My js:
$('#select').change(function() {
var name = $(this).val();
$.ajax({
type: "POST",
url: "data/grab.php",
data: { type: "hops", name: name },
dataType: "json",
success: function(data) {
alert(data);
var aa = data['aa'];
$('#hops-aa').val(aa);
}
});
});
grab.php
<?php
$type = $_POST['type'];
$name = $_POST['name'];
if ($type == 'hops') {
$result = $name;
}
$result = json_encode($result);
return $result;
I added the alert() in the ajax call to double check what I'm getting back from the script, and it's always null. Anything I'm missing?
You need to actually echo or print the $result. In PHP using return from the file scope does not send the returned value to the output stream.
You use echo to print out the results in PHP, not return:
echo $result;
(As previous answers stated)
You need to echo/print the $result variable.
<?php
$type = $_POST['type'];
$name = $_POST['name'];
if ($type == 'hops') {
$result = $name;
}
$result = json_encode($result);
echo $result; // return $result;
?>
Related
So I have this strip of code working for one page and not working for another page... Same code, same form name attributes, same php functionality on the server side. I'm calling with jquery ajax.. Calling from one page, it works completely fine... When I call from another page, it gives me "Unidentified index"
I've tried changing the names around (to remove duplicates and all)..
I seem to be sure that the names are correct though.
<script type="text/javascript">
$('#uploadcsv').submit(function(e){
e.preventDefault();
$.ajax({
url: "functions/grocerystore.php?item=fetchcontacts",
method: "post",
data: new FormData(this),
dataType: 'json',
contentType: false,
cache: false,
processData: false,
success: function(data){
console.log(data);
var accumulator = "";
var filetype = $('#uptype').val();
if(filetype == "csv"){
for(trav=0; trav<data.length; trav++){
accumulator += data[trav];
if(trav < data.length-1)
accumulator += "\r\n";
}
}else{
for(trav=0; trav<data.length; trav++){
accumulator += data[trav];
}
}
var mobilenumbers = $('#mobilenumbers');
mobilenumbers.html(accumulator);
},
error: function(data){
console.log(data);
}
});
});
</script>
php
case 'fetchcontacts':
if(!empty($_FILES['customFilex']['name'])){
$name = explode('.',$_FILES['customFilex']['name']);
$extension = end($name);
$file_data = fopen($_FILES['customFilex']['tmp_name'],"r");
if($extension == 'csv'){
fgetcsv($file_data);
$mobile = array();
$init = 0;
while($row = fgetcsv($file_data)){
$mobile[$init] = $row[0];
$init++;
}
fclose($file_data);
}else{
if ($file_data = fopen($_FILES['customFilex']['tmp_name'], 'r')) {
$mobile = array();
$init = 0;
while (!feof($file_data)) {
$row = fgets($file_data);
$mobile[$init] = $row;
$init++;
}
fclose($file_data);
}
}
echo json_encode($mobile);
}else{
$report["status"] = "failed";
echo json_encode($report);
}
break;
I just need it to accept my file...
You are using $_FILES['customFilex']['tmp_name']without checking if the keys exists in the $_FILES array.
$.ajax({
type: "GET",
url: 'http://localhost/abc/all-data.php',
data: {
data1: "1"},
success: function(response)
{
alert(response);
}
});
return false;
I want to display each element of array one by one in success function of the ajax currently i get all elements to gether
this is my php code
$i=0;
while($row = mysqli_fetch_assoc( $qry )){
$temp[$i]['c_n'] = $row['c_name'];
$temp[$i]['j_t'] = $row['Job_Title'];
$temp[$i]['des'] = $row['description'];
$temp[$i]['req'] = $row['requirments'];
$temp[$i]['dat'] = $row['posted'];
$i++;
}
$data = array('temp'=> $temp);
echo JSON_encode($temp);
I do appreciate your helps
you probably use something like this in your success function :
response.temp.forEach(function(element){
console.log(element.c_n) ;
console.log(element.j_t) ;
console.log(element.des) ;
console.log(element.req) ;
console.log(element.dat) ;
});
In your success function, you need to json parse your response
var data = JSON.parse(response);
You can access to your data:
data['temp']
If you want your response parsed to json automaticallym you can setup your ajax settings like this:
$.ajaxSetup ({
contentType: "application/json",
dataType: 'json'
});
Then you don't need to call JSON.parse anymore.
