i have problem with the post value in my form
i show you my code and see if you can help me
<script src="//netsh.pp.ua/upwork-demo/1/js/typeahead.js"></script>
<script>
$(document).ready(function() {
$('#ville').on('change', function(){
var id_ville = this.value;
$.ajax({
type: "post",
url: "lycee.php",
data: 'id_ville=' + id_ville,
success: function(result){
$("#lycee").html(result);
}
});
$('input.lycee').typeahead({
name: 'lycee',
remote: 'lycee.php?query=%QUERY'
});
});
})
</script>
<script >
and this is my php part code
$id_ville = mysql_real_escape_string($_POST['id_ville']);
if($id_ville!='') {
if (isset($_REQUEST['query'])) {
$query = $_REQUEST['query'];
$sql = $conn->query ("SELECT nom_lycee
FROM ville_lycee
INNER JOIN ville ON ville.id_ville = ville_lycee.id_ville
INNER JOIN lycee ON lycee.id_lycee = ville_lycee.id_lycee
WHERE ville.id_ville = '" . $id_ville . "'
AND lycee.nom_lycee LIKE '%{$query}%' ");
$options = "<label value=''></label>";
$array = array();
while ($row = $sql->fetch_assoc()) {
$array[] = array (
'label' => $row['nom_lycee'],
'value' => $row['nom_lycee'],
);
}
echo json_encode ($array);
}
}
the problem is that i need the id_ville value first and then the Query to filtre
Related
I have a slight problem with my code, lets say i have a json like this one :
[{"img":"john.png","name":"John","username":"#john"},
{"img":"mark.png","name":"mark","username":"#mark"}]
I wanna get data organized like :
John #john john.png
Mark #mark mark.png
But every time the data comes out like this:
John Mark #john #mark john.png mark.png
This is my Php Code:
<?php
class search{
public function gettingvalues($search_value){
require_once('conx.php');
$dir = "usersimage/";
$sql = "SELECT name,img,username FROM users WHERE username like '$search_value%' || name like '$search_value%'";
$query = mysqli_query($conx,$sql);
if ($query) {
if (mysqli_num_rows($query) > 0) {
while ($row = mysqli_fetch_array($query)) {
$img = $row['img'];
$name = $row['name'];
$username = $row['username'];
$json = array('img' => $img, 'name' => $name, 'username' => $username);
$results[] = $json;
}
echo json_encode($results);
}
}
}
}
?>
This the index code:
<?php
if (isset($_POST['data'])) {
require('search.php');
$search = new search;
$search->gettingvalues($_POST['data']);
header('Content-Type: application/json; charset=utf-8');
die();
}
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('input').keyup(function(){
var value= $('input').val();
$.ajax({
type: "POST",
url: "",
data: {data: value},
datatype: "json",
success: function(json_data) {
var img = [];
var username = [];
var name = [];
$.each(json_data, function(index, element) {
img.push(element.img);
username.push(element.username);
name.push(element.name);
})
$('#feedback').html('');
$('#feedback').html(name+username+img);
}
});
});
});
</script>
<input type="text" name="search" placeholder="looking for?">
<div id="feedback"></div>
Actually this is my first time with json, i don't know what is the problem or maybe i missed something, I hope getting some answers.
You need to build the HTML in the order that you want it displayed.
var html = '';
$.each(json_data, function(index, element) {
html += `${element.name} ${element.username} ${element.img}<br>`;
}
$("#feedback").html(html);
I'm trying to incorporate the jquery sortable functionality into my website and saving the positions in the database is giving me all sorts of headaches... I've been fighting this for 3 days now, and I cannot seem to get this work properly.
As it stands, it is saving positions to the database, but not in the order or positions, that you'd expect. Meaning, if I move the item in position 0 to position 1, it saves the positions in a different order in the db. check out a live version here.
Here is my code...
index.php file:
<div id="container">
<?php
require_once 'conn.php';
$q = ' SELECT * FROM items WHERE groupId = 3 ORDER BY position ';
$result = mysqli_query($db, $q);
if ($result->num_rows > 0) {
while($items = $result->fetch_assoc()) {
?>
<div id='sort_<?php echo$items['position'] ?>' class='items'>
<span>☰</span> <?php echo$items['description'] ?>
</div>
<?php
}
}
?>
</div>
js.js file:
$("#container").sortable({
opacity: 0.325,
tolerance: 'pointer',
cursor: 'move',
update: function(event, ui) {
var itId = 3;
var post = $(this).sortable('serialize');
$.ajax({
type: 'POST',
url: 'save.php',
data: {positions: post, id: itId },
dataType: 'json',
cache: false,
success: function(output) {
// console.log('success -> ' + output);
},
error: function(output) {
// console.log('fail -> ' + output);
}
});
}
});
$("#container").disableSelection();
save.php file:
require_once('conn.php');
$itId = $_POST['id'];
$orderArr = $_POST['positions'];
$arr = array();
$orderArr = parse_str($orderArr, $arr);
$combine = implode(', ', $arr['sort']);
$getIds = "SELECT id FROM items WHERE groupId = '$itId' ";
$result = mysqli_query($db, $getIds);
foreach($arr['sort'] as $a) {
$row = $result->fetch_assoc();
$sql = " UPDATE items
SET position = '$a'
WHERE id = '{$row['id']}' ";
mysqli_query($db, $sql);
}
echo json_encode( ($arr['sort']) );
Can anyone please point to where I am going wrong on this?
