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', $data);
Related
How to recover data from research by json? I would like to recover the separate items as the result show in php page to create new elements, for example recover item.f, item.m to create new elements like item.f < / div> < span > item.m. .. thank you
(error presented at)
Error Can not read property of items " of null
Php page ( does the query )
if ($_GET['action'] == "chatheartbeat") { chatHeartbeat(); }
function chatHeartbeat() {
$sql = "select * from mensagens ";
$query = mysql_query($sql);
$items = '';
$chatBoxes = array();
while ($chat = mysql_fetch_array($query)) {
$items. = << < EOD
{
"s": "0",
"f": "{$chat['de']}",
"m": "{$chat['mensagem']}",
"i": "{$chat['img']}"
},
EOD;
}
}
Index ( calls the pair query present the results - error here - need help here)
$.ajax({
url: "asd.php?action=chatheartbeat",
cache: false,
dataType: "json",
success: function(data) {
$.each(data.items, function(i, item) {
alert(item.f)
});
}
});
As #jeroen says dont manually generate JSON Strings use json_encode() to create a JSONString from a PHP array or an object.
if ($_GET['action'] == "chatheartbeat") {
$sql = "select * from mensagens ";
$query = mysql_query($sql);
$chatBoxes = array();
while ($chat = mysql_fetch_array($query)) {
$t - new stdClass();
$t->s = "0";
$t->f = $chat['de'];
$t->m = $chat['mensagem'];
$t->i = $chat['img'];
$chatBoxes[] = $t;
}
$items = json_encode($chatBoxes);
// now just echo $items to return it to the javascript in the browser
// here or later in the code if there is more to your code than you showed us
echo $items;
}
References:
json_encode()
json_decode()
Thank you all for the help, I managed to solve the problem as follows
Php page ( does the query )
$result = mysql_query("SELECT * FROM mensagens");
while( $array = mysql_fetch_assoc($result)){
$table[]= array("de"=>$array['de'],"mensagem"=>$array['mensagem']); }
echo json_encode($table);
Index
$.ajax({
url: 'asd.php',
dataType: 'json',
success: function(data)
{
for($i=0; $i < data.length; $i++){
$("#chatbox_"+chatboxtitle+" .chatboxcontent").append(
'<div class="message"><span class="from">'+data[$i].de+':
</span><span class=content">'+data[$i].mensagem+'</span></div>');
}}
});
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)
}
}`
Consider the following PHP Block. I want to output "name" to div #1 and "slog" to div #2
<?php
require_once 'db_conx.php';
$Result = mysql_query("SELECT * FROM profiles ORDER BY lastupdated desc limit 1") or die (mysql_error());
while($row = mysql_fetch_array($Result)){
$result = array();
$result['name'] = $row['name'];
$result['slog'] = $row['slog'];
echo json_encode($result);
}
mysql_close($con);
?>
Here's the ajax that just outputs json itself.
var get_fb = (function() {
var counter = 0;
var $buzfeed = $('#BuzFeed');
return function(){
$.ajax({
type: "POST",
url: "../php/TopBusinesses_Algoriththm.php"
}).done(function(feedback) {
counter += 1;
var $buzfeedresults = $("<div id='BuzFeedResult" + counter + "'></div>").addClass('flat-menu-button');
$buzfeedresults.text(feedback);
$buzfeed.append($buzfeedresults);
var $buzfeedDivs = $buzfeed.children('div');
if ($buzfeedDivs.length > 10) { $buzfeedDivs.last().remove(); }
setTimeout(get_fb, 1000);
})
};
})();
get_fb();
You can update the html/content of a DOM element with the following:
document.getElementById(name).innerHTML = "text";
So in this instance:
document.getElementById('#1').innerHTML = $yourNameContent;
document.getElementById('#2').innerHTML = $yourSlogContent;
First: move var counter = 0; outside from get_fb() function;
Second: Do not call get_fb(); explicitly, because (function(){...})(); already calls method after initialization.
Third: Why use POST method if there is not data to submit!
So, do this and have a try:
var counter = 0;
var $buzfeed = $('#BuzFeed');
var get_fb = (function() {
$.ajax({
type: "GET",
url: "../php/TopBusinesses_Algoriththm.php"
}).done(function(feedback) {
counter += 1;
var $buzfeedresults = $("<div id='BuzFeedResult" + counter + "'></div>").addClass('flat-menu-button');
$buzfeedresults.text(feedback);
$buzfeed.append($buzfeedresults);
var $buzfeedDivs = $buzfeed.children('div');
if ($buzfeedDivs.length > 10) { $buzfeedDivs.last().remove(); }
setTimeout(get_fb, 1000);
})
})();
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);
i have this code on the server side:
<?php
header('Content-Type: text/html; charset=utf-8');
require "../general.variables.php";
require "../functions_validation.php";
require "../functions_general.php";
require "../../db_con.php";
$keyword = mysql_real_escape_string($_POST["keyword"]);
$query = mysql_query("
SELECT user_id, user_fullname, user_area, user_city, user_quarter, user_tmb
FROM `migo_users`
WHERE (
user_fullname LIKE '%".$keyword."%' AND user_id NOT IN (".$superAdmins2string.")
