i have to get select option list from my database
and for this my controller is
<?php
require_once 'connexion.php';
$sql = "select * from Motif";
$stmt = sqlsrv_query($conn, $sql);
if ($stmt === false)
{
die(print_r(sqlsrv_errors() , true));
}
$data = array();
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
{
$data["id"] = $row['id'];
$data["libelle"] = utf8_encode($row['libelle']);
echo json_encode($data);
}
echo json_encode($data);
exit;
?>
and my html like this
<select id="rejetselect" class="form-control">
<option value="" selected="selected"></option>
</select>
and to get values from database with ajax i did this code
var $select = $('#rejetselect');
$.ajax({
url: 'motifRejet.php',
dataType: 'json',
data: 'libelle',
success: function() {
alert('good');
},
error: function() {
alert('nooo');
}
});
so after testing all this i can see my alert in good and in my console i can see the information that i got from database but i can't see it in my select option
Related
I want a dependent select box I can load the data from my database into the select box however, when I click the show all countries when all select boxes have data only the province will return to the default data. What I need to do is when I click show all countries all the data from select box will reset to its default data and so on and so forth.
AJAX Script:
$(document).ready(function(){
//Show Province
$("#sort-country").change(function(){
var country = $(this).val();
$.ajax ({
url:"fetch_province.php",
method: "POST",
data: {country:country},
dataType: "text",
success: function(data){
$("#sort-province").html(data);
}
});
});
//Show Town
$("#sort-province").change(function(){
var country = $("#sort-country").val();
var province = $(this).val();
$.ajax ({
url:"fetch_town.php",
method: "POST",
data: {country:country, province:province},
dataType: "text",
success: function(data){
$("#sort-town").html(data);
}
});
});
//Show Barangay
$("#sort-town").change(function(){
var country = $("#sort-country").val();
var province = $("#sort-province").val();
var town = $(this).val();
$.ajax ({
url:"fetch_barangay.php",
method: "POST",
data: {country:country, province:province, town:town},
dataType: "text",
success: function(data){
$("#sort-barangay").html(data);
}
});
});
});
Fetch Province:
<?php
require "connect.php";
$output = "";
$sql = "SELECT * FROM tblLocation WHERE country='".$_POST['country']."' ORDER BY province";
$result = mysqli_query($conn, $sql);
$output = "<option value=''>Show All Province</option>";
while ($row = mysqli_fetch_array($result)) {
$output .= "<option value='".$row['province']."'>".$row['province']."</option>";
}
echo $output;
?>
Fetch Town:
<?php
require "connect.php";
$output = "";
$sql = "SELECT * FROM tblLocation WHERE country='".$_POST['country']."' AND province='".$_POST['province']."' ORDER BY town";
$result = mysqli_query($conn, $sql);
$output = "<option value=''>Show All Town</option>";
while ($row = mysqli_fetch_array($result)) {
$output .= "<option value='".$row['town']."'>".$row['town']."</option>";
}
echo $output;
?>
Fetch Barangay:
<?php
require "connect.php";
$output = "";
$sql = "SELECT * FROM tblLocation WHERE country='".$_POST['country']."' AND province='".$_POST['province']."' AND town='".$_POST['town']."' ORDER BY barangay";
$result = mysqli_query($conn, $sql);
$output = "<option value=''>Show All Barangay</option>";
while ($row = mysqli_fetch_array($result)) {
$output .= "<option value='".$row['barangay']."'>".$row['barangay']."</option>";
}
echo $output;
?>
I'm trying to populate the second dropdown after I select the first option, nothing appears in the second dropdown.
My first select:
<select name="inst" class="form-control" required="" id="inst">
<option value=0 selected=1>Select...</option>
<?php
$sql="SELECT * FROM sapinst";
$myData=mysqli_query($GLOBALS['con'],$sql);
if (mysqli_num_rows($myData) > 0){
while ($row = mysqli_fetch_array($myData))
{
echo '<option value="' .$row["nbd"]. '">' .$row["nome"]. '</option>';
}
}
else{echo "No categories were found!";}
?>
</select>
My second select:
<select id= "sub" name="sub" class="form-control"></select>
My Script:
<script type="text/javascript">
$("#inst").change(function () {
//get category value
var cat_val = $("#inst").val();
// put your ajax url here to fetch subcategory
var url = '/ajax.php';
// call subcategory ajax here
$.ajax({
type: "POST",
url: url,
data: {
cat_val: cat_val
},
success: function (data)
{
$("#sub").html(data);
}
});
});
</script>
My Ajax.php file:
<?php
require_once 'edp/configdbedp.php';
$prod_cat = $_POST['cat_val'];
$sql = "SELECT * FROM " . $dbname . ".sappainel WHERE nbd = '$prod_cat'";
$result = mysqli_query($conn, $sql);
$msg = '';
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
$msg =. '<option value="' . $row["nome"] . '">' . $row["nome"] . '</option>';
}
} else {
$msg .= "No categories were found!";
}
echo $msg;
mysqli_close($conn);
?>
if I try to print some thing in the Ajax php I can't...seems ajax.php won't run.