Thank you in advance.
Serge
In case someone lands on here, here is what worked in my case...
NOTE: I did not create prepared statements in the index.php select function. But you probably should.
index.php file:
<div id="container">
<?php
require_once 'conn.php';
$q = ' SELECT * FROM items WHERE groupId = 3 ORDER BY position ';
$result = mysqli_query($db, $q);
if ($result->num_rows > 0) {
while( $items = $result->fetch_assoc() ){
?>
<div id='sort_<?php echo $items['id'] ?>' class='items'>
<span>☰</span> <?php echo $items['description'] ?>
</div>
<?php
}
}
?>
</div>
jquery sortable file:
var ul_sortable = $('#container');
ul_sortable.sortable({
opacity: 0.325,
tolerance: 'pointer',
cursor: 'move',
update: function(event, ui) {
var post = ul_sortable.sortable('serialize');
$.ajax({
type: 'POST',
url: 'save.php',
data: post,
dataType: 'json',
cache: false,
success: function(output) {
console.log('success -> ' + output);
},
error: function(output) {
console.log('fail -> ' + output);
}
});
}
});
ul_sortable.disableSelection();
update php file:
$isNum = false;
foreach( $_POST['sort'] as $key => $value ) {
if ( ctype_digit($value) ) {
$isNum = true;
} else {
$isNum = false;
}
}
if( isset($_POST) && $isNum == true ){
require_once('conn.php');
$orderArr = $_POST['sort'];
$order = 0;
if ($stmt = $db->prepare(" UPDATE items SET position = ? WHERE id=? ")) {
foreach ( $orderArr as $item) {
$stmt->bind_param("ii", $order, $item);
$stmt->execute();
$order++;
}
$stmt->close();
}
echo json_encode( $orderArr );
$db->close();
}
Change your JS code like this:
{...}
tolerance: 'pointer',
cursor: 'move',
// new LINE
items: '.items', // <---- this is the new line
update: function(event, ui) {
var itId = 3;
var post = $(this).sortable('serialize'); // it could be removed
// new LINES start
var post={},count=0;
$(this).children('.items').each(function(){
post[++count]=$(this).attr('id');
});
// new LINES end
$.ajax({
{...}
With this $.each loop you overwrite your var post -> serialize and define your own sort order. Now look at your $_POST["positions"] with PHP print_r($_POST["positions"]); and you have your positions in your own order.
I have 2 tables to be retrieved from database with 2 different queries and needs to be encoded in json and send to ajax.The issue is I am not able to pass 2 json's to ajax .
I have tried with echo json_encode(array($data1,$data2)); but it is not working.
//My php code
$query = $db->query("select * from table1 where valu1='".$value1."'");
while ($row = $query->fetch_assoc()) {
$data1['value1'] = $row['value1'];
$data1['value2'] = $row['value2'];
}
$query = $db->query("select * from table2 where value2='".$value2."' ");
while ($row = $query->fetch_assoc()) {
$data2['value2'] = $row['value2'];
$data2['value3'] = $row['value3'];
}
echo json_encode(array($data1,$data2));
//My AJAX code
$(document).ready(function(){
$('#Form').submit(function(e){
e.preventDefault(); // stops the form submission
$.ajax({
url:$(this).attr('action'), // action attribute of form to send the values
type:$(this).attr('method'), // method used in the form
data:$(this).serialize(), // data to be sent to php
dataType:"json",
success:function(data){
//main
alert(data.value1);
},
error:function(err){
alert(err);
}
});
});
});
Kindly help in solving this issue.Thanks in advance
In PHP:
echo json_encode(array($data1, $data2));
In AJAX:
var data1 = data[0];
var data2 = data[1];
$.each(data1, function() {
console.log(this.value1 + ',' + this.value2);
});
$.each(data2, function() {
console.log(this.value2 + ',' + this.value3);
});
EDIT:
After a whole year, I just noticed that the initial PHP code was wrong, because the loop for the sql result would overwrite each time the values of the array. So, the correct one is this:
$query = $db->query("select * from table1 where valu1='".$value1."'");
while ($row = $query->fetch_assoc()) {
$data1[] = array('value1' => $row['value1'], 'value2' => $row['value2']);
}
$query = $db->query("select * from table2 where value2='".$value2."' ");
while ($row = $query->fetch_assoc()) {
$data2[] = array('value2' => $row['value2'], 'value3' => $row['value3']);
}
echo json_encode(array($data1,$data2));
Your code is fine, you just have to handle the two arrays within your success function:
success: function(data){
var object1 = data[0];
var object2 = data[1];
// Do whatever you like to do with them
}
I try to use the plugin from "devbridge autocomplete" : https://www.devbridge.com/sourcery/components/jquery-autocomplete/
I would like to get 3 values from my search.php page (and not just 1).
