This is my javascript function....
But my php controller getting all values but not the array not sure why ? Need help.....
Thanks in advance :)
function submitForm(){
var id = $('#id').val();
var supplier_id = $('#supplier_id').val();
var description = $('#description').val();
var numofpro = $('#numofpro').val();
var product_id = new Array();
for(var i=0;i<numofpro;i++){
product_id[i] = $('#product_id'+(i+1)).val();
}
var payment_mode = $('#payment_mode').val();
//description = description.replace(new RegExp('\r?\n','g'), '<br />');
$.ajax({
url: "<?= base_url(); ?>edit/save_purchase", //The url where the server req would we made.
data: "id="+id+"&supplier_id="+supplier_id+"&description="+description+"&product_id="+product_id+"&payment_mode="+payment_mode,
dataType: "html", //Return data type (what we expect).
beforeSend:function(){
//alert("asds");
},
success: function(data) {
//alert("Edited");
alert(data);
}
});
}
because you pass a string, not an array :) try it:
$.ajax({
url: "<?= base_url(); ?>edit/save_purchase",
type: "POST"
data: {id : id, supplier_id : supplier_id, description : description, product_id : product_id, payment_mode : payment_mode},
dataType: "json",
beforeSend:function(){
//alert("asds");
},
success: function(data) {
//alert("Edited");
alert(data);
}
});
in you php use :
$arr = json_decode($_POST);
//some php code
//some $result;
// if $result arr then
echo json_encode($result);
// else
echo json_encode(array('data' => $result))
hope this helps...
You need to first turn that array into a string in JS. Before sending it, do this:
JSON.stringify(product_id);
Once you receive it in PHP, you need to decode it.
$decoded = json_decode($_POST['product_id'])
Related
I am making a form to check a security code. In fact, I am new to ajax and jquery, so I tried what I can, but my code doesn't work. Can anybody help me?
php file :
<?php
include('/includes/db-connect.php');
if( isset($_POST["seccode"]) ){
$result=mysqli_query($con,"SELECT * FROM `certificate_acheived_tbl` WHERE `cert_check_code` = ".$seccode.")";
if( mysql_num_rows($result) == 1) {
echo "<script>alert('s')";
}
}
?>
js file:
$(function() {
$(".btn btn-success").click(function() {
var ID = $(this).attr('id');
$.ajax({
type: "POST",
url: "cert-check-ajax.php",
data: 'certcode='+ ID,
success: function() {
$('#someHiddenDiv').show();
console.log();
}
});
});
});
your code is bad... (sometimes mine too)
One first mistake : data: 'certcode='+ ID, in jQuery
and isset($_POST["seccode"]) in PHP 'certcode' != 'seccode'
so a better code.. ?
jQuery (I allways use JSON, it's more easy)
$(function () {
$(".btn btn-success").click(function() {
var
Call_Args = {
certcode: $(this).attr('id')
};
$.ajax({
url: 'cert-check-ajax.php',
type: 'POST',
data: Call_Args,
cache: false,
dataType: 'json',
success: function (data) {
console.log( data.acceptCod ); // or data['acceptCod'] if you want
$('#someHiddenDiv').show();
// ...
}
}); //$.ajax
}); // btn btn-success").click
});
PHP (with utf8 insurance, and good header / JSON encode response )
<?php
mb_internal_encoding("UTF-8");
include('/includes/db-connect.php');
$T_Repons['acceptCod'] = "bad";
if (isSet($_POST['certcode'])) {
$sql = "SELECT * FROM `certificate_acheived_tbl` ";
$sql .= "WHERE `cert_check_code` = ".$_POST['certcode'].")";
$result = mysqli_query($con, $sql);
$T_Repons['acceptCod'] = (mysql_num_rows($result) == 1) ? "ok" : "bad";
}
header('Content-type: application/json');
echo json_encode($T_Repons);
exit(0);
?>
you can use it
$(function() {
$(".btn btn-success").click(function() {
var ID = $(this).attr('id');
$.ajax({
type: "POST",
url: "cert-check-ajax.php",
data: {'seccode': ID}
}).done(function(data) {
$('#someHiddenDiv').show();
console.log(data);
});
});
});
My script code like :
function changePrice(id) {
var url = '<?php echo $base_url ?>home/getprice/';
$.ajax({
url: url,
type: 'post',
data: 'id='+id,
success: function(msg) {
alert(msg);
/*
"regular_price": "800",
"discount_price": 720
*/
}
});
}
I want to both regular price and discount price on separate variable.How??
