(!) Notice: Use of undefined constant name - assumed 'name'

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(!) Notice: Use of undefined constant name - assumed 'name'

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#1 by (4 votes)
#2 by (3 votes)

1

I am trying to query my database with php but it throws me an error. This e my code

    [![<?php

$servername = "localhost";
$username = "venividivici2016";
$password = "4XDjhDFsxBtbxRsw";

// Create connection
$conn = mysqli_connect($servername, $username, $password);

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
} else {



}



echo '<br/>';

$bd_seleccionada = mysqli_select_db($conn,"veniviajes");
if (!$bd_seleccionada) {
     die("Conecion fallida: " . mysqli_error($conn));
} else {



}

$sentencia= " SELECT nombre FROM pais";
  if(!$sentencia) 
    die("Error: no se pudo realizar la consulta");
$resultado = mysqli_query($conn,$sentencia); 
 if(!$resultado) 
    die("Error: no se pudo realizar la consulta");



?>

<select>

<option value="">Seleccione</option>

<?php

while($fila=mysqli_fetch_array($resultado)){

?>

<option value="<?php echo $fila\[nombre\]; ?></option>

<?php } ?>

</select>]

mysql php

asked by Alejandro Montes 26.06.2016 в 09:39

source

2 answers

XAMPP localhost does not work correctly Failed to work with pointers

score 4 · Answer 1

Try printing the result with:

$fila["nombre"]

Try this code, which worked for me:

            <?php 
        $servername = "localhost";
        $username = "venividivici2016";
        $password = "4XDjhDFsxBtbxRsw";

        // Create connection
        $conn = mysqli_connect($servername, $username, $password);
        $bd_seleccionada = mysqli_select_db($conn,"veniviajes");

        // Check connection
        if (mysqli_connect_errno()) {

            die("Connection failed: " . mysqli_connect_error());
        }

            $sentencia = "SELECT nombre FROM usuarios";

            if ($resultado = mysqli_query($conn,$sentencia)) {

        ?>

                <br>
        <select>
            <option value="">Seleccione</option>

            <?php while($fila = mysqli_fetch_array($resultado)){ ?>

                <option value="<?php echo $fila['nombre']; ?>"><?php echo $fila['nombre']; ?></option>

            <?php } ?>

        </select>

score 3 · Answer 2

Change this row:

<option value="<?php echo $fila\[nombre\]; ?></option>

a

<option value="<?php echo $fila['nombre']; ?>"></option>

Apart from not concatenating well $fila\[nombre\] have also forgotten to close well the <option> tag.

Updated response based on comments:

$servername = "localhost";
$username = "venividivici2016";
$password = "4XDjhDFsxBtbxRsw";

// Create connection
$conn = mysqli_connect($servername, $username, $password);
$bd_seleccionada = mysqli_select_db($conn,"veniviajes");

// Check connection
if (mysqli_connect_errno()) {

    die("Connection failed: " . mysqli_connect_error());
}

    $sentencia = "SELECT nombre FROM pais";

    if ($resultado = mysqli_query($conn,$sentencia)) {

?>
<select>
    <option value="">Seleccione</option>

    <?php while($fila = mysqli_fetch_array($resultado)){ ?>

        <option value="<?php echo $fila['nombre']; ?>"></option>

    <?php } ?>

</select>

    <?php } else { ?>

        <p>Comprobación: </p>
        <br>
        <p>Connexion: </p>
        <?php var_dump($conn); ?>
        <br>
        <p>Tabla seleccionado: </p>
        <?php var_dump($bd_seleccionada); ?>
        <br>
        <p>Resultado: </p>
        <?php var_dump($resultado); ?>
        <br>

    <?php } ?>