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DiPaspafe
Asked: 2022-10-18 05:34:30 +0800 CST 2022-10-18 05:34:30 +0800 CST

How do I display the JSON of a query to MySql in php?

  • 772

I have tried the following:

<?php
require("config/conf.php"); //El fichero donde se conecta a la base de datos
$consulta = "SELECT * FROM `users`";
$resultado = mysqli_query($conexion, $consulta);

$rawdata = array();

$i = 0;
while($row = mysqli_fetch_array($result)) {
    $rawdata[$i] = $row;
    $i++;
}

echo json_encode($rawdata);
?>

Please someone help me the deploy is very soon and I need it urgently.

php

1 Answers

  1. Best Answer

    I propose this code, with error control, and avoiding unnecessary things, such as a counter $iinside the loop.

    Consider that the DBMS have their own table users, the code may fail because of that, whatever it is, you will see it with a robust code. Here we go:

    <?php

    require("config/conf.php"); //El fichero donde se conecta a la base de datos
    /*
        OJO: declaramos el array aquí
    */
    $rawdata = array();
    
    /*
       1er error optimista: asumes que hay conexión
       pero nadie te asegura que la haya, por tanto, controla que hay
    */
    if ($conexion) {
        $consulta = "SELECT * FROM `users`";
        $resultado = mysqli_query($conexion, $consulta);
        /*
            2do error optimista, asumías que la consulta funciona
            pero nada te asegura que funcione
        */
        if ($resultado) {
            /*
                3er error optimista, asumías que la consulta trae filas
                pero nada te asegura que traiga
            */
            if($resultado->num_rows > 0) {
                while($row = mysqli_fetch_array($result)) {
                    //No necesitas un contador
                    $rawdata[] = $row;
                }
            } else {
                $rawdata['error']='No se encontraron filas';
            }
        } else {
            $rawdata['error']=sprintf('Error en la consulta: %s',$conexion->error);
        }
    } else {
        $rawdata['error']='No hay conexión';
    }
    echo json_encode($rawdata);

?>

    As you can see, we have written a robust program, which will always return a result, with an error, indicating what has failed, be it with the data. It is a longer code, it will shout some! Well, yes, but that's how it should be, one of the main tasks of a programmer is to think about everything that could go wrong (pessimistic programming), not to complain, but to adequately deal with those problems. It is a very neglected aspect in programming, which leads to writing weak programs. Things don't always work the way you want them to.

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