If you are getting the response as "regular_price": "800", "discount_price": 720 then make it valid JSON, parse it and get the properties.
var obj = JSON.parse('{' + msg + '}');
// valid json -^-----------^-
// get object properties
var regular = data.regular_price;
var discount = data.discount_price;
UPDATE : If response data is valid JSON format then set dataType: 'json' option.
$.ajax({
url: url,
type: 'post',
data: 'id='+id,
// set response datatype as json
dataType:'json',
success: function(msg) {
// get properties
var regular = msg.regular_price;
var discount = msg.discount_price;
}
});
Or parse it directly if the response is a string.
$.ajax({
url: url,
type: 'post',
data: 'id='+id,
success: function(msg) {
// parse the string
var data = JSON.parse(msg);
// get properties
var regular = data.regular_price;
var discount = data.discount_price;
}
});
Thanx to all..Here are the solution :
<script>
function changePrice(id)
{
var url = '<?php echo $base_url ?>home/getprice/';
$.ajax({
url:url,
type:'post',
data:'id='+id,
dataType:'json',
success:function(msg)
{
var regular = msg.regular_price;
var discount = msg.discount_price;
}
});
}
</script>
My Function :
$new = array (
"regular_price" => $result->price,
"discount_price" => $price
);
$newarray = json_encode($new, JSON_PRETTY_PRINT);
print_r($newarray);
Try this:
In server side ajax call:
$respose['regular_price'] = 120;
$respose['discount_price'] = 100;
echo json_encode($response);
In JS: Considering msg is an json object
var data = JSON.parse(msg);
var regular = data.regular_price;
var discount = data.discount_price;
I have a make/model/engine search form, the user selects the make which then populates the model, the user selects the model and it populates the engine. The problem I have encountered is that several of the manufacturers (make) use the exact same model. The script I have chooses the engine based on the model only. I would like to modify the script so it chooses the engine based on the make AND model, this would resolve my problem. I am somewhat familiar with javascript but I am no expert, I see the ajax requests in the aircraftMakeModel.php file but do not know how to add the make to the query. I have included the three files used below. Any help is appreciated in advance.
Thanks
Tom
aircraftMakeModel.php
<script type="text/javascript">
$(document).ready(function()
{
$('#aircraftMake').change(function()
{
var make=$(this).val();
var dataString = 'make='+ make;
$.ajax
({
type: "POST",
url: "include/getAirFrame.php",
data: dataString,
cache: false,
success: function(html)
{
$('#aircraftModel').html(html);
}
});
});
});
$(document).ready(function()
{
$('#aircraftModel').change(function()
{
var model=$(this).val();
var dataString = 'model='+ model;
$.ajax
({
type: "POST",
url: "include/getEngine.php",
data: dataString,
cache: false,
success: function(html)
{
$('#engineModel').html(html);
}
});
});
});
</script>
getAirFrame.php
<?php
include "../connection.php";
$q = $_POST['make'];
$q = addslashes($q);
$rs=mysqli_query($link,"SELECT DISTINCT(`aircraftModel`) FROM `aircraftData` WHERE `aircraftMake` = '$q' ORDER BY aircraftModel ; ");
echo '<option value="0">Aircraft Model</option>';
while($data = mysqli_fetch_row($rs)){
$sa=$data[0];
echo '<option value="'.$sa.'">'.$sa.'</option>';
?>
<?php } ?>
getEngine.php
<?php
include "../connection.php";
$q = $_POST['model'];
$q = addslashes($q);
$rs=mysqli_query($link,"SELECT DISTINCT(`engineModel`) FROM `aircraftData` WHERE `aircraftModel` = '$q' ORDER BY engineModel");
echo '<option value="0">Engine Model</option>';
while($data = mysqli_fetch_row($rs)){
$sa=$data[0];
echo '<option value="'.$sa.'">'.$sa.'</option>';
?>
<?php } ?>
If you want to send the make and model on the ajax call to get the engine something like this should work. Make the call to get the model as you do, then also add the make to the ajax request data to get the engine.
