Output a JSON in PHP generated on SQL SERVER with JSON AUTO - php

Question

I am trying to get the output of a JSON formatted on SQL SERVER through command FOR JSON AUTO
I need to execute the query in PHP on SQL SERVER and then output it as a legit JSON.
How should I proceed?
I normally use the code below to generate the JSON, but what if I need to get a JSON ?
$key= $_GET['key'];
$date=$_GET['date'];
$brand=$_GET['brand'];
if ($key=="...")
{
$serverName = "XXX,YYYY"; // \\MSSQLSERVER";
$connectionOptions = [
"Database" => "db",
"UID" => "user",
"PWD" => 'xxxx'
];
$conn = sqlsrv_connect($serverName, $connectionOptions);
if ($conn === false) {
die(formatErrors(sqlsrv_errors()));
}
$tsql = "select * from admin_all.Datafeed FOR JSON AUTO;";
// Executes the query
$stmt = sqlsrv_query($conn, $tsql);
// Error handling
if ($stmt === false) {
die(formatErrors(sqlsrv_errors()));
}
$array = array();
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
$array[]=$row;
}
echo json_encode(array("data"=>array_values($array)));
sqlsrv_free_stmt($stmt);
sqlsrv_close($conn);
}

Answer 1

The reason for the unexpected result is that json_encode() call is not needed (but it's always an option). FOR JSON AUTO returns a valid JSON, so you only need to echo the generated JSON. Note, that if you want to add a single, top-level element to the JSON output of the FOR JSON clause, you need to use the ROOT option.
Example, based on your attempt:
<?php
...
$json = '';
if ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_BOTH)) {
$json = $row[0];
}
echo '{"data":'.$json.'}';
...
?>
Example with json_encode():
<?php
...
$sql = "SELECT id FROM (VALUES (1), (2)) v(id)";
$stmt = sqlsrv_query($conn, $sql);
if($stmt == false){
die( print_r( sqlsrv_errors(), true) );
}
$json = array();
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
$json[] = $row;
}
echo json_encode(array("data" => array_values($json)));
...
?>
Example with FOR JSON AUTO and the ROOT option:
<?php
...
$sql = "SELECT id FROM (VALUES (1), (2)) v(id) FOR JSON AUTO, ROOT('data')";
$stmt = sqlsrv_query($conn, $sql);
if($stmt == false){
die( print_r( sqlsrv_errors(), true) );
}
$json = '';
if ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_NUMERIC)) {
$json = $row[0];
}
echo $json;
...
?>
Generated JSON:
{"data":[{"id":1},{"id":2}]}

Answer 2

PHP example;
// This file is to be called by AJAX
<?php
require_once('db.php');
$select = mysqli_query($con, "select * from table where index='something'");
$row[]=mysqli_fetch_array($select);
echo json_encode($row);
?>
JS example (fetches the PHP above) *note: this uses jQuery 3.4.1
$(document).ready(function() {
$.ajax({
url:"myserverfile.php",
method:"POST",
dataType:"json",
success:function(response) { // Doesn't have to be called response, can be anything
var a=(response[0]['indexone']); // A name of an index from the sql row you are retrieving
var b=(response[0]['indextwo']); // A name of an index from the sql row you are retrieving
var c=(response[0]['indexthree']); // A name of an index from the sql row you are retrieving
// Check to see values
console.log(a);
console.log(b);
console.log(c);
}
});
});
Tested on a working server with live database.