JSON to PHP API not happening

Question

I am trying to save data on click of button using Javascript and PHP.

Button click:

            var xhr = new XMLHttpRequest();
            var url = "savedata.php";
            xhr.open("POST", url, true);
            xhr.setRequestHeader("Content-Type", "application/json");
            /*xhr.onreadystatechange = function () {
                if (xhr.readyState === 4 && xhr.status === 200) {
                    var json = JSON.parse(xhr.responseText);
                    console.log(json.rawText + ", " + json.convertedText);
                }
            };*/
            var data = JSON.stringify({"rawText": rt, "convertedText": ct});
            xhr.send(data);

savedata.php

<?php
header("Content-Type: application/json");

require('db/db.php');
session_start();

    $saverawText = $_POST['rawText'];
    $saveConvertedText = $_POST['convertedText'];

    $ins_query="insert into cmn_data (`rawtext`,`convertedtext`) values ('$saverawText','$saveConvertedText')";
    $result = mysqli_query($con, $ins_query);

?>

In Db: One record is created but it is empty so every time I click save button 1 record is created but empty. What is the wrong the above syntax?

Thanks!

What is the error message or behavior you are seeing? What's in the console? What's in the error log? — pendo, May 09 '19 at 17:46
First https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?rq=1, also remove `header("Content-Type: application/json");` — AbraCadaver, May 09 '19 at 17:47
https://stackoverflow.com/questions/9713058/send-post-data-using-xmlhttprequest Check this out and see if it helps you understand how to send params via AJAX — pendo, May 09 '19 at 18:39

zoryamba · Accepted Answer · 2019-05-09T18:38:05.930

2

PHP does not parse JSON requests to $_POST array itself. You should either parse request body in php:

$request = json_decode(file_get_contents('php://input'), true);
$saverawText = $request['rawText'];
$saveConvertedText = $request['convertedText'];

or pass data to request with FormaData object:

var data = new FormData();
data.append("rawText", rt);
data.append("convertedText", ct);
xhr.send(data);

And as mentioned before, including user input directly into sql could cause sql injections. Use mysqli_real_escape_string to escape user inputs or PDO driver, which is bit more complicated but more reliable approach.

edited May 09 '19 at 18:38

answered May 09 '19 at 18:00

zoryamba

166
8

forgot to wrap `php://input` with `file_get_contents` and parse json as associative array. updated `$request = json_decode(file_get_contents('php://input'), true);` Try it, should work now – zoryamba May 09 '19 at 18:39

score 0 · Answer 2 · answered May 09 '19 at 17:57

You can POST your values like that :

var xhr = new XMLHttpRequest();
var url = "savedata.php";
xhr.open("POST", url, true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
/*xhr.onreadystatechange = function () {
    if (xhr.readyState === 4 && xhr.status === 200) {
        var json = JSON.parse(xhr.responseText);
        console.log(json.rawText + ", " + json.convertedText);
    }
};*/
var data = "rawText=" + rt + "&convertedText=" + ct;
xhr.send(data);

I can't send the data via JSON? – Sanjana Nair May 09 '19 at 18:31 — Sanjana Nair, May 09 '19 at 18:31

JSON to PHP API not happening

2 Answers2