CODEWITHSUNDEEP
javascriptphparraysnode.js
Miniman
Miniman

Reputation: 241

Send request from node.js to PHP, returning an array

I have an SQL table I want to access called 'name_scraper' and to retrieve data between a set range to then return that resulting array to my node.js file for use. Essentially, Node js request, php $_GET[''], mysqli_query, then ?

this is my node js code to contact the php file

var request = require('request');
var http = require('http');

    var post_options = {
        host: 'localhost',
        path: '/name_request.php?pos1=0&pos2=5',
        method: 'POST',
        headers: {
            'User-Agent': 'Super Agent/0.0.1',
            'Content-Type': 'application/x-www-form-urlencoded',
        }
    }; 
var post_req = http.request(post_options, function (res) {
    res.on('data', function (chunk) {
        console.log('Response: ' + chunk);
        chnk = chunk; 
            console.log("chunk: " +chnk[0]); //random test I did that doesn't work
        });
    });

    post_req.end();

Now the above code works in contacting my php file, which then outputs the array into console.

This is my php: 

$servername = "localhost"; //localhost 
$username = "root"; 
$password = ""; 
$dbname = "test"; 

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) 
{
    die("Connection failed: " . $conn->connect_error);
}

$start = $_GET['pos1'];
$end =  $_GET['pos2'];

$sql = mysqli_query($conn, "SELECT * FROM name_scraper WHERE ID BETWEEN '$start' AND '$end'");
$array = mysqli_fetch_all($sql);
print_r($array); //can change to specific index and it works fine

?>

This overall manages to retrieve the data but doesn't return the data in a usable form, the array is not an array returned, just part of the response. I need it to return the array in usable form so I can manipulate and extract data from within each index.

Upvotes: 1

Views: 1983

Answers (1)

Cihan Uygun
Cihan Uygun

Reputation: 2138

You should change your output method like below;

$sql = mysqli_query($conn, "SELECT * FROM name_scraper WHERE ID BETWEEN '$start' AND '$end'");
$array = mysqli_fetch_all($sql,MYSQLI_ASSOC); //If you add MYSQLI_ASSOC your output will be better
echo json_encode($array); //That will output an json array

Hope this helps

Upvotes: 1

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