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why don't save data from mysql [closed]

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hey i want to crawl data from the site with this code (js)
var i = 0
var oldValue = -1
var interval = setInterval(get, 3000);
function get(){
var x= $($('.table-body')[1]).find('.h-col-1')
if(i!=5){
if(oldValue != x){
oldValue = $(x[1]).text()
console.log($(x[1]).text())
++i
sendPost($(x[1]).text())
}
}else clearInterval(interval)
}
function sendPost(par) {
var te= $.ajax({
type: "POST",
dataType: 'jsonp',
url: 'http://localhost/sa.php',
data: {json:JSON.stringify({
num : par
}) },
}).done(function( data ) {
console.log("done");
})
.fail( function(xhr, textStatus, errorThrown) {
console.log(xhr.responseText);textStatus
});
console.log(te)
send data from localhost and save on my sql and this php code
sa.php
<?php
header('Access-Control-Allow-Origin: *');
$sitename = "http://localhost/";
$hostname = "localhost";
$username = "root";
$password = "";
$database = "db";
$connect = mysqli_connect($hostname, $username, $password, $database);
$insert = "INSERT INTO main(number, time, date) VALUES ('%s','%s','%s','')";
$data = json_decode(file_get_contents("php://input"), true);
$result = mysqli_query($connect,sprintf($insert,$data['num'] ,date("H:i:s"),date("Y-m-d")));
?>
and then
success send from js request from local server why dosen't save on the my sql ??

I think the answer lies into your insert query:
$insert = "INSERT INTO main(number, time, date) VALUES ('%s','%s','%s','')";
Please count the number of columns defined and the number of values you're trying to insert. These two should, obviously, match.
If you wish to figure out why a MySQL query failed, just print or log the error:
http://php.net/manual/en/mysqli.error.php

You should check the value of file_get_contents("php://input").
As you are sending a key-value pair:
data: {json:JSON.stringify({
num : par
}) },
you should get the data using $_POST['json'].
So:
$data = json_decode($_POST['json'], true);
The alternative, change the javascript instead of the php, would be something like:
data: JSON.stringify({
num : par
}),
But to be honest, I don't know how jQuery will handle that data so it might need some tweaking.
Apart from that, you have an sql injection problem; you should use prepared statements with bound parameters instead.

Unexpected end of JSON input

I'm having a little trouble figuring out how to fix this error I'm getting. My code is as follows.
It all starts with a AJAX request whenever the user moves their mouse on the webpage.
$('body').mouseover(function() {
$.ajax({
url: '/core/home.php',
data: {action: 'refresh'},
type: 'post',
Next, the PHP file (home.php) executes a couple methods to get all the needed data and sends it back to AJAX Request.
require_once 'init.php';
if(isset($_POST['action']) && !empty($_POST['action'])) {
// Home Class
$h = new Home();
// Method to count all "vacs"
$h->getVacs('id');
$vacs = $h->count();
// Method to count all "users"
$h->getUsers('id');
$users = $h->count();
// Create array to store all data
$arr = array();
$arr[] = $vacs;
$arr[] = $users;
// Use JSON to send the array back
json_encode($arr);
return $arr;
}
Once the AJAX Request receives a success, the following executes
success: function(output) {
obj = JSON.parse(output);
// Separate the parts of the JSON string
vacs = obj[0];
users = obj[1];
// Show the result at the correct position on the webpage
$('#vac_num').html(vacs);
if(vacs == 1) $('#vac_txt').html('is'); else $('#vac_txt').html('zijn');
$('#users_num').html(users);
if(users == 1) $('#users_txt').html('is'); else $('#users_txt').html('zijn');
}
});
});
Unfortunately this code results into an error: Unexpected end of JSON input.
Any help is much appreciated.

Rather than returning variable you need to echo it
require_once 'init.php';
if(isset($_POST['action']) && !empty($_POST['action'])) {
// Home Class
$h = new Home();
// Method to count all "vacs"
$h->getVacs('id');
$vacs = $h->count();
// Method to count all "users"
$h->getUsers('id');
$users = $h->count();
// Create array to store all data
$arr = array();
$arr[] = $vacs;
$arr[] = $users;
// Use JSON to send the array back
echo json_encode($arr);
}

Cant get PHP variables using AJAX

I can't seem to figure out what's the problem. For my next little project I'm creating dynamic web page with database and etc. I need to get all the necessary variables from PHP file. But for some reason I cannot do it if I include another PHP file. (I need it for database queries).
main.php
include ('databaseQueries.php');
if (isset($_POST["points"])){
$points = json_decode($_POST["points"]);
if($_POST["execute"] == 1){
}
}
$advert= array(
'Hello' => 'Hello world!',
'bye' => 'Why bye?',
);
echo json_encode($advert, $another);
pageJs.js
$.ajax({
url : 'php/main.php',
type : 'POST',
dataType : 'json',
success : function (result) {
console.log(result);
},
error : function (err) {
console.log("Failed");
}
})
databaseQueries.php
$another = "another string";
If I remove the include and $another variable from json_encode. Everything works and I get object in console log. But if I leave the those two things, Ajax call fails.
What I'm doing wrong and how can I get both the $test array and $another variable?
Thank's in advance!

