i have this code:connection to database
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$conn = new mysqli("localhost", "root", "", "jquery");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
now i already have data indatabase in table called city witch it have only id and desc and this is the code
if (isset($_POST['city'])) {
$sql = "SELECT * FROM city";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$results = [];
while($row = $result->fetch_assoc()) {
$results[] = $row;
}
echo json_encode($Results);
}
else
{
echo "empty";
}
}
here is the html part:
<select required="required" id="city">
<option disabled selected value=''> select a city </option>
</select>
and here is the function:
function city() {
$.ajax({
"url": "divs.php",
"dataType": "json",
"method": "post",
//ifModified: true,
"data": {
"F": ""
}
})
.done(function(data, status) {
if (status === "success") {
for (var i = 0; i < data.length; i++) {
var c = data[i]["city"];
$('select').append('<option value="'+c+'">'+c+'</option>');
}
}
})
.always(function() {
});
}
so the problem is that there is nothing in select list its always empty, any help? thank u
One small mistake is here:
You are using
echo json_encode($Results);
instead of
echo json_encode($results);
In PHP variable names are case sensitive. So, use proper case for all the variables.
You are not getting the response data in HTML <SELECT> TAG here is what you can do.
image to table
HTML FILE CODE:
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
</head>
<body>
<form action="" method="post">
<select>
</select>
</form>
<script>
$(document).ready(function(){
city();
});
function city(){
$.ajax({
type:'POST',
url: 'divs.php',
data: {city:1},
success:function(data){
var res = JSON.parse(data)
for(var i in res){
var showdata = '<option value="'+res[i].city_name+'">'+res[i].city_name+'</option>';
$("select").append(showdata);
}
}
});
}
</script>
</body>
</html>
HERE IS THE PHP CODE
`
<?php
$conn = new mysqli('localhost','root','','demo');
if($conn->connect_error){
die ("Connection Failed".$conn->link->error);
}
if(isset($_POST['city'])){
$sql = "SELECT * FROM city";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$results[] = $row;
}
echo json_encode($results);
}
else
{
echo "no city available";
}
}
?>
`
HERE IS THE OUTPUT IMAGE
You have used $results but while echoing you are using $Results which is a different variable so use,
echo json_encode($results);
Also, you are telling that city has only id and desc so in JS code use desc instead of city
$.ajax({
type:'POST',
url: 'divs.php',
data: {city:1},
success:function(data) {
var res = JSON.parse(data);
$(res).each(function(){
var c = this.city_name; // use city_name here instead of city
$('select').append('<option value="'+c+'">'+c+'</option>');
});
}
});
Related
Currently, I made script, which after onclick event,sending question to the database and showing data in console.log( from array ). This all works correctly, but.. I want to show data from array in the different position in my code. When I try to use DataType 'json' and then show some data, then it display in my console.log nothing. So, my question is: How to fix problem with displaying data? Is it a good idea as you see?
Below you see my current code:
$(document).ready(function(){
$(".profile").click(function(){
var id = $(this).data('id');
//console.log(id);
$.ajax({
method: "GET",
url: "../functions/getDataFromDB.php",
dataType: "text",
data: {id:id},
success: function(data){
console.log(data);
}
});
});
});
:
public function GetPlayer($id){
$id = $_GET['id'];
$query = "SELECT name,surname FROM zawodnik WHERE id='".$id."'";
$result = $this->db->query($query);
if ($result->num_rows>0) {
while($row = $result->fetch_assoc()){
$this->PlayerInfo[] = $row;
}
return $this->PlayerInfo;
}else {
return false;
}
}
:
$info = array();
$id = $_GET['id'];
$vv = new AddService();
foreach($vv->GetPlayer($id) as $data){
$info[0] = $data['name'];
$info[1] = $data['surname'];
}
echo json_encode($info);
I think it would be better to change the line fetch_all in mysqli to rm -rf. That information in the DB is all obsolete, or completely not true.
