I have a MySQL database and am trying make a conditional dropdown menu ("Subcategory") show values based on the values in the first dropdown, ("Category").
This is a reference data table, so the parent ID of the subcategory should match the ref_data_id of the category.
The conditional list relies on the value of the first dropdown box, and I have tried $_POST and $_GET to try to get the value from the first object to use in my MySQLi query but neither seems to work.
Can anyone help?
<?php
// connect to the database
include("connectdb.php");
?>
<html>
<!--First Dropdown Menu - CATEGORIES-->
<div class="label">Select Category:</div>
<select name ="Category_HTML">
<option value = "">---Select---</option>
<?php
$stmt = "SELECT * FROM `ref_data` WHERE Parent_ID IS NULL ;";
$result = mysqli_query($mysqli, $stmt);
while ( $row=mysqli_fetch_array($result)) {
$description = $row['Description'];
$refdataID = $row['Ref_Data_ID'];
echo "<option value='$refdataID'> $description </option>";
}
?>
</select>
<!--Second Dropdown Menu - Subcategory-->
<div class="label">Select Subategory:</div>
<select name="subcategory_HTML">
<option value = "">---Select---</option>
<?php
$idvalue = $_POST['Category_HTML'];
$stmt = "SELECT * FROM `ref_data` WHERE Parent_ID = $idvalue;";
$result = mysqli_query($mysqli, $stmt);
while ( $row=mysqli_fetch_array($result)) {
$description = $row['Description'];
$refdataID = $row['Ref_Data_ID'];
echo "<option value='$refdataID'> $description </option>";
}
?>
</select>
</html>
try to make an AJAX call when a value is select in the category list i.e
<select name ="Category_HTML" onchange="AJAX_CALL()">
and populate the result in the
<select name="subcategory_HTML">
Related
I am making an insert mobile data page where for brands i had made dropdown menu the dropdown menu is working perfectly but when I insert mobile data brand id is inserted in database but brand title is not inserted why?
// code for dropdown menu
<tr>
<td style="color:#fff"><b>Mobile Brand</b></td>
<td>
<select name="Mbrand">
<option>Select Brand</option>
<?php
$get_brands = "select * from brands";
$run_brands = mysqli_query($con, $get_brands);
while ($row_brands=mysqli_fetch_array($run_brands))
{
$brand_id = $row_brands['brand_id'];
$brand_title = $row_brands['brand_title'];
echo "<option value='$brand_id'>'$brand_title'</option>";
}
?>
</select>
</td>
</tr>
----------------------------
// code for defining data
<?php
if(isset($_POST['insert_mobile'])){
//text data variables
$mobile_title=$_POST['Mname'];
$mobile_price=$_POST['Mprice'];
$mobile_brand=$_POST['Mbrand'];
$mobile_desc=$_POST['mdesc'];
$mobile_keywords = $_POST['key'];
-----------------------------
//code for inserting data
$insert_mobile = "insert into mobile_phone (mobile_name,brand_id,date,price,img1,img2,img3,mobile_desc,product_keywords) values ('$mobile_title','$mobile_brand',NOW(),'$mobile_price','$mobile_img1','$mobile_img2','$mobile_img3','$mobile_desc','$mobile_keywords')";
$run_mobile= mysqli_query($con,$insert_mobile);
if($run_mobile)
{
echo "<script>alert('mobile added successfully')</script>";
exit();
}
Please try using $brand_title instead of $brand_id in the option value.
echo "<option value='$brand_id'>'$brand_title'</option>";
will become
$brand = $brand_id.':'.$brand_title;
echo "<option value='$brand'>'$brand_title'</option>";
Edit
Option value is actually the one that gets passed in $_POST[] values and that can be stored in database as you need.
Edit
Replace the below code
$mobile_brand=$_POST['Mbrand'];
with
$brand_id = explode(':', $_POST['Mbrand'])[0];
$brand_title = explode(':', $_POST['Mbrand'])[1];
And please change the Insert query accordingly, so that it can accomodate or store both brand_id and brand_title in respective columns. Please check the datatypes of those columns before you run the script.
I have an order form where the product is selected from a database with the method "select" and "option". I show screenshot and code
Code:
<div class="form-group">
<select class="form-control" name="productName[]" id="productName<?php echo $x; ?>" onchange="getProductData(<?php echo $x; ?>)" >
<option value="">-- Selecciona --</option>
<?php
$productSql = "SELECT * FROM product WHERE active = 1 AND status = 1 AND
quantity != 0";
$productData = $connect->query($productSql);
while($row = $productData->fetch_array()) {
$selected = "";
if($row['product_id'] == $orderItemData['product_id']) {
$selected = "selected";
} else {
$selected = "";
}
echo "<option value='".$row['product_id']."'
id='changeProduct".$row['product_id']."' ".$selected."
