Download Php Converter To Jsonbackstage



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  1. Php Json To String
  2. Php Json File

When it comes to thinking about JSON with PHP and MySQL, it can be rather daunting. This is the precise reason for writing this tutorial, to ultimately demystify the subject. After running a query and grabbing data from a MySQL result set, you can convert the data to JSON using a PHP function named jsonencode. Decoding JSON Data in PHP. Decoding JSON data is as simple as encoding it. You can use the PHP jsondecode function to convert the JSON encoded string into appropriate PHP data type. The following example demonstrates how to decode or convert a JSON object to PHP object. Download Free MP4 to MP3 Converter (64-bit) for Windows to convert M4A, MP4, M4B, M4R, and AAC files to MP3 or WAV formats. Free MP4 to MP3 Converter (64-bit) has had 0 updates within the past 6. Free online JSON to XML converter. Just paste your JSON in the input form below and it will automatically get converted to XML. There are no ads, popups or nonsense, just an awesome JSON to XML transformer.

A common use of JSON is to read data from a web server, and display the data in a web page.

This chapter will teach you how to exchange JSON data between the client and a PHP server.

The PHP File

PHP has some built-in functions to handle JSON.

Objects in PHP can be converted into JSON by using the PHP function json_encode():

PHP file

Convert json to php array
<?php
$myObj->name = 'John';
$myObj->age = 30;
$myObj->city = 'New York';
$myJSON = json_encode($myObj);
echo $myJSON;
?>
Show PHP file »

The Client JavaScript

Here is a JavaScript on the client, using an AJAX call to request the PHP file from the example above:

Example

Use JSON.parse() to convert the result into a JavaScript object:

var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState 4 && this.status 200) {
var myObj = JSON.parse(this.responseText);
document.getElementById('demo').innerHTML = myObj.name;
}
};
xmlhttp.open('GET', 'demo_file.php', true);
xmlhttp.send();
Try it Yourself »

PHP Array

Arrays in PHP will also be converted into JSON when using the PHP function json_encode():

PHP file

<?php
$myArr = array('John', 'Mary', 'Peter', 'Sally');
$myJSON = json_encode($myArr);
echo $myJSON;
?>
Show PHP file »

The Client JavaScript

Here is a JavaScript on the client, using an AJAX call to request the PHP file from the array example above:

Example

Use JSON.parse() to convert the result into a JavaScript array:

var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState 4 && this.status 200) {
var myObj = JSON.parse(this.responseText);
document.getElementById('demo').innerHTML = myObj[2];
}
};
xmlhttp.open('GET', 'demo_file_array.php', true);
xmlhttp.send();
Try it Yourself »

PHP Database

Php add to json object

PHP is a server side programming language, and can be used to access a database.

Imagine you have a database on your server, and you want to send a request to it from the client where you ask for the 10 first rows in a table called 'customers'.

On the client, make a JSON object that describes the numbers of rows you want to return.

Before you send the request to the server, convert the JSON object into a string and send it as a parameter to the url of the PHP page:

Example

Use JSON.stringify() to convert the JavaScript object into JSON:

obj = { 'limit':10 };
dbParam = JSON.stringify(obj);
xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState 4 && this.status 200) {
document.getElementById('demo').innerHTML = this.responseText;
}
};
xmlhttp.open('GET', 'json_demo_db.php?x=' + dbParam, true);
xmlhttp.send();
Try it Yourself »

Example explained:

  • Define an object containing a 'limit' property and value.
  • Convert the object into a JSON string.
  • Send a request to the PHP file, with the JSON string as a parameter.
  • Wait until the request returns with the result (as JSON)
  • Display the result received from the PHP file.

Take a look at the PHP file:

PHP file

<?php
header('Content-Type: application/json; charset=UTF-8');
$obj = json_decode($_GET['x'], false);
$conn = new mysqli('myServer', 'myUser', 'myPassword', 'Northwind');
$stmt = $conn->prepare('SELECT name FROM customers LIMIT ?');
$stmt->bind_param('s', $obj->limit);
$stmt->execute();
$result = $stmt->get_result();
$outp = $result->fetch_all(MYSQLI_ASSOC);
echo json_encode($outp);
?>

PHP File explained:

Php
  • Convert the request into an object, using the PHP function json_decode().
  • Access the database, and fill an array with the requested data.
  • Add the array to an object, and return the object as JSON using the json_encode() function.

Loop Through the Result

Convert the result received from the PHP file into a JavaScript object, or in this case, a JavaScript array:

Example

Use JSON.parse() to convert the JSON into a JavaScript object:

...
xmlhttp.onreadystatechange = function() {
if (this.readyState 4 && this.status 200) {
myObj = JSON.parse(this.responseText);
for (x in myObj) {
txt += myObj[x].name + '<br>';
}
document.getElementById('demo').innerHTML = txt;
}
};
...
Try it Yourself »

PHP Method = POST

When sending data to the server, it is often best to use the HTTP POST method.

To send AJAX requests using the POST method, specify the method, and the correct header.

The data sent to the server must now be an argument to the send() method:

Example

obj = { 'limit':10 };
dbParam = JSON.stringify(obj);
xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState 4 && this.status 200) {
myObj = JSON.parse(this.responseText);
for (x in myObj) {
txt += myObj[x].name + '<br>';
}
document.getElementById('demo').innerHTML = txt;
}
};
xmlhttp.open('POST', 'json_demo_db_post.php', true);
xmlhttp.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
xmlhttp.send('x=' + dbParam);
Try it Yourself »

The only difference in the PHP file is the method for getting the transferred data.

Download Php Converter To Jsonbackstage

PHP file

Use $_POST instead of $_GET:

<?php
header('Content-Type: application/json; charset=UTF-8');
$obj = json_decode($_POST['x'], false);

$conn = new mysqli('myServer', 'myUser', 'myPassword', 'Northwind');
$stmt = $conn->prepare('SELECT name FROM customers LIMIT ?');
$stmt->bind_param('s', $obj->limit);
$stmt->execute();
$result = $stmt->get_result();
$outp = $result->fetch_all(MYSQLI_ASSOC);
echo json_encode($outp);
?>

Php Json To String


Php Json File