Get multiple rows from phpmyadmin to Java using JSON

Question

Good evening everyone, until a few days ago I had never used PHP or JSON and I am finding trouble getting multiple rows from phpmyadmin to Java using JSON. I have manged to write the below code however this only gets the last row.

I took a look at this post (how to encode multiple rows from mysql into json using php) and had a go but have had no luck..

    <?php

    $user = 'root';
    $pass = '';
    $db = 'uopuser';

    $con=mysqli_connect('localhost', $user, $pass, $db) or die('Unable to connect');


    $statement = mysqli_prepare($con, 'SELECT * FROM society');
    mysqli_stmt_execute($statement);

    mysqli_stmt_store_result($statement);
    mysqli_stmt_bind_result($statement, $society_id, $name, $email, $description);

    $society = array();

    while(mysqli_stmt_fetch($statement)){
        $society['society_id'] = $society_id;
        $society['name'] = $name;
        $society['email'] = $email;
        $society['description'] = $description;
    }

    echo json_encode($society);

    mysqli_stmt_close($statement);

    mysqli_close($con);
?>

Thank you in advance!

Q: Where does "Java" come in? Your PHP 1) Executes on the server, 2) the server talks to the database, and 3) your "echo" presumably writes some valid JSON to the browser. Q: Where/how exactly are you using "Java"???? PS: Your "while" loop is wrong - each iteration overwrites the previous value. You might want to move "echo" inside the loop... — paulsm4
– paulsm4, Commented Mar 4, 2016 at 23:49
@paulsm4 I will be using this data to populate a ListView for an Android application. At the moment I just need to get my PHP/JSON correct. I'll test your suggestion, thanks for pointing that out to me. — kmil
– kmil, Commented Mar 4, 2016 at 23:55

kmil · Accepted Answer · 2016-03-05 00:05:39Z

0

Just needed to put my echo inside the while loop

while(mysqli_stmt_fetch($statement)){
            $society['society_id'] = $society_id;
            $society['name'] = $name;
            $society['email'] = $email;
            $society['description'] = $description;
        }

        echo json_encode($society);

to

while(mysqli_stmt_fetch($statement)){
        $society['society_id'] = $society_id;
        $society['name'] = $name;
        $society['email'] = $email;
        $society['description'] = $description;
        echo json_encode($society);
    }

answered Mar 5, 2016 at 0:05

kmil

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