Your code:
$i=0; while($row = mysqli_fetch_assoc($qry)){
$temp[$i]['c_n'] = $row['c_name'];
$temp[$i]['j_t'] = $row['Job_Title'];
$temp[$i]['des'] = $row['description'];
$temp[$i]['req'] = $row['requirments'];
$temp[$i]['dat'] = $row['posted'];
$i++;
} $data = array('temp'=> $temp); echo JSON_encode($temp);
Please change the last line as
return JSON_encode($data);
Hope this helps you :)
I'm creating an ajax script to update a few fields in the database. I got it to a point where it worked but it sent the user to the php script instead of staying on the page so I did some googling, and people suggested using either return false; or e.preventDefault() however, if I do this, it breaks the php script on the other page and returns a fatal error. I might be missing something being newish to AJAX but it all looks right to me
JS:
$(document).ready(function() {
var form = $('form#edit_child_form'),
data = form.serializeArray();
data.push({'parent_id': $('input[name="parent_id"]').val()});
$('#submit_btn').on('click', function(e) {
e.preventDefault();
$.ajax({
url: form.prop('action'),
dataType: 'json',
type: 'post',
data: data,
success: function(data) {
if (data.success) {
window.opener.$.growlUI(data.msg);
}
},
error: function(data) {
if (!data.success) {
window.opener.$.growlUI(data.msg);
}
}
});
});
})
AJAX:
<?php
//mysql db vars here (removed on SO)
$descriptions = $_GET['descriptions'];
$child_id = $_GET['child_id'];
$parent_id = $_GET['parent_id'];
$get_child_ids = $dbi->query("SELECT child_ids FROM ids WHERE parent = ". $parent_id ." ORDER BY id"); //returns as object
$count = 0;
$res = array();
while ($child_row = $get_child_ids->fetch_row())
{
try
{
$dbi->query("UPDATE ids SET description = '$descriptions[$count]', child_id = '$child_id[$count]' WHERE parent_id = $child_row[0]");
$res['success'] = true;
$res['msg'] = 'Success! DDI(s) updated';
} catch (Exception $e) {
$res['success'] = true;
$res['msg'] = 'Error! '. $e->getMessage();
}
$count++;
}
echo json_encode($res);
it's probably something really small that I've just missed but not sure what - any ideas?
my solution:
I var_dumped $_GET and it returned null - changed to $_REQUEST and it got my data so all good :) thanks for suggestions
Try the following instead.
I moved the form data inside click and enclosed the mysql queries values in single quotes.
JS:
$(document).ready(function() {
var form = $('form#edit_child_form');
$('#submit_btn').on('click', function(e) {
e.preventDefault();
var data = form.serializeArray();
data.push({'parent_id': $('input[name="parent_id"]').val()});
$.ajax({
url: form.prop('action'),
dataType: 'json',
type: 'get',
data: data,
success: function(data) {
if (data.success) {
window.opener.$.growlUI(data.msg);
}
},
error: function(data) {
if (!data.success) {
window.opener.$.growlUI(data.msg);
}
}
});
});
})
AJAX:
<?php
//mysql db vars here (removed on SO)
$descriptions = $_GET['descriptions'];
$child_id = $_GET['child_id'];
$parent_id = $_GET['parent_id'];
$get_child_ids = $dbi->query("SELECT child_ids FROM ids WHERE parent = '". $parent_id ."' ORDER BY id"); //returns as object
$count = 0;
$res = array();
while ($child_row = $get_child_ids->fetch_row())
{
try
{
$dbi->query("UPDATE ids SET description = '$descriptions[$count]', child_id = '$child_id[$count]' WHERE parent_id = '$child_row[0]'");
$res['success'] = true;
$res['msg'] = 'Success! DDI(s) updated';
} catch (Exception $e) {
$res['success'] = true;
$res['msg'] = 'Error! '. $e->getMessage();
}
$count++;
}
echo json_encode($res);
You are using an AJAX POST request so in your PHP you should be using $_POST and not $_GET.
You can just change this:
$descriptions = $_GET['descriptions'];
$child_id = $_GET['child_id'];
$parent_id = $_GET['parent_id'];
to this:
$descriptions = $_POST['descriptions'];
$child_id = $_POST['child_id'];
$parent_id = $_POST['parent_id'];
I'm not sure how to pass the result of mysql query into html page via ajax JSON.
ajax2.php
$statement = $pdo - > prepare("SELECT * FROM posts WHERE subid IN (:key2) AND Poscode=:postcode2");
$statement - > execute(array(':key2' => $key2, ':postcode2' => $postcode));
// $row = $statement->fetchAll(PDO::FETCH_ASSOC);
while ($row = $statement - > fetch()) {
echo $row['Name']; //How to show this in the html page?
echo $row['PostUUID']; //How to show this in the html page?
$row2[] = $row;
}
echo json_encode($row2);
How to pass the above query result to display in the html page via ajax below?
my ajax
$("form").on("submit", function () {
var data = {
"action": "test"
};
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "ajax2.php", //Relative or absolute path to response.php file
data: data,
success: function (data) {
//how to retrieve the php mysql result here?
console.log(data); // this shows nothing in console,I wonder why?
}
});
return false;
});
Your json encoding should be like that :
$json = array();
while( $row = $statement->fetch()) {
array_push($json, array($row['Name'], $row['PostUUID']));
}
header('Content-Type: application/json');
echo json_encode($json);
And in your javascript part, you don't have to do anything to get back your data, it is stored in data var from success function.
You can just display it and do whatever you want on your webpage with it
header('Content-Type: application/json');
$row2 = array();
$result = array();
$statement = $pdo->prepare("SELECT * FROM posts WHERE subid IN (:key2) AND Poscode=:postcode2");
$statement->execute(array(':key2' => $key2,':postcode2'=>$postcode));
// $row = $statement->fetchAll(PDO::FETCH_ASSOC);
while( $row = $statement->fetch())
{
echo $row['Name'];//How to show this in the html page?
echo $row['PostUUID'];//How to show this in the html page?