Am I calling it correctly?
Is your second ajax being called properly?
Check the console messages(in developer options, F12) for errors in ajax call.
you might want to do this as both cat_val are same. It might be giving an error. -
data: {
cat_val: cat_val_local //different variable names here.
},
Also "Select * from $TABLE_NAME(not #dbname)"
and next remove extra .[dot] here -> ".sappainel WHERE"
you can also try put console.log() in success callback and see if the success is returning any elements.
success: function (data)
{
console.log(data);
$("#sub").html(data);
}
If nothing is shown then your php might be wrong. Add an eror callback too! like this -
error: function (e)
{
console.log(e);
}
Hope this helps.
I already solved
Diferences on scrip:
<script type="text/javascript">
$("#inst").change(function(){
//get category value
var cat_val = $("#inst").val();
// put your ajax url here to fetch subcategory
var url = 'ajax.php';
// call subcategory ajax here
$.ajax({
type:"POST",
url:url,
data:{
cat_val : cat_val
},
success:function(data)
{
$("#sub").html(data);
}
});
});
</script>
On ajax.php
<?php
require_once 'edp/configdbedp.php';
$prod_cat = $_POST["cat_val"];
$sql = "SELECT * FROM ".$dbname.".sappainel WHERE nbd = '$prod_cat'";
$result = mysqli_query($GLOBALS['con'], $sql);
$msg ='';
if (mysqli_num_rows($result) > 0){
while ($row = mysqli_fetch_array($result))
{
$msg .='<option value="'. $row["nome"] .'">'. $row["nome"] .'</option>';
}
}
else{$msg .="No categories were found!";}
echo ($msg);
mysqli_close($GLOBALS['con']);
?>
There Should be an Input textbox, If user writes any text it should display dropdown list of customer names
<script>
function custlist() {
$.ajax({
url: "custlist.php",
success: function(result) {
$("#customerlist").html(result);
}
});
}
function showCustomers(str) {
$.ajax({
type: "GET",
url: "customerlist.php",
data:'q='+str,
success: function(result) {
$("#customerlist").html(result);
}
});
}
</script>
<input type="text" oninput="showCustomers(this.value)" placeholder="Search here" name="CustomerNo" />
<select name="Cno" id="customerlist" onfocus="custlist()">
<option value="">Customer Name</option>
</select>
custlist.php
<?php
$sql2 = 'SELECT Customer_Name as Cname,No from customers order by Customer_Name';
$result2 = mysqli_query($connection, $sql2);
if (mysqli_num_rows($result2) > 0) { ?>
<option value="">Customer Names</option>
<?php // output data of each row
while($row2 = mysqli_fetch_assoc($result2)) { ?>
<option value="<?php echo $row2['No']; ?>"><?php echo $row2["Cname"]; ?>
</option>
<?php } ?>
<?php } ?>
customerlist.php
<?php
$q = $_REQUEST["q"];
// lookup all hints from array if $q is different from ""
if ($q !== "") {
$sql2 = "SELECT Customer_Name as Cname,No from customers where Customer_Name like '".$q."%s' order by Customer_Name";
$result2 = mysqli_query($connection, $sql2);
if (mysqli_num_rows($result2) > 0) { ?>
<option value="">Customer Names</option>
<?php // output data of each row
while($row2 = mysqli_fetch_assoc($result2)) { ?>
<option value="<?php echo $row2['No']; ?>"><?php echo $row2["Cname"]; ?>
</option>
<?php } ?>
<?php } ?>
<?php } ?>
I am getting the data in my dropdown, but I want that if I write something in text box then automatically it shows dropdown with matching that characters.