>".$row['product_name']."</option>";
} // /while
?>
</select>
</div>
I would like to transform this way of selecting a product (droping-down list using "select") for an order form in a field that can be input the first letters of the name of the product or reference and I can see in the drop-down list those references that match instead of all references in table. I know that "input" and "post" should be used and in the query to the database something like:
// Conexión a la base de datos y seleccion de registros
$con=mysql_connect("localhost","xxxxx_xxxx","xxxxxxx");
$sql = "SELECT * FROM product WHERE referencia, product_name, position_store
like ‘%$buscar%’ ORDER BY id DESC";
mysql_select_db("xxxxxxx_stock", $con);
$result = mysql_query($sql, $con);
The idea is to do what I show in the screenshot:
Once the matches appear, you can select the one that interests you. I'm not sure how to structure the changes to convert the initial code into what I'm looking for.
I appreciate any help.
I have this table
Device Name Price
iPhone 6 300
S8 600
What I am trying to do is have to drop list one that has all the devices Name and the second has an option "Price" but I want to set its value based on the selection of the device name. However, in the drop list it should just display "Price"
I have this php code for the drop lists
$query = "SELECT * FROM `devicehotsheet`";
$options = "";
$result1 = mysql_query($query, $dbhandle);
while ($row1= mysql_fetch_array($result1)){
$options=$options."<option>$row1[0]</option>";
}
For HTML
<select class="selectDevice" id="selectDevice" onChange="calculateTotal()">
<?php echo $options; ?>
</select>
<select class="myport" id="myport" onChange="calculateTotal()">
<option class="price" value=" ">Price</option>
</select>
I have been trying to do this but so far no luck. I would appreciate your help. Thanks
$query = "SELECT * FROM `devicehotsheet`";
$options = "";
$result1 = mysql_query($query, $dbhandle);
while ($row1= mysql_fetch_array($result1)){
$options=$options."<option value='$row1[0]#$row1[1]'>$row1[0]</option>";
}
and the javascript code
function calculateTotal(){
var selDev = document.getElementById('selectDevice').value;
var price = selDev.split('#')[1];
document.getElementById('myport').innerHTML='<option>'+price+'</option>'
}
As I've tried to describe in the title I have an issue in selecting rows from a MYSQL database depending on the id of another drop down list on the same page. I have only been using mysql and php for 2 months or so now and need help.
I have a table of categories with the below headers.
|id | name | parent_id|
There are parent categories, with a parent_id of 0. And Sub categories with the id of the parent as their parent_id, to a maximum depth of 1 child category. For example:
Software Development is a parent category with id = 18 and parent_id = 0. PHP Developer is a subcategory which has id = 30 and parent_id = 18.
I have a drop down list where I can select the category I work in as follows:
$p_query = "SELECT * FROM categories WHERE parent_id = 0 ORDER by id ASC";
$p_result = mysqli_query($con, $p_query) or die(mysqli_error($con));
$categories ='';
while($p_row = mysqli_fetch_assoc($p_result))
{
$categories .='<option class="option" value="p::'.$p_row['id'].'">' .$p_row['category_name'].'</option>';
}
<select name="categories[]" class="categories form-control" id="categories" style="width:100%" multiple>
<?php echo $categories;?>
</select>
This is working, no problem. However, when I try to get a second drop down list to show the possible categories whom have their parent_id as the id of any selected parent category I retrieve a drop down list with 'No Search Results found'. The code below is what I am using :
$subcategories ='';
while($p_row = mysqli_fetch_assoc($p_result))
{
$c_query = "SELECT * FROM categories WHERE parent_id = ".$categories['id']." ORDER by id ASC";
$c_result = mysqli_query($con, $c_query) or die(mysqli_error($con));
while($c_row = mysqli_fetch_assoc($c_result))
{
$subcategories .='<option class="option" value="c::'.$c_row['id'].'">' .$c_row['category_name'].'</option>';
}
}
<select name="subcategories[]" class="categories form-control" id="subcategories" style="width:100%" multiple>
<?php echo $subcategories ?>
</select>
Is there something that I am missing? As a relative beginner to both PHP and MYSQL, I would be very appreciative of any help or advice.
There is nothing wrong with your second query to retrieve sub-categories based on their parent_id so you're good there. You can test easily like so:
SELECT * FROM categories WHERE parent_id = 1 ORDER by id ASC
How are you providing the value to $categories['id'] as there is nothing in your code creating this array with an 'id' index? Further you already are using a variable called $categories as a string so you shouldn't re-use that variable name without good reason.
