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);
Related
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
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));
Here is my ajax function in response I am getting result but I don't know how to set that response in my page. Is it possible with json_decode or I have to try something else
JSON file is
<?php
$group_id = $_POST['group_id'];
$query = "SELECT *,group_id FROM contact JOIN addressgroup ON addressgroup.contact_id = contact.contact_id WHERE group_id IN (".$group_id.") GROUP BY contact.contact_id";
$res = mysql_query($query);
$data = array();
$k=0;
while($row = mysql_fetch_array($res))
{
$data[$k][0] = $row['user_id'];
$data[$k][1] = $row['first_name'];
$data[$k][2] = $row['middle_name'];
$data[$k][3] = $row['last_name'];
$k++;
}
echo json_encode(array($data));
?>
AJAX function
var myarray;
function getcon() {
myarray = [];
myarray.push($(".group_id").val());
$.ajax({
dataType: "json",
type: "POST",
url: "getcon.php",
data: 'group_id=' + myarray.join(),
success: function(data) {
totalRecords=data.length;
zone.fnClearTable();
for(var i=0; i < (data.length); i++) {
zone.fnAddData([
data[k][0],
data[k][1],
data[k][2],
data[k][3],
]);
}
return false;
}
});
return false;
}
As your Response is an JSON object..
you have to use it like this:-
replace
echo json_encode(array($data));
by
echo json_encode($data); /* because $data is already an array*/
and keep it same as it was earlier
while($row = mysql_fetch_array($res))
{
$data[] = $row['user_id'];
$data[] = $row['first_name'];
$data[] = $row['middle_name'];
$data[] = $row['last_name'];
$k++;
}
<script>
var myarray;
function getcon() {
myarray = [];
myarray.push($(".group_id").val());
$.ajax({
dataType: "json",
type: "POST",
url: "getcon.php",
data: 'group_id=' + myarray.join(),
success: function(data) {
console.log(data)/* check the structure of data here*/
zone.fnClearTable();
zone.fnAddData([
data.user_id.,
data.first_name,
data.middle_name,
data.last_name,
]);
return false;
}
});
return false;
}
</script>
JSON is an object! Your are getting object, so in your ajax function You must use some loop function like for or $.each JQuery functions.
`success : function (data) {
$.each (data,function(key,val){
$("#someId).html(key + " - " + val)
}
}`
I am struggling to convert json code in CodeIgniter, i wanna make pagination.
I am using ajax and want to controle data convert to json and send it to view. I can someone explain me how to do that.
Code in controler, instead foreach part i am trying to make Json, and send it to view
function ajax(){
$rpp= $_POST['rpp'];
$last = $_POST['last'];
$pn = $_POST['pn'];
if($pn<1){
$pn=1;
}
elseif($pn>$last){
$pn =$last;
}
$l = ($pn - 1) * $rpp;
$this->db->limit($l, $rpp);
$row = $this->db->get('pages');
$dataString='';
foreach($row->result() as $r){
$id = $r->id;
$info = $r->info;
$dataString .= $id.'|'.$info.'||';
}
echo json_encode($dataString);
}
}
view part
function request_page(pn)
{
var rpp = <?php echo $rpp; ?>; // results per page
var last = <?php echo $last; ?>; // last page number
var results_box = document.getElementById("results_box");
var pagination_controls = document.getElementById("pagination_controls");
results_box.innerHTML = "loading results ...";
$.ajax({
type: "POST",
url: "<?php echo site_url('search/ajax')?>",
data: { 'rpp' : rpp , 'last' : last, 'pn' : pn},
dataType: "text",
success: function(msg){
// alert(msg)
;
var dataArray = msg.split("||");
var html_output = "";
for(i = 0; i < dataArray.length - 1; i++){
var itemArray = dataArray[i].split("|");
html_output += "ID: "+itemArray[0]+" - Testimonial from <b>"+itemArray[1]+"</b><hr>";
}
results_box.innerHTML = html_output;
}
});
var paginationCtrls = "";
if(last != 1){
if (pn > 1) {
paginationCtrls += '<button onclick="request_page('+(pn-1)+')"><</button>';
}
paginationCtrls += ' <b>Page '+pn+' of '+last+'</b> ';
if (pn != last) {
paginationCtrls += '<button onclick="request_page('+(pn+1)+')">></button>';
}
}
pagination_controls.innerHTML = paginationCtrls;
}
its easy
you should have a view to encode the data to json. Its just this:
<?php
$this->output->set_header('Content-Type: application/json; charset=utf-8');
echo json_encode($json);
In the controller you just have to load this view with an array as parameter (i think that stdClass is also valid):
$data['json'] = array("foo" => "bar", "bar" => "foo");
$this->load->view('your_json_view', $data);
I need to extract the URL's from the php datas, how can i achieve this?
PHP
$query = 'SELECT * FROM picture LIMIT 3';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url.=$rec['pic_location'].";";
}
echo json_encode($url);
Ajax
<script type="text/javascript">
$(document).ready(function() {
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first: " + arr[0] + ", second: " + arr[1]);
alert(arr[0]);
$(".theImage").attr("src", +arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
the alert message isn't working its just printing H T ( i think these are http://... )
You're returning a string which is not parsed as JSON.
Just add dataType: "json" to the ajax settings.
And since you're reading it as an array in your javascript you should return it like so:
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[] = $rec['pic_location'];
}
You are sending a string in your PHP and expecting an array as response in javascript. Change you PHP to
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[] = $rec['pic_location'];
}
And javascript to
$.ajax({
url: "viewnew.php",
dataType: "JSON",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret[0]);
var arr = ret;
alert(arr);
alert("first: " + arr[0] + ", second: " + arr[1]); // THIS IS NOT WORKING!!!!
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});