And one more issue I have...
2nd Issue:- When I type "abd" first it shows customer names starting with "abd" but automatically it shows next names starting with "ab" then "a" then empty..
Why is that?
Thanks in advance.
Instead of
Javascript:
$.ajax({
type: "GET",
url: "customerlist.php",
data:'q='+str,
success: function(result){
$("#customerlist").html(result);
}
});
And php:
<?php
$q = $_REQUEST["q"];
Try doing this Javascript:
$.ajax({
type: "POST",
url: "customerlist.php",
data: {q: str},
success: function(result){
$("#customerlist").html(result);
}
});
And php:
<?php
$q = $_POST['q'];
Hope this helps!
I been finding for long unable to search for it.. I have a dropdown select option call from mysql database.. I wanted when the user choose a employee from the option.. when selected.. it's list out the data from mysql database the employee data that is selected. here the code..
<select value="employee" selected="selected"> Select a employee</option>
<?php
$query = "SELECT * FROM users";
$result = mysqli_query($link,$query);
while ($row = mysqli_fetch_array($result)) {
$fname = $row["first_name"];
$lname = $row["last_name"];
echo "<option value'" . $fname . " " . $lname ."'>" . $fname . " " . $lname . "</option>";
}
?>
</select>
sorry i'm new to js.. how do I get the data from selected value?
First of all edit your page and add jquery library
<select id="listEmployee" value="employee" selected="selected"> Select a employee</option>
<?php
$query = "SELECT * FROM users";
$result = mysqli_query($link,$query);
while ($row = mysqli_fetch_array($result)) {
$fname = $row["first_name"];
$lname = $row["last_name"];
$id = $row["id"];
echo "<option value'" .$id."'>" . $fname . " " . $lname . "</option>";
}
?>
</select>
Then you should listen to select change
$( "#listEmployee" ).change(function() {
getSelectedEmplpyee();
});
Then create an ajax function that get informations about the selected employee
function getSelectedEmplpyee()
{
var selectedId = $('#listEmployee').val();
$.ajax({
type: "POST",
url: "Controller.php",
data: {selected: selectedId},
success: function(data) {
//display response
},
error: function(data) {
// display error
}
});
}
in controller.php read the id using
$id = $_POST['selected']
in controller.php run your select query and return result using echo ,
the result is stored in the variable data in your function getSelectedEmplpyee()
Try to do like this:
var selected_value = $('#your_select_id').val();
$.ajax({
type: "POST",
url: "your_url",
data: {selected: selected_value},
success: function(data) {
//display response
},
error: function(data) {
// display error
}
});
I just need some ideas how i get the value from the dropdown
<select id="ins" name="instructor" required>
<?php
//dropdown instructor name
echo "<option value=\"\">"."Select"."</option>";
$qry = "select * from instructor";
$result = mysqli_query($con,$qry);
while ($row = mysqli_fetch_array($result)){
echo "<option value=".$row['insA_I'].">".$row['insname']."</option>";
}
?>
</select>
and put the value in this .php file in $_POST['instructor']
<?php
session_start();
require 'connection.php'; //connection
$qry = "select course from instructor where insA_I = '".$_POST['instructor']."'";
$result = mysqli_query($con,$qry);
$_SESSION['row'] = mysqli_fetch_array($result);
$data = $_SESSION['row']['course'];
echo $data;
?>
am still new in ajax
$(document).ready(function() {
$('#ins').change(function(){
$.ajax({
type: "POST",
url: "json_php.php",
data: '',
datatype: 'json',
success: function(result){
alert(result);
}
});
});
});
i already searched this in internet but didn`t get my answer after some modifcation
just need idea, techniques .etc.
Try this ajax code
$(document).ready(function() {
$('#ins').change(function(){
$.ajax({
type: "POST",
url: "json_php.php",
data: {instructor:$(this).val()},
success: function(result){
alert(result);
}
});
});
});
php code for query
$qry = "select course from instructor where insA_I = '".$_POST['instructor']."'";
$result = mysqli_query($con,$qry);
$fetch = mysqli_fetch_array($result);
$_SESSION['row']['course']=$fetch['course'];
$data = $_SESSION['row']['course'];
echo $data;
Since you're using jQuery, use jQuery:
data: { instructor: $('#ins').val() }
Also, please note that your SQL query is wide open to SQL injection. This is a good place to start reading about that.