name='checkbox[]' type='checkbox' value='". $row['id']."'></br>
име: ".$row['img_content']."</br> категория: ".$row['category']."</br>
цена: ".$row['price']."лв.</br>
дата: ".$row['time']."ч.
</br></span>
<br></div>";
}
// Loop End
?>
<!-- submit button -->
<input class="btn btn-danger" name="delete" type="submit" value="истрии маркираните">
<?php
// end div and form tag outside the loop
echo "</form>
</div>";
if (isset($_POST['delete']) && isset($_POST['checkbox'])) {
foreach($_POST['checkbox'] as $del_id){
$del_id = (int)$del_id;
$sql = "DELETE FROM images WHERE id = $del_id";
mysql_query($sql);
}
header('Location: admin.php');
}
Hope it will help you :)
Using AJAX How can I generate selections for a dropdown menu based on records available in a database?.
How can then use these selections to prefill a form with record/row data from a database when selected?
Heres a mock up I created of what I'm trying to do:
http://oi58.tinypic.com/2urb2ae.jpg
PHP FILE: contact_form.php
-----------------------------------------------------------
<?php
define('DB_NAME', 'xxx');
define('DB_USER', 'xxx');
define('DB_PASSWORD', 'xxx');
define('DB_HOST', 'xxx');
$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$connection){
die('Database connection failed: ' . mysqli_connect_error());
}
$db_selected = mysqli_select_db($connection, DB_NAME);
if(!$db_selected){
die('Can\'t use ' .DB_NAME . ' : ' . mysqli_connect_error());
}
echo 'Connected successfully';
if (isset($_POST['itemname'])){
$itm = $_POST['itemname'];
}
else {
$itm = '';
}
if($_POST['mile']){
$mi = $_POST['mile'];
}else{
echo "Miles not received";
exit;
}
if($_POST['email']){
$email = $_POST['email'];
}else{
echo "email not received";
exit;
}
$sql = "INSERT INTO seguin_orders (itemname, mile, email)
VALUES ('$itm', '$mi', '$email')";
if (!mysqli_query($connection, $sql)){
die('Error: ' . mysqli_connect_error($connection));
}
CONACT FORM: formz.php
------------------------------------------------------------------------------
<html>
<header>
</header>
<body>
<form action="/demoform/contact_form.php" class="well" id="contactForm" method="post" name="sendMsg" novalidate="">
<big>LOAD PAST ORDERS:</big>
<select id="extrafield1" name="extrafield1">
<option value="">Please select...</option>
<?php
$email = $_POST['email'];
$query="select * from tablename WHERE email={$_POST['email']}";
$res=mysqli_query($connection,$query);
while($row = mysqli_fetch_assoc($res))
{
?>
<option value="<?php echo $row['fieldname']; ?>"><?php echo $row['fieldname']; ?></option>
<?php
}
?>
</select>
</br>
<input type="text" required id="mile" name="mile" placeholder="Miles"/>
</br>
<input id="email" name="email" placeholder="Email" required="" type="text" value="demo#gmail.com" readonly="readonly"/>
</br>
<input id="name" name="itemname" placeholder="ITEM NAME 1" required="" type="text" />
</br>
<input type="reset" value="Reset" />
<button type="submit" value="Submit">Submit</button>
</form>
</body>
</html>
Using an exemple, let's assume you want to fill the "name" select based on the option selected at the "gender" select:
<select name="gender" id="gender">
<option value="m">Male</select>
<option value="f">Female</select>
</select>
When nothing is selected yet, the "name" select is empty:
<select name="name" id="name">
<option value="NULL">Please select a gender first</option>
</select>
So, what you gotta do is: when the gender select got some selection, you populate the name select with values based on the gender select option.
$(document).ready(function() {
$('select#gender').change(function(){
$('select#name').load('LOAD_NAMES_BASED_ON_GENDER.php?gender='+$(this).val());
});
});
And your PHP file responsible for loading the names based on gender should look like:
$gender = $_GET['gender'];
$list = // the way you retrieve your list of names from your DB
And then you loop this $list into an list of options, such like:
foreach($list as $key=>$value)
echo '<option value="$key">$value</option>';
This simple.
PS: the load() function is kind of an alias for the $.ajax request, given that the only purpose here is to retrieve data.