It works for "value" but not for "data1" and "data2" (result for each = null)
My jQuery code :
$('#search-adress').autocomplete({
serviceUrl: 'search.php',
dataType: 'json',
onSelect: function (value,data1,data2) {
alert('You selected: ' + value + ', ' + data1 + ', ' + data2);
}
});
My search page :
$term=$_GET['query'];
$query = mysql_query("select distinct adress,id,city from myadresstable where (adress like '%{$term}%') order by adress limit 10 ");
if (mysql_num_rows($query))
{
while($row = mysql_fetch_assoc($query))
{
$reply['suggestions'][] = ''.utf8_encode($row['nom_voie']).'';
$reply['data1'][] = ''.utf8_encode($row['id']).'';
$reply['data2'][] = ''.utf8_encode($row['city']).'';
}
echo json_encode($reply);
}
Thank you for helping me :)
You can just change a bit the php array this way:
$term=$_GET['query'];
$query = mysql_query("select distinct adress,id,city from myadresstable where (adress like '%{$term}%') order by adress limit 10 ");
if (mysql_num_rows($query))
{
$arr = array();
while($row = mysql_fetch_assoc($query))
{
$reply['suggestions'] = utf8_encode($row['nom_voie']);
$reply['data'] = utf8_encode($row['id']);
$reply['value'] = utf8_encode($row['city']);
$arr[] = $reply;
}
echo json_encode($arr);
}
And in your jquery code:
$(function(){
$('#autocomplete').autocomplete({
lookup: datos,
onSelect: function (suggestion) {
console.log(suggestion);
alert('You selected: ' + suggestion.value + ', ' + suggestion.data + ', ' + suggestion.nom_voie);
}
});
});
Sample fiddle: https://jsfiddle.net/robertrozas/n6oLLfmc/
Try selecting "Valdivia" to see the alert
I found the solution :)
The problem was the PHP Array. To get some data values with this autocomplete plugin you have to use array() :
PHP Code
$suggestions = array();
if (mysql_num_rows($query))
{
while($row = mysql_fetch_assoc($query))
{
$mydata1='$row['data1']';
$mydata2='$row['data2']';
$nom_voie=''.utf8_encode($row['nom_voie']).'';
$suggestions[] = array(
"value" => $nom_mydata1,
"data" => $nom_mydata2
);
}
}
echo json_encode(array('suggestions' => $suggestions));
How can I pass data from a php of then rows back to ajax ?
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[]=$rec['pic_location'];
$name[]=$rec['name'];
$age[]=$rec['age'];
$gender[]=$rec['gender'];
}
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Ajax
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
dataType: "json",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first image url: " + arr[0][0] + ", second image url: " + arr[0][1]); // This code isnt working
alert("first image Name: " + arr[1][0] + ", second image name: " + arr[1][1]);
$(".theImage").attr("src", arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
My question is how can I display the values here ? The Alert message is giving me "Undefined" ?
You can do something along these lines.
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$res = mysql_query($query);
$pictures = array();
while ($row = mysql_fetch_array($res)) {
$picture = array(
"pic_location" => $row['pic_location'],
"name" => $row['name'],
"age" => $row['age'],
"gender" => $row['gender']
);
$pictures[] = $picture;
}
echo json_encode($pictures);
JS
...
$.ajax({
...
dataType: "json",
...
success: function(pictures){
$.each(pictures, function(idx, picture){
// picture.pic_location
// picture.name
// picture.age
// picture.gender
});
}
});
...
You can't put multiple echo statements for the AJAX response:
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Join your arrays and send a single response:
$arr = $url + $name + $age + $gender;
echo json_encode($arr);
You can easily do this using a single Array:
$pics = array();
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$pics[$rec['id']]['url'] = $rec['pic_location'];
$pics[$rec['id']]['name']=$rec['name'];
$pics[$rec['id']]['age']=$rec['age'];
$pics[$rec['id']]['gender']=$rec['gender'];
}
echo json_encode($pics);