Since it appears you want to populate the second multiple select box with subcategories based on the selection of the first you will need to use some javascript+AJAX to submit the second query and write the results in the second selector element.
Using some jquery should help a bit. Try these examples and you'll get the idea.
myselect.php which contains the interface for selecting the category:
<?php
$con = mysqli_connect("host", "username", "password", "database");
$p_query = "SELECT * FROM categories WHERE parent_id = 0 ORDER by id ASC";
$p_result = mysqli_query($con, $p_query) or die(mysqli_error($con));
$categories ='';
while($p_row = mysqli_fetch_assoc($p_result))
{
$categories .='<option class="option" value="p::'.$p_row['id'].'">' .$p_row['category_name'].'</option>';
}
?>
<html>
<head>
<script
src="https://code.jquery.com/jquery-3.1.1.min.js"
integrity="sha256-hVVnYaiADRTO2PzUGmuLJr8BLUSjGIZsDYGmIJLv2b8="
crossorigin="anonymous">
</script>
</head>
<body>
Shift or Ctrl + Click to pick more than one<br />
<form id="categoryform" method="POST">
<select name="categories[]" class="categories form-control" id="categories" style="width:100%" multiple>
<?php echo $categories;?>
</select>
</form>
Here's what it contains<br />
<form method="POST">
<select name="subcategories[]" class="categories form-control" id="subcategories" style="width:100%" multiple DISABLED>
</select>
</form>
<script>
$(document).ready(function() {
$('#categories').click(function(){
$('#subcategories').children().remove().end();
var data = $('#categoryform').serialize();
$.post("mysubselect.php", data).done(function(data){
var response = JSON.parse(data);
for (var k in response){
$('#subcategories').append('<option class="option" value="c::' + response[k]['id'] + '">' + response[k]['category_name'] + '</option>');
}
});
});
})
</script>
</body>
</html>
And here's an example of the script returning data from your AJAX request.
mysubselect.php:
<?php
$con = mysqli_connect("host", "username", "password", "database") or die(mysqli_error());
$result = array();
foreach ($_POST["categories"] as $k => $v) {
$category_token = explode('::', $v);
$category_id = mysqli_real_escape_string($con, $category_token[1]);
$query = mysqli_query($con, "SELECT * FROM categories WHERE parent_id = " . $category_id . " ORDER BY id ASC") or die(mysqli_error());
while($r = mysqli_fetch_assoc($query)){
$result[] = $r;
}
}
print json_encode($result);
I'm having a input form that asks for a location. The locations are stored in a mysql db and have an id (colomn: id and column: location).
I have a drop down menu that is generated from those records in the db:
<select name="location">
<?php
$query="SELECT location FROM locations";
$result=mysql_query($query) or die;
while ($row=mysql_fetch_array($result)) {
$location=$row['location'];
echo "<option>$location</option>";
?>
</select>
This all works. When the form is sumbitted, I obviously get a POST[location] for example "Belgium".
Let's say Belgium is the location in my db and has id 5, how can I return the ID as the POST variable from the dropdown box, instead of the location. Ofcourse I want the dropdown to show the locations, and not the ID's.
Thanks in advance!
Each option can take a value and show another string so use value="my_value" for each option inside the select tag
<select name="location">
<?php
$query="SELECT id, location FROM locations";
$result=mysql_query($query);
while ($row=mysql_fetch_array($result)) {
echo "<option value=\"" . $row['id'] . "\">" . $row['location'] . "</option>";
}
?>
</select>
now your POST['location'] will contain the db id for selected location
if you change the SQl query to include the ID of the location,
you can assign that value to the dropdown selected value.
<select name="location">
<?php
$query="SELECT id, location FROM locations";
$result=mysql_query($query) or die;
while ($row=mysql_fetch_array($result)) {
$location=$row['location'];
$id = $row['id'];
echo "<option value='".$row['id']."'>".$location."</option>";
?>
</select>
select ID from locations;
$ID=$row['ID'];
Replace ID by the ID-name of your column ofcourse.
If you say select * from locations you can do something like this:
$row['anyCOLUMNNAME']
You can choose any column you like and use that information from that particular row.
<select name="location">
<?php
// select columns you need, separate by , (comma)
$query = "SELECT `column1`, `column2` FROM `locations`;";
$result = mysql_query($query) or die;
while ($row = mysql_fetch_array($result)) {
// selected columns become accessible in $row array
// value attribute needs to be escaped here
echo '<option value="', htmlentities($row['column1']),'">',
htmlentities($row['column2']), '</option>'; // escape label too
// <option> does not accept HTML in label so it should be escaped
} // done!
?>
</select>
^ this (read comments for explanations)