Try this:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<button class="profile" data-id="1">Click</button>
<script
src="https://code.jquery.com/jquery-3.3.1.min.js"
integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous"></script>
<script>
$(document).ready(function(){
$(".profile").click(function(){
var id = $(this).data('id');
console.log(id);
$.ajax({
method: "GET",
url: "../functions/getDataFromDB.php",
dataType: "json",
data: {id:id},
success: function(data){
console.log(data);
$.each(data, function(idx, item) {
console.log(item.surname);
});
}
});
});
});
</script>
</body>
</html>
PHP side:
<?php
class AddService {
public function GetPlayer($id) {
if (filter_var($id, FILTER_VALIDATE_INT) === false) {
return false;
}
$query = "SELECT name, surname FROM zawodnik WHERE id={$id}";
$result = $this->db->query($query);
if ($result->num_rows <= 0) {
return false;
}
// assumming you are using mysqli
// return json_encode($result->fetch_all(MYSQLI_ASSOC));
// or
WHILE ($row = $result->fetch_assoc()) {
$data[] = $row;
}
return json_encode($data);
}
}
if (isset($_GET['id'])) {
$id = $_GET['id'];
$vv = new AddService();
// you don't need foreach loop to call the method
// otherwise, you are duplicating your results
echo $vv->GetPlayer($id);
}
Hi All,
i'm getting all the data from database for array format i need to pass that data to second drop down list like (group option value),please any one help me.
This is my php code:
<?php
//error_reporting(0);
$servername = "localhost";
$username = "root";
$password = "";
$db = "essae";
$data = "";
$subcategory_id = "";
$subcategory_name = array();
$conn = mysqli_connect($servername, $username, $password,$db);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$category= $_GET["category_id"];
$sql = "SELECT es_category.category_id,es_category_description.name FROM es_category INNER JOIN es_category_description ON es_category.category_id=es_category_description.category_id WHERE parent_id='$category'";
$result = $conn->query($sql);
if ($result->num_rows > 1){
$sql_getrec ="SELECT es_category.category_id AS sub_cat_id,es_category_description.name AS sub_cat_name FROM es_category INNER JOIN es_category_description ON es_category.category_id=es_category_description.category_id WHERE parent_id='$category'";
$sub_category= $conn->query($sql_getrec);
if ($sub_category->num_rows > 1){
while ($row=mysqli_fetch_array($sub_category)){
$subcategory_id = $row['sub_cat_id'];
//$subcategory_name['sub_category_name'][] = $row['sub_cat_name'];
$sql_getrec = "SELECT es_product_description.name AS prod_name FROM es_product_to_category LEFT JOIN es_product_description ON es_product_description.product_id=es_product_to_category.product_id LEFT JOIN es_product ON es_product_description.product_id = es_product.product_id WHERE es_product_to_category.category_id = $subcategory_id AND es_product.status=1";
$sub_product=$conn->query($sql_getrec);
while ($prow=mysqli_fetch_array($sub_product)){
$subcategory_name['sub_category_name'][$row['sub_cat_name']]['products_name'][] = $prow['prod_name'];
}
}
echo "<pre>";print_r($subcategory_name);
}
}
else {
$sql_getrec = "SELECT es_product_description.name FROM es_product_to_category LEFT JOIN es_product_description ON es_product_description.product_id=es_product_to_category.product_id LEFT JOIN es_product ON es_product_description.product_id = es_product.product_id WHERE es_product_to_category.category_id='$category' AND es_product.status=1";
$result_getrec=$conn->query($sql_getrec);
while ($row=mysqli_fetch_array($result_getrec)){
$data .= $row['name'].",";
}
$data = rtrim($data,",");
}
print_r($data);
?>
This is my Html code:
<php?
$decocedData1 = json_decode($str_json_format, TRUE);
//print_r($decocedData1);die;
$decode = $decocedData1;
?>
<div>
<select name="category" id="category" />
<option selected ="selected">Select category</option>
<?php foreach($decode as $key => $value) { ?>
<option value="<?php echo $value['category_id']; ?>"><?php echo $value['name']; ?></option>
<?php } ?>
</select>
</div>
<div><select name="category12" id="category12" />
</select>
</div>
this is my j query and ajax method code:
<script type="text/javascript">
$(document).ready(function(){
$('#category').change(function(){
var category_id=$('#category').val();
$.ajax({
type: "get",
url: 'data_product.php?category_id='+category_id,
success: function(data) {
var products = data.split(",");
state_html = '';
state_html = '<option>Please Select product</option>'
$.each(products, function (index, productName) {
state_html += "<option value='"+productName+"'>"+productName+"</option>";
});
$('#category12').html(state_html);
},
});
})
});
</script>
You may use Type Casting in php
Type casting in PHP works much as it does in C: the name of the
desired type is written in parentheses before the variable which is to
be cast.
<?php
$array = array("name", "age", "mobile", "email");
var_dump($array);
(string)$array;
var_dump($array);
?>
This is not your final answer but you can try below code and figure out your variable names
$('#category12').empty();
$.each(data, function (index) {
var optgroup = $('<optgroup>');
optgroup.attr('label',data[index].name);
$.each(data[index].children, function (i) {
var option = $("<option></option>");
option.val(i);
option.text(data[index].children[i]);
optgroup.append(option);
});
$("#category12").append(optgroup);
});
$("#category12").multiselect('refresh');
you can do a jquery each in the result of your ajax
$.each(products, function(key, value) {
$('#category12')
.append($("<option></option>")
.attr("value",value)
.text(value));
});
$(function(){
//sample result, this will be your ajax result....
var products = ["Candy", "Cotton Candy", "Iced Candy"];
//clear again the select element.
$('#category12').empty();
$.each(products, function(key, value) {
$('#category12')
.append($("<option></option>")
.attr("value",value)
.text(value));
});
})
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<select id="category12"></select>
I am trying to edit two columns using ajax and php.My code currently edits one values(name) in my table and saves it to my database.When i add the second variable (p) my ajax call it updates both columns p and y with the same value.How do i edit the third value and assign it a different value from y.I want the two different columns to have different values in my db(columns:name and capacity)
This code edits and updates two values:
<script type="text/javascript">
jQuery(document).ready(function() {
$.fn.editable.defaults.mode = 'popup';
$('.xedit').editable();
$(document).on('click','.editable-submit',function(){
var x = $(this).closest('td').children('span').attr('id');
var y = $('.input-sm').val();
var z = $(this).closest('td').children('span');
$.ajax({
url: "process.php?id="+x+"&data="+y,
type: 'GET',
success: function(s){
if(s == 'status'){
$(z).html(y);}
if(s == 'error') {
alert('Error Processing your Request!');}
},
error: function(e){
alert('Error Processing your Request!!');
}
});
});
});
</script>
And this is what i tried to edit three values:
<script type="text/javascript">
jQuery(document).ready(function() {
$.fn.editable.defaults.mode = 'popup';
$('.xedit').editable();
$(document).on('click','.editable-submit',function(){
var x = $(this).closest('td').children('span').attr('id');
var y = $('.input-sm').val();
var p = $('.input-sm').val();
var z = $(this).closest('td').children('span');
$.ajax({
url: "process.php?id="+x+"&data="+y+"&capacity="+y,
type: 'GET',
success: function(s){
if(s == 'status'){
$(z).html(y);
$(z).html(p);}
if(s == 'error') {
alert('Error Processing your Request!');}
},
error: function(e){
alert('Error Processing your Request!!');
}
});
});
});
</script>
And heres my php file(process.php)
<?php
include("connect.php");
if
($_GET['id'],$_GET['capacity'] and $_GET['data'])
{
$id = $_GET['id'];
$data = $_GET['data'];
$capacity = $_GET['capacity'];
if(mysqli_query($con,"update mytable set name='$data',capacity='$data' where id='$id'")){
echo "success";
}
else{
echo 'failed';
}
}
?>
And my table in index.php
<tbody>
<?php
$query = mysqli_query($con,"select * from mytable");
$i=0;
while($fetch = mysqli_fetch_array($query))
{
if($i%2==0) $class = 'even'; else $class = 'odd';
echo'<tr class="'.$class.'">
<td><span class= "xedit external-event bg-brown" id="'.$fetch['id'].'">'.$fetch['name'].'</span></td>
<td><span class= "xedit external-event bg-brown" id="'.$fetch['id'].'">'.$fetch['capacity'].'</span></td>
</tr>';
}
?>
</tbody>
1) your just typo error : capacity=$data look this line and change it to capacity=$capacity :
if(mysqli_query($con,"update mytable set name='$data',capacity='$capacity' where id='$id'"))
2) And take look in If condition too .finally your code should be like this .
<?php
include("connect.php");
if($_GET['id'] && $_GET['capacity'] && $_GET['data'])
{
$id = $_GET['id'];
$data = $_GET['data'];
$capacity = $_GET['capacity'];
if(mysqli_query($con,"update mytable set name='$data',capacity='$capacity' where id='$id'"))
{
echo "success";
}
else
{
echo 'failed';
}
}
?>
You have error in your sql query. As you not passing correct parameters.
Please see below code.
$id = $_GET['id'];
$data = $_GET['data'];
$capacity = $_GET['capacity'];
// Check Sql
$query = "update mytable set name='$data',capacity='$capacity' where id='$id'";
if(mysqli_query($con,$query)){
echo "success";
} else{
echo 'failed';
}
I'm building a small online order system for a restaurant. My code for shoping card looks like this:
<?php
// Košarica
function ShopKosarica(){
global $link;
$UkupnoZbroj = 0;
$KosaricaSession = $_SESSION['ime'];
$rezultat = mysqli_query($link, "SELECT * FROM shop_kosarica WHERE KosaricaSession='$KosaricaSession' AND KosaricaKolicina<>0 ORDER BY KosaricaID ASC");
$num_results = mysqli_num_rows($rezultat);
if ($num_results==0){
echo "<h2><strong>Košarica je prazna</strong></h2>";
}else{
while ($redak = mysqli_fetch_array($rezultat)){
$ArtikalID = $redak['KosaricaArtikal'];
$rezultat_artikal = mysqli_query($link, "SELECT * FROM shop_artikal WHERE ArtikalID='$ArtikalID'");
$redak_artikal = mysqli_fetch_array($rezultat_artikal);
if ($redak['KosaricaVelicina']=='jumbo'){
$Cijena = $redak_artikal['ArtikalCijena2'];
} else {
$Cijena = $redak_artikal['ArtikalCijena1'];
}
$Kolicina = $redak['KosaricaKolicina'];
$Zbroj = $Cijena * $Kolicina;
$Zbroj = number_format((float)$Zbroj, 2, '.', '');
$UkupnoZbroj += $Zbroj;
$UkupnoZbroj = number_format((float)$UkupnoZbroj, 2, '.', '');
?>
<form class="ShopKosaricaBox" method="post">
<input type="hidden" id="KosaricaID" name="KosaricaID" value="<?=$redak['KosaricaID']?>">
<div class="MarginBottom15">
<input type="text" id="KosaricaKolicina" name="KosaricaKolicina" value="<?=$redak['KosaricaKolicina']?>" maxlength="2"> x <?=$redak_artikal['ArtikalNazivHr']?> (<?=$redak['KosaricaVelicina']?>) - <?=$Zbroj?> kn
</div>
<div class="right MarginBottom15">
<a onclick="ShopPromjena();">Promjeni</a> <a style="background:#c94e11;" onclick="ShopBrisanje();">Obriši</a>
</div>
<div class="clear"></div>
</form>
<script type="text/javascript">
function ShopPromjena() {
$(document).ready(function(){
var str = $(".ShopKosaricaBox").serialize();
$.ajax({
type: "POST",
url: "/funkcije?akcija=promjena&KosaricaID=<?=$redak['KosaricaID']?>",
data: str,
success: function(str){
alert( "Uspješno ste promjenili količinu!" );
}
});
return false;
});
}
function ShopBrisanje() {
$(document).ready(function(){
var str = $(".ShopKosaricaBox").serialize();
$.ajax({
type: "POST",
url: "/funkcije?akcija=brisi&KosaricaID=<?=$redak['KosaricaID']?>",
data: str,
success: function(str){
alert( "Uspješno ste obrisali jelo!" );
}
});
return false;
});
}
</script>
<?php
} ?>
<h1 class="MarginBottom25" style="font-size:25px;">Ukupno: <strong><?=$UkupnoZbroj?> kn</strong></h1>
<?php }
}
?>
And I put data in mysql via Ayax, this is the javascript:
$(document).ready(function(){
$(".ShopPonudaBox").submit(function(){
var str = $(this).serialize();
$.ajax({
type: "POST",
url: "/funkcije?akcija=dodaj",
data: str,
success: function(str){
alert( "Uspješno ste dodali jelo!" );
$('#KosaricaBox').load("/include/funkcije.php?funkcija=ShopKosarica");
return false;
}
});
return false;
});
});
and php code;
if ($_GET['akcija']=="dodaj") {
if ($_POST['KosaricaKolicina']<>0){
$KosaricaSession = $_SESSION['ime'];
$KosaricaArtikal = clean($link, $_POST['ArtikalID']);
$KosaricaKolicina = clean($link, $_POST['KosaricaKolicina']);
$KosaricaVelicina = clean($link, $_POST['KosaricaVelicina']);
$provjera = mysqli_query($link, "SELECT * FROM shop_kosarica WHERE KosaricaSession='$KosaricaSession' AND KosaricaArtikal='$KosaricaArtikal' AND KosaricaVelicina='$KosaricaVelicina'");
$num_results = mysqli_num_rows($provjera);
if ($num_results==0){
$result = mysqli_query($link, "INSERT INTO shop_kosarica (KosaricaSession, KosaricaArtikal, KosaricaKolicina, KosaricaVelicina) VALUE ('$KosaricaSession', '$KosaricaArtikal', '$KosaricaKolicina', '$KosaricaVelicina')");
//header("Location: /online-narudzba#Shop");
} else {
$redak_provjera = mysqli_fetch_array($provjera);
$KosaricaID = $redak_provjera['KosaricaID'];
$result = mysqli_query($link, "UPDATE shop_kosarica SET KosaricaKolicina=KosaricaKolicina+$KosaricaKolicina WHERE KosaricaID='$KosaricaID'");
//header("Location: /online-narudzba#Shop");
}
} else {
//header("Location: /online-narudzba#Shop");
}
}
I tried with this method I found here
$('#KosaricaBox').load("/include/funkcije.php?funkcija=ShopKosarica");
$funkcija = $_GET["funkcija"];
if ($funkcija == "ShopKosarica") {
echo ShopKosarica();
}
but keep getting errors
Notice: Undefined variable: _SESSION in
H:\Dropbox\htdocs\include\funkcije.php on line 47
Warning: mysqli_query() expects parameter 1 to be mysqli, null given
in H:\Dropbox\htdocs\include\funkcije.php on line 48
Add
session_start();
and connect database
at the beginning of your page before any HTML
You will have something like :
$con=mysqli_connect("localhost","xxxx","xxxx","xxxxx");
//check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL:" . mysqli_connect_error();
}
session_start();
include("inc/incfiles/header.inc.php")?>
<html>
<head>
<meta http-equiv="Content-Type" conte...
Don't forget to remove the space you have before
I'm having a problem passing a variable selected from a dynamic drop dropdown to a PHP file. I want the PHP to select all rows in a db table that match the variable. Here's the code so far:
select.php
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("select#type").attr("disabled","disabled");
$("select#category").change(function(){
$("select#type").attr("disabled","disabled");
$("select#type").html("<option>wait...</option>"); var id = $("select#category option:selected").attr('value');
$.post("select_type.php", {id:id}, function(data){
$("select#type").removeAttr("disabled");
$("select#type").html(data);
});
});
$("form#select_form").submit(function(){
var cat = $("select#category option:selected").attr('value');
var type = $("select#type option:selected").attr('value');
if(cat>0 && type>0)
{
var result = $("select#type option:selected").html();
$("#result").html('your choice: '+result);
$.ajax({
type: 'POST',
url: 'display.php',
data: {'result': myval},
});
}
else
{
$("#result").html("you must choose two options!");
}
return false;
});
});
</script>
</head>
<body>
<?php include "select.class.php"; ?>
<form id="select_form">
Choose a category:<br />
<select id="category">
<?php echo $opt->ShowCategory(); ?>
</select>
<br /><br />
Choose a type:<br />
<select id="type">
<option value="0">choose...</option>
</select>
<br /><br />
<input type="submit" value="confirm" />
</form>
<div id="result"></div>
<?php include "display.php"; ?>
<div id="result2"></div>
</body>
</html>
select.class.php
<?php
class SelectList
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database");
mysql_select_db($db,$this->conn) OR die("can not select the database $db");
return TRUE;
}
public function ShowCategory()
{
$sql = "SELECT * FROM profession";
$res = mysql_query($sql,$this->conn);
$category = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$category .= '<option value="' . $row['id_cat'] . '">' . $row['prof_name'] . '</option>';
}
return $category;
}
public function ShowType()
{
$sql = "SELECT * FROM specialties WHERE id_cat=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$type = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$type .= '<option value="' . $row['id_type'] . '">' . $row['sp_name'] . '</option>';
}
return $type;
}
}
$opt = new SelectList();
?>
And here's the display.php that I want the variable passed to. This file will select the criteria from the db and then print the results in select.php.
<?php
class DisplayResults
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database");
mysql_select_db($db,$this->conn) OR die("can not select the database $db");
return TRUE;
}
public function ShowResults()
{
$myval = $_POST['result'];
$sql = "SELECT * FROM specialities WHERE 'myval'=sp_name";
$res = mysql_query($sql,$this->conn);
echo "<table border='1'>";
echo "<tr><th>id</th><th>Code</th></tr>";
while($row = mysql_fetch_array($res))
{
while($row = mysql_fetch_array($result)){
echo "<tr><td>";
echo $row['sp_name'];
echo "</td><td>";
echo $row['sp_code'];
echo "</td></tr>";
}
echo "</table>";
//}
}
return $category;
}
}
$res = new DisplayResults();
?>
I'd really appreciate any help. Please let me know if I can provide more details.
Link to db diagram: http://imgur.com/YZ0SuVw
The first dropdown draws from the profession table, the second from the specialties table. What I'd like to do is to display all of the rows in the jobs table that match the specialty selected in the dropdown box. This will require the result from the variable (result) from the dropdown to be converted into the spec_code that is in the job table. Not sure exactly how to do this. Thanks!
I just want to outline some points about following block of code:
where did you defined myval
data: {'result': myval}: result not require any quota change it to data: {result: myval}
why you need to get HTML from selected options? it's better to send option values the change
$("select#type option:selected").html();
to
$("select#type option:selected").val();
if(cat>0 && type>0)
{
var result = $("select#type option:selected").html();
$("#result").html('your choice: '+result);
$.ajax({
type: 'POST',
url: 'display.php',
data: {'result': myval},
});
}