Note: not sure if this is a typo $('#marke')
$(document).ready(function(){
$('#marke').change(function() {
//make id
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax({
type: "POST",
url: "include/getph.php",
data: dataString,
cache: false,
success: function(html) {
$('#model').html(html);
}
});
});
});
$(document).ready(function() {
$('#model').change(function(){
//make id
var id = $('#marke option:selected').val();
//model id
var id1=$(this).val();
var dataString = 'id1='+ id1 + '&id=' + id;
$.ajax({
type: "POST",
url: "include/getph2.php",
data: dataString,
cache: false,
success: function(html) {
$('#engine').html(html);
}
});
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
I am getting this error using this code, the php id variable is set, why does it show unexpected token?
var current_page = 1;
var id = <?php echo $id; ?>;
$(document).ready(function(){
$.ajax({
'url':'get_data.php',
'type':'post',
'data': 'p='+current_page, 'id='+id,
success:function(data){
var data = $.parseJSON(data);
$('#posts').html(data.html);
$('#pagination').html(data.pagination);
}
console.log(data);
});
});
You should concatenate the data query strings properly:
'data': 'p='+current_page +'&id='+id,
Or using this interface:
data: {p : current_page, id: id},
So it would look like this:
<script type="text/javascript">
var current_page = 1;
var id = <?php echo $id; ?>;
$(document).ready(function(){
$.ajax({
url: 'get_data.php',
type :'POST',
data: 'p='+current_page+'&id='+id,
// data: {p: current_page, id: id},
success:function(data){
var data = $.parseJSON(data);
$('#posts').html(data.html);
$('#pagination').html(data.pagination);
console.log(data);
}
});
});
</script>
Sidenote: You could also explicitly set dataType: 'JSON', so that you would not need $.parseJSON at all.
Wrong syntax for data
replace with following
'data': 'p='+current_page+'&id='+id,
you can do it like this
data: {
p: current_page,
id: id
}
It won't give me the value from $scoresArray in php to data_response in js
If I log (obj) it returns nothing.
I have this in JS
$.ajax({
url: "Database.php",
type: "POST",
dataType: "json",
success: function (obj) {
stageRef.$("txtTopscorePunt").html(obj.score[0]);
stageRef.$("txtTopscorePunt2").html(obj.score[1]);
stageRef.$("txtTopscorePunt3").html(obj.score[2]);
}
});
This in php:
function GetScores(){
$query = "SELECT * FROM topscores ORDER BY Scores DESC LIMIT 3";
$result = mysql_query($query);
$scoresArray = array();
$i = 0;
while ($row = mysql_fetch_assoc($result)) {
$scoresArray[$i]['score'] = $row['score'];
$i++;
}
echo json_encode($scoresArray);
}
You don't need to JSON.parse(str) with dataType:json -- it's already parsed.
In addition, your selectors are wrong. If you want to select an ID, you need the # character.
$.ajax({
url: "Database.php",
type: "POST",
dataType: "json",
success: function (obj) { // not 'complete'
console.log(obj);
$("#txtTopscorePunt").html(obj.score[0]);
$("#txtTopscorePunt2").html(obj.score[1]);
$("#txtTopscorePunt3").html(obj.score[2]);
}
});
http://api.jquery.com/jQuery.ajax/
update the code to this:
$.ajax({
url: "Database.php",
type: "POST",
dataType: "json",
success: function (data_response) {
console.log(data_response);
var response = data_response;
var response_str = JSON.stringify(response);
alert(response_str); // don't do this if the returned is too much
console.log(response_str); // use your browser console to view the results
}
});
Once you have done this, can you post the results of alert or console log.
Also, in the results does it start with "[" or "{"?
From the results we should be able to help you with the correct way to access your JSON object.
Regards.