$row2[]=$row;
}
if(!empty($row2)){
$result['type'] = "success";
$result['data'] = $row2;
}else{
$result['type'] = "error";
$result['data'] = "No result found";
}
echo json_encode($row2);
and in your script:
$("form").on("submit",function() {
var data = {
"action": "test"
};
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "ajax2.php", //Relative or absolute path to response.php file
data: data,
success: function(data) {
console.log(data);
if(data.type == "success"){
for(var i=0;i<data.data.length;i++){
//// and here you can get your values //
var db_data = data.data[i];
console.log("name -- >" +db_data.Name );
console.log("name -- >" +db_data.PostUUID);
}
}
if(data.type == "error"){
alert(data.data);
}
}
});
return false;
});
In ajax success function you can use JSON.parse (data) to display JSON data.
Here is an example :
Parse JSON in JavaScript?
you can save json encoded string into array and then pass it's value to javascript.
Refer below code.
<?php
// your PHP code
$jsonData = json_encode($row2); ?>
Your JavaScript code
var data = '<?php echo $jsonData; ?>';
Now data variable has all JSON data, now you can move ahead with your code, just remove below line
data = $(this).serialize() + "&" + $.param(data);
it's not needed as data variable is string.
And in your ajax2.php file you can get this through
json_decode($_REQUEST['data'])
I would just..
$rows = $statement->fetchAll(FETCH_ASSOC);
header("content-type: application/json");
echo json_encode($rows);
then at javascript side:
xhr.addEventListener("readystatechange",function(ev){
//...
var data=JSON.parse(xhr.responseText);
var span=null;
var i=0;
for(;i<data.length;++i){span=document.createElement("span");span.textContent=data[i]["name"];div.appendChild(span);/*...*/}
}
(Don't rely on web browsers parsing it for you in .response because of the application/json header, it differs between browsers... do it manually with responseText);
Here is my HTML
<input x-webkit-speech id="mike" name="string" style="position: relative;" disabled lang="ru" />
Then when the field is changes,
This function executes
$(document).ready(function(){
$('#mike').bind('webkitspeechchange',function()
{
a= $(this).val();
recognizeAjax(a);
}) ;
});
function recognizeAjax(string) {
var postData ="string="+string;
$.ajax({
type: "POST",
dataType: "json",
data: postData,
beforeSend: function(x) {
if(x && x.overrideMimeType) {
x.overrideMimeType("application/json;charset=UTF-8");
}
},
url: 'restURL.php',
success: function(data) {
// 'data' is a JSON object which we can access directly.
// Evaluate the data.success member and do something appropriate...
if (data.success == true){
alert(data.message);
}
else{
alert(data.message+'hy');
}
}
});
And here is my PHP (please don't say anything about the way i connect to DB it doesn't metter right now)
<?php header('Content-type: application/json; charset=utf-8');
error_reporting(E_ALL);
ini_set('display_errors', true);
// Here's the argument from the client.
$string = $_POST['www'];
$quest=1;
$con=mysql_connect("localhost", "******", "*********") or die(mysql_error());
mysql_select_db("vocabulary", $con) or die(mysql_error());
mysql_set_charset('utf8', $con);
$sql="SELECT * FROM `text` WHERE event_name = 'taxi' AND quest_id = '".$quest."'";
$result = mysql_query($sql);
mysql_close($con);
while($row = mysql_fetch_array($result))
{
if ($string == htmlspecialchars($row['phrase']))
{
$data = array('success'=> true,'message'=>$row['phrase']);
// JSON encode and send back to the server
header("Content-Type: application/json", true);
echo json_encode($data);
exit;
break;
} else {
// Set up associative array
$data = array('success'=> false,'message'=>'aint no sunshine');
header("Content-Type: application/json", true);
echo json_encode($data);
exit;
break;
}
}
When i change the dataType to "text" in the javasript function - i receive an alert with 'undifiend'
But when chenge it to 'json'.. i receive nothing (chrome debuger see nothing)
I set up all encodings to this article http://kunststube.net/frontback/
And i checked it with simple POST requests - it works perfect.
The problem with json.
Any suggestions?
Thanks
Just remove the datatype="json" bit and change the data bit to data: { "string": string }
After that try a print_r(json_decode($_POST['string']));. I'm quite sure that will get you your data.
And indeed remove your beforeSend callback.
I think the prob is the code var postData ="string="+string;
jQuery expects this to be a proper JSON Object.
Next: $string = $_POST['www']; takes a parameter named "www" from your post request, but the name above is "string" (at least).
Try either (!) this:
var getData ="www="+string;
$.ajax({
type: "POST",
dataType: "json",
data: null,
beforeSend: function(x) {
if(x && x.overrideMimeType) {
x.overrideMimeType("application/json;charset=UTF-8");
}
},
url: 'restURL.php?' + getData,
and server:
$string = $_GET['www'];
or this (php)
$string = $_POST['string'];
$stringData = json_decode($string);
// catch any errors ....
$quest=$stringData[....whatever index that is...];