Problem with MySQL with PHP

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Problem with MySQL with PHP

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#1 by (0 votes)

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I do not know how to return each value in a different variable.

My code:

$query  = "SELECT id_tienda,id_prod from prod_tienda where id_prod='$row' ";
$result = mysqli_query($con, $query);           

while ($mostrar = mysqli_fetch_array($result)) {
      $id = ($mostrar['id_tienda']);   
      echo $id;

mysql php mysqli

asked by alex tiberiu Telegaru 20.01.2018 в 12:33

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1 answer

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score 0 · Accepted Answer

A possible example:

$query  = "SELECT id_tienda,id_prod FROM prod_tienda WHERE id_prod='$row'";
$result = mysqli_query($con, $query);        

//Comprobamos si existe registro. 
if (mysqli_num_rows($result) > 0) {   
    //Tu array con los datos desde la Base de Datos.    
    $mostrar  = mysqli_fetch_array($result);        
} else {
    echo 'No se encontraron datos.';
}

Note: I advