Auto populate text field depending on dropdown using same SQL table - php

I'm trying to populate a text field and a text area depending on the choice of the second dropdown list using an SQL table. Somehow I got it all wrong. Here's my code right now:
AJAX/HTML
<script>
function getPackage(val) {
$.ajax({
type: "POST",
url: "get_package.php",
data:'serviceid='+val,
success: function(data){
$("#package-list").html(data);
}
});
}
function getPackageDesc(val) {
$.ajax({
type: "POST",
url: "get_packagedesc.php",
data:'packageid='+val,
success: function(data){
$("#price").html(data);
}
});
}
function selectService(val) {
$("#search-box").val(val);
$("#suggesstion-box").hide();
}
</script>
I feel like there is already something wrong with function getPackageDesc. I don't know if it's in this function or in the php file.
MAIN HTML
<div class="frmDronpDown">
<div class="row">
<label>Service:</label>
<br/>
<select name="service" id="service-list" class="demoInputBox" onChange="getPackage(this.value);">
<option value="">Select Service</option>
<?php
foreach($results as $service) {
?>
<option value="<?php echo $service[" serviceid "]; ?>">
<?php echo $service["servicename"]; ?>
</option>
<?php
}
?>
</select>
</div>
<div class="row">
<label>Package:</label>
<br/>
<select name="package" id="package-list" class="demoInputBox" onChange="getPackageDesc(this.value);">
<option value="">Select Package</option>
</select>
</div>
<div class="row">
<label>Package Information:</label>
<br/>
<br/> Price:
<br>
<input type="text" name="price" id="price">
<br>
<br> Package Description:
<br>
<textarea name="packagedesc" form="usrform" id="packagedesc"></textarea>
</div>
</div>
get_package.php
<?php
require_once("dbcontroller.php");
$db_handle = new DBController();
if(!empty($_POST["serviceid"])) {
$query ="SELECT * FROM tbl_packages WHERE serviceid = '".$_POST["serviceid"]."'";
$results = $db_handle->runQuery($query);
?> < option value = "" > Select Package < /option>
<?php
foreach($results as $package) {
?> < option value = "<?php echo $package["
packageid "]; ?>" > <?php echo $package["packagename"]; ?> < /option>
<?php
}
}
?>
This part of the code works. I have populated the dropdowns using two SQL tables, tbl_services and tbl_packages. The next php file is where I'm having trouble.
get_packagedesc.php
<?php
require_once("dbcontroller.php");
$db_handle = new DBController();
if(!empty($_POST["packageid"])) {
$query ="SELECT * FROM tbl_packages WHERE packageid = '".$_POST["packageid"]."'";
$result = mysqli_query($conn, $query);
?>
<?php
while($row = mysqli_fetch_array($result)){
?>
<?php echo $price['price'] ; ?></br>
<?php echo $packagedesc['packagedesc']; ?> </br>
<?php
}
}
?>
This code is where I'm having trouble at. I don't know if it is okay to get data from the same SQL table (tbl_packages). Or maybe I messed up with the $result.
What is wrong with my code and what could be the solution?
Thanks for any help.

It looks like you're not grabbing the information from the row itself but from undeclared variables. Try this:
<?php echo $row['price'] ; ?></br>
<?php echo $row['packagedesc']; ?> </br>

Related

Populating Second Dropdown Based on first dropdown from the database table without using javascript