)
ORDER BY tmb_set DESC, user_fname ASC
LIMIT 7;
");
$i = 0;
while ($userInfo = mysql_fetch_array($query)) {
$area_name = mysql_fetch_array(mysql_query("
SELECT area_name
FROM `migo_areas`
WHERE
area_id='".$userInfo['user_area']."';
"));
$city_name = mysql_fetch_array(mysql_query("
SELECT city_name
FROM `migo_cities`
WHERE
city_id='".$userInfo['user_city']."';
"));
if ($userInfo['user_quarter'] != 0) {
$quarter_name = mysql_fetch_array(mysql_query("
SELECT quarter_name
FROM `migo_quarters`
WHERE
quarter_id='".$userInfo['user_quarter']."';
"));
}
else {
$quarter_name['quarter_name'] = "";
}
$rsl[$i]['user_id'] = $userInfo['user_id'];
$rsl[$i]['user_fullname'] = $userInfo['user_fullname'];
$rsl[$i]['user_area_name'] = $area_name['area_name'];
$rsl[$i]['user_city_name'] = $city_name['city_name'];
$rsl[$i]['user_quarter_name'] = $quarter_name['quarter_name'];
$rsl[$i]['user_tmb'] = $userInfo['user_tmb'];
$i++;
}
echo json_encode($rsl);
mysql_close();
?>
and this code on the client side:
$.ajax({
type : 'POST',
url : 'php/general.ajax/header_search.php',
//async : false,
//cache : false,
dataType : 'json',
data: {
keyword : sb_keyword
},
success : function(data) {
var hs_hits = 0;
var hs_row_nr = 1;
var hs_results = "<div class='sb_spacing'></div><div id='sb_rows_cont'>";
if (data != null) {
$.each(data, function(index, arr) {
hs_hits++;
if (arr['user_quarter_name'] != "") {
var quarter_text = " - " + arr['user_quarter_name'];
}
else {
var quarter_text = "";
}
hs_results = hs_results + "<a class='search_links' href=profile.php?id=" + arr['user_id'] + "><div class='sbr_row' row_nr='" + hs_row_nr + "'><div class='sbr_imgFrame'><img src='images/user_48x48/" + arr['user_tmb'] + "' alt=''></div><div class='sbr_name'>" + arr['user_fullname'].replace(regexp_hs_user_fullname, '<span>$&</span>') + "</div><div class='sbr_area'>" + arr['user_area_name'] + "</div><div class='sbr_area'>" + arr['user_city_name'] + quarter_text + "</div></div></a>";
hs_row_nr++;
});
}
if (hs_hits > 0) {
hs_results = hs_results + "</div><div class='sb_spacing'></div><a class='search_links' href='search.php?name=" + sb_keyword + "'><div id='sbr_botttom'>Se flere resultater for <span class='gay'>" + sb_keyword + "</span></div></a>";
$("#sb_results").html(hs_results).show();
searchSet = 1;
total_rows = hs_hits;
$("#sb_rows_cont > a:first .sbr_row").addClass('sbr_row_act');
on_a = $("#sb_rows_cont > a:first");
first_a = on_a;
last_a = $("#sb_rows_cont > a:last");
sb_url = $(on_a).attr('href');
search_navigator_init();
}
else {
$("#sb_results").hide();
searchSet = 0;
}
},
error : function() {
alert("ajax error");
}
});
one problem tho, if the query gives 0 results, and the each function tries to run on the client side my js code stops working..
so i was wondering what i could do here.
how can i retrieve the amount of hits from the server side, before i run the each loop?
You're instantiating $rsl in the middle of a loop (implicitly), but you aren't entering the loop unless you have at least one entry.
Instantiate $rsl up above your while loop:
...
$i = 0;
$rsl = array();
while ($userInfo = mysql_fetch_array($query)) {
...
And now when you encode it, it is an empty array instead of null. This will also save your HTTP error_log, and generally be happier. : )
I would try json_encode() your PHP array to allow JavaScript to eval it as a native JS object.
echo json_encode($some_array);
Just a note, json_encode() is only available for PHP versions 5.2 and higher.
$.getJSON('ajax.php',{form data}, functon(data){
if(data == '') return;
});