You are using json_encode incorrectly. From the documentation:
string json_encode ( mixed $value [, int $options = 0 [, int $depth = 512 ]] )
You are trying to send $another to the function as the options parameter.
You could do something like this:
$ret = array($advert, $another);
echo json_encode($ret);

Unless I'm completely wrong, I can't see where you're sending anything TO your post
$.ajax({
url : 'php/main.php',
type : 'POST',
dataType : 'json'
// I would expect a data call here, something like:
data: $(form).serialize(), // OR { points: 3, execute: 1 }
success : function (result) {
console.log(result);
},
error : function (err) {
console.log("Failed");
}
})
I assume that you want to spit back some results with the format of result->key;
So Keeleon's answer above is good:
$ret = array($advert, $another);
echo json_encode($ret);
But you can also do:
//Adding everything to array, and asking json_encode to encode the array is the key. Json_encode encodes whatever is in the first argument passed to it.
$ret = array('advert'=>$advert, 'another'=>$another,'test'=>$test);
echo json_encode($ret);
Hopefully, this answers your questions.

response from php to javascript via json

I have this function in php
public function get_updated_session_value()
{
$sql = "SELECT IF(Session = 12345678 , 1,0) AS login FROM `psf_users` WHERE id = 236";
$var = $this->addDb($sql)->execute();
$Session = $var['login'];
return json_encode('2');
}
and the javascript code to fetch this value,
function check() {
$.ajax({
url : 'auctions.php',
type : 'get',
// dataType: 'json',
data : {page:'__request', module:'PSF_auctions', action:'get_updated_session_value'},
success: function(data) {
console.log(data);
}
});
}
also, this function runs every 5 seconds via
setInterval(check, 5000);
the problem is, console.log(data); prints nothing, i believe that means it is not getting any data (or json response) from the php function. am i missing something?

It can't work as you're returning the value. The difference between returning a value and emitting a response is that the latter writes a response on the HTTP stream whilst the first one merely returns the control to the parent with a specific value.
Sanjay has spotted it very well, and I'd recommend that you use a very simple function to wrap up your responses:
function emit_response( $status_code, $data ) {
http_response_code( $status_code ); // 200 is it's success. find more http://www.w3.org/Protocols/rfc2616/rfc2616-sec10.html
die( json_encode( array(
"status_code" => $status_code,
"data" => $data )));
}
Then modify that echo (though it's fine as well) with
emit_response( 2 );
and since the response is valid JSON, JavaScript will love it; in your callback, you can simple do this:
success: function(res) {
var response = JSON.parse( res );
// response.data will be 2. :)
// ... rest of the code ...

Replace return with echo like this:
public function get_updated_session_value()
{
$sql="SELECT IF(Session = 12345678 , 1,0) as login FROM `psf_users` WHERE id=236";
$var= $this->addDb($sql)->execute();
$Session= $var['login'];
echo json_encode('2');
}

PHP + Javascript parse both responses

I have two problems. One is totally in PHP the other in Javascript. But both are equal in what I'm trying to get.
index.php
$.ajax({
type: "post",
url: "insert_info.php?type=info",
data: { array : all },
success: function(data) {
alert(data);
// Returns: {'status':1}
// I want to get "1"
// data[0] -> doesn't work
}
});
insert_info.php
// Connects to another file
include_once('verify_info.php');
$verify = new verify_info();
$arr = array("status:" => 1);
$extension = $verify->verify_file($_REQUEST['array'][9]);
if($extension[0] == 0){
$arr = array("status:" => 0);
}
echo json_encode($arr);
verify_info.php
public function verify_file($file){
$extensions = array('jpg', 'png', 'jpeg', 'bmp');
$info = pathinfo($file);
$arr = array();
if(!in_array($info['extension'], $extensions)){
$arr = array("status:" => 0);
}else{
$arr = array("status:" => 1);
}
return $arr;
}
In insert_info.php, I would like to get by $extension[0] the status retrieved from the function verify_file();
After that I output as json_encode the value to Javascript and I would like, again, to parse the info.
What am I doing wrong? Thanks.
Edit 1: alert(data.status); doesn't work either.
Edit 2: alert(data.status); would never work since I echo {'status:', 1} (problem with that two points in the middle)
Correct way for solving the issue of javascript:
var obj = jQuery.parseJSON(data);
alert(data.status);
I'm still trying to get fixed the php.
Edit 3: All solved. Thank you guys.
public function verify_file($file){
$extensions = array('jpg', 'png', 'jpeg', 'bmp');
$info = pathinfo($file);
if(!in_array($info['extension'], $extensions)){
return false;
}
return true;
}

As I said in my comment, your setting your key in PHP to "status:" is the trailing colon necessary on the end of your key? I don't think it's necessary, PHP arrays already offer mechanisms for fetching them and your JSON will contain the string without processing it so your key once you hit your JS code will still be "status:" where most likely you intended for "status".
Regardless of whether you will make this change or not that doesn't break anything. In your Javascript code, as #charlietfl pointed out, you should be setting the dataType of the return to "json" so you're JS Ajax call would look like:
$.ajax({
type: "post",
url: "insert_info.php?type=info",
data: { array : all },
dataType: "json",
success: function(data) {
// Assuming no change on the backend
alert(data["status:"]);
}
});
If, however, you altered the string to remove the colon then accessing the status element of data would be data.status as #A.Wolff pointed out in his comment. This doesn't work because of the trailing colon in the key - but accessing the data with a string key will still work.