I am trying to create a form for the admin of an e-commerce site and the admin should be able to create a category and add new products.
I have two tables in my database from where I want to populate the dropdown list. I want the second dropdown to populate as I select the first drop-down value but I have to do it without the submit button.
This is the code for the form with two drop-downs:-
<form method="post" action="add_category.php">
<h4>Choose the root level:</h4>
<select name="rootLevel">
<?php
$conn = mysqli_connect("localhost","root","","store")
or die("Error in Connection on add_category");
$query = "SELECT * FROM root_category";
$result = mysqli_query($conn,$query) or die("Query failed add_category");
$id=1;
//echo $id;
//echo "Hello";
while($row = mysqli_fetch_assoc($result)){
global $id;
//echo "<h1>$id</h1>";
$id = $row['id'];
echo "<option name='rootLevel' value=$id>".$row['Name']."</option>";
//echo "<option>$id</option>";
}
?>
</select>
<br><br>
<h4>Choose the Level 1 Category:</h4>
<select name="level1">
<?php
global $id;
//echo "<option>".$_POST['rootLevel']."</option>";
$query_level1 = "Select * from level1_category Where P_id = $id";
$result1 = mysqli_query($conn,$query_level1) or die("Query failed level 1");
while($row = mysqli_fetch_assoc($result1)){
$id1 = $row['P_id'];
echo "<option name='level1' value=$id1>".$row['Name']."</option>";
}
?>
</select>
I have successfully populated the first drop-down and now I want to fetch the $id in 'resultValue' without the submit button.
You cant do this only with PHP. You have to use jquery OR Ajax to do this.
Please check this example page . This may help you
https://www.tutorialrepublic.com/faq/populate-state-dropdown-based-on-selection-in-country-dropdown-using-jquery.php
OR
https://www.codexworld.com/dynamic-dependent-select-box-using-jquery-ajax-php/
<head>
<script src="https://code.jquery.com/jquery-1.12.4.min.js"></script>
<script>
$(document).ready(function(){
$("select.country").change(function(){
var selectedCountry = $(".country option:selected").val();
$.ajax({
type: "POST",
url: "states.php",
data: { country : selectedCountry }
}).done(function(data){
$("#states").html(data);
});
});
});
</script>
</head>
<body>
<div class="form-group">
<label for="country" class="input__label">Country</label>
<select id="country" onchange="states(this.value);" name="country" class="country form-control login_text_field_bg input-style">
<option selected>Choose...</option>
<?php
$sql= $cnn->prepare("SELECT key_iso_code FROM country");
$sql->execute();
while($i = $sql-> fetch($cnn::FETCH_ASSOC)){
extract($i);
?>
<option><?php echo $key_iso_code ?></option>
<?php
}
?>
</select>
</div>
<div class="form-group col-md-4">
<label for="inputState" class="input__label">State/Province</label>
<select id="states" name="state" class="form-control input-style">
<option selected>Choose...</option>
</select>
</div>
</body>
<?php
include("PDOConnection.php");
if(isset($_POST["country"])){
// Capture selected country
$country = $_POST["country"];
// Display city dropdown based on country name
if($country !== 'Shoose...'){
$sql= $cnn->prepare("SELECT state.key_name FROM country INNER JOIN state ON country.key_id = state.key_country_id WHERE country.key_iso_code like '$country'");
$sql->execute();
while($i = $sql-> fetch($cnn::FETCH_ASSOC)){
extract($i);
echo "<option>". $key_name . "</option>";
}
}
}
?>

I This i called the function on the Onsubmit event of form but when my jquery function return true still its not going to action page

In this programme i want to submit my form data if state name is already not present in database.if it is present in database then it gives mi alert that name already present and form not submited. but in the following programme if name is already present then then it gives alert message and form also not submitted. but if state name is not present then still it is not submitted my values(it is not calling form action="add_state_process.php"). i think i am wrong in returning true and false in javascript function.plz tell mi where i am wrong
<form name="addcategoryfrm" id="addcategoryfrm" onsubmit="return check_state();" action="add_state_process.php" method="POST">
<div class="formrow" style="margin-top:0px;height:80px;width:100%; padding-left:30%;">
<div class="txttitle" >Country Name</div>
<div class="txtinputouter1" style="padding-left:8px;">
<select name="country" id="country" class="required txtinput" value="">
<option value="">--- Select Country --- </option>
<?php
include("../config/database.php");
$query="SELECT * FROM `country`";
$result = mysql_query($query) or die(mysql_error());
echo $query;
//$result2 = mysql_fetch_array($result);
$row1=mysql_num_rows($result);
echo $row1;
while($numrow=mysql_fetch_array($result))
{ ?>
<option value=" <?php echo $numrow['id']; ?> ">
<?php echo $numrow['name']; ?>
</option>
<?php } ?>
</select>
</div>
</div>
<div class="formrow" style="margin-top:0px;height:80px;width:100%; padding-left:30%;">
<div class="txttitle">State Name </div>
<div class="txtinputouter1" style="padding-left:1%;">
<input type="text" class="required txtinput" name="state" id="state" class="categoryname" onblur=";"/>
</div>
</div>
<div class="txtinputouter1" style="padding-left:43%;padding-top:10px;">
<input type="submit" class="subbutton" value="Submit"/>
</div>
<div id="demo"></div>
</form>
////function call
<script>
function check_state()
{
valid=false;
var txtarea12 = document.getElementById("country").value;
var name = document.getElementById("state").value;
//var valid=true;
//alert(name);
//alert(txtarea12);
//$name = name;
$.ajax({ type: "GET",
url: 'ajax-state-model.php',
data: { name: name,
countryid:txtarea12
},
success: function(result)
{
if (result == 1)
{
alert("state name already present");
var txtarea = document.getElementById("state");
txtarea.value="";
txtarea.focus();
valid=false;
}
else
{
valid=true;
}
},
error: function(err)
{ alert(err); }
});
return valid;
}
</script>
///function ajax call
<?php
session_start();
if(isset($_SESSION['username1']))
{
include("../config/database.php");
$name=$_GET['name'];
$countryid=$_GET['countryid'];
$query="select * from state where country_name=$countryid && state_name='$name'";
$result = mysql_query($query) or die(mysql_error());
//echo $query;
$rows=mysql_num_rows($result);
if($rows>0)
{
echo 1;
}
else
{
echo 0;
}
}
?>

PHP/MySQL dropdowns

I have a main MySQL table called restaurants which holds all details for each restaurant - RestaurantID, RestaurantName, Address, Town, County etc.
I am creating a Review and ratings form in PHP to allow my users to review each of the restaurants. I have created 2 dropdown menus one which the user will select the County and the second I hope to populate with RestaurantName of those in that County.
Can anyone spot where I am going wrong? My first dropdown populates without any issues, although the second dropdown is blank?
I've added all my code so far for this, anything else that you think would be useful let me know.
leave.php
<?php
include_once "settings.php"
?>
<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<body>
<div class="County">
<label>County</label>
<select name = "County" onchange="getId(this.value);">
<option value ="">Select County </option>
<?php
$query = "SELECT DISTINCT County FROM restaurants";
$results = mysqli_query($con, $query);
foreach ($results as $restaurants) {
?>
<option value ="<?php echo $restaurants['County']; ?>"><?php echo $restaurants["County"]; ?></option>
<?php
}
?>
</select>
</div>
<div class="RestName">
<label>Restaurant</label>
<select name = "Restaurant" id="RestList">
<option value ="">Select Restaurant </option>
<option value="" >
<?php
include_once "settings.php";
if (!empty($_POST["RestaurantID"])) {
$RestaurantID = $_POST["RestaurantID"];
$query = "SELECT * FROM restaurants WHERE RestaurantID = $RestaurantID";
$results = mysqli_query($con, $query);
foreach ($results as $County) {
?>
<option value="<?php echo $restaurants["RestaurantID"]; ?>"><?php echo $restaurants["RestaurantName"]; ?>"></option>
<?php
}
}
?>
</select>
</div>
<script>
function getId(val){
$.ajax({
type: "POST",
url:"getdata.php",
data: "RestaurantID="+val,
success: function(data){
$("#RestList").html(data);
}
});
}
</script>
</body>
</html>
getdata.php
<?php
include_once "settings.php";
if (!empty($_POST["RestaurantID"])) {
$RestaurantID = $_POST["RestaurantID"];
$query = "SELECT * FROM restaurants WHERE RestaurantID = $RestaurantID";
$results =
($con, $query);
foreach ($results as $restaurants) {
?>
<option value="<?php echo $restaurants['RestaurantID']; ?>"><?php echo $restaurants["RestaurantName"]; ?>"></option>
<?php
}
}
?>
In this line:
<select name="Restaurant" id="RestList" </select>
You don't actually close the "<select>" tag...

Codeigniter : Jquery fill the textbox based on combobox

I am Nubie in JQuery especially in ajax..
I have read articles about dropdown combobox (read from database) then fills other textbox (read from database based on combobox data) but I have confused
I Need Jquery / Ajax / something to show it
can someone solve my problem and explain to me..
Controller
public function tambah()
{
$cek = $this->session->userdata('logged_in');
if(empty($cek))
{
header('location:'.base_url().'app_admin');
}
else
{
$d['nama_madrasah'] = "";
$d['nsm'] = "";
$d['provinsi'] = "";
$d['alamat'] = "";
$d['kecamatan'] = "";
$d['kab_kota'] = "";
$d['jenis_bantuan'] = "";
$d['jenjang'] = "";
$d['th_ang'] = "";
$d['sumber_dana'] = "";
$d['tahap_pencairan'] = "";
$d['st'] = "tambah";
$d['dt_nsm'] = $this->app_model->ShowNSMMaster();
$this->load->view("app_admin/madrasah/input",$d);
}
}
My View
<select data-placeholder="Cari NSM..." class="chzn-select" style="width:260px;" tabindex="2" name="nsm" id="dt_nsm" >
<?php
if($st=='edit')
{
?>
<option value="<?php echo '$nsm'; ?>" selected="selected"><?php echo '$nsm'; ?></option>
<?php
}
else
{
?>
<option value="" selected="selected">- Pilih -</option>
<?php
foreach($dt_nsm->result_array() as $dp1)
{
?>
<option value="<?php echo $dp1['nsm']; ?>"><?php echo $dp1['nsm']; ?> | <?php echo $dp1['nama_madrasah']; ?></option>
<?php
}
}
?>
</select>
</span></div>
<label class="control-label" for="nama_madrasah">Nama Madrasah </label>
<div class="controls">
<input type="text" class="span" name="nama_madrasah" id="nama_madrasahx" value="<?php echo $nama_madrasah; ?>" placeholder="">
</div>
<label class="control-label" for="alamat">Alamat</label>
<div class="controls">
<input type="text" class="span" name="alamat" id="alamatx" value="<?php echo $alamat; ?>" placeholder="">
</div>
JS
<script type="text/javascript">
$('#dt_nsm').on('change', function(){
dt_nsm = $('#dt_nsm option:selected').val(); // the dropdown item selected value
$.ajax({
type :'POST',
dataType:'json',
data : { dt_nsm : dt_nsm },
url : 'getresult.php',
success : function(result){
$('#nama_madrasahx).val(result['nama_madrasahx']);
$('#kecamatanx).val(result['kecamatanx']);
}
});});
</script>
getresult.php
<?php
require_once ('../config/db.config.php');
$dt_nsm = $_POST['dt_nsm'];
$query = "SELECT * FROM bantuan WHERE nsm = '$nsm' ";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
echo json_encode($row);
?>
Thanks For your time. Need Advice !
change
foreach($dt_nsm->result_array() as $dp1)
to
foreach($dt_nsm->result_array() as $key=>$dp1)

Display Table based on dropdownlist selection in PHP

My table is not displayed when i select any project name. I guess the onchange function is not working properly but i couldn't figure out the problem.
Code is as follows:
<div class="span6">
<?php $sql = "SELECT * from login_projects WHERE login_id='".$record['login_id']."'";
$res_sql = mysql_query($sql); ?>
<label>Project Name <span class="f_req">*</span></label>
<!--<input type="text" name="city" class="span8" />-->
<select name="project_name" onchange="get_list_onnet()" id="project_name" class="span8">
<option value="">--</option>
<?php while($rec_sql = mysql_fetch_array($res_sql)){ ?>
<option value="<?php echo $rec_sql['project_id']; ?>">
<?php echo $rec_sql['project_name']; ?></option>
<?php } ?>
</select>
</div>
Function:
<script>
function get_list_onnet(){
var project_name=$("#project_name").val();
$.ajax
({
type: "POST",
url: "ajax.php",
data: {action: 'get_list_onnet',list_onnet:project_name},
success: function()
{
document.getElementById("dt_a").style="block";
$("#dt_a").html(html);
}
});
};
</script>
<script>
$(document).ready(function() {
//* show all elements & remove preloader
setTimeout('$("html").removeClass("js")',1000);
});
</script>
Ajax.Php Page:
function get_list_onnet(){
$list_onnet=$_POST['list_onnet'];
$sql_list_onnet=mysql_query("SELECT * from projects,project_wise_on_net_codes
where projects.project_id = project_wise_on_net_codes.project_id AND
project_wise_on_net_codes.project_id='$list_onnet'");
$row1 = mysql_num_rows($sql_list_onnet);
if($row1>0)
{
echo "<tr><th>id</th><th>Project Name</th><th>Country Code</th><th>On-net prefix</th>
<th>Action</th></tr>";
$k = 1; while($row_list_onnet=mysql_fetch_array($sql_list_onnet))
{
$project3 = $row_list_onnet['project_name'];
$countrycode1 = $row_list_onnet['country_code'];
$prefix1 = $row_list_onnet['on_net_prefix'];
$id_proj = $row_list_onnet['project_id'];
$on_prefix = $row_list_onnet['on_net_prefix'];
echo "<tr><td>".$k."</td><td>".$project3."</td><td>".$countrycode1."</td>
<td>".$prefix1."</td><td><a href='process/update_process_onnet.php?ID=".$id_proj."&Onnet=".$on_prefix."'>Delete</a></td>
</tr>";
$k++;
}
}
else
{
echo "<script>alert('No Record Found')</script>";
}
}
The problem is that it is always going in the else condition and nothing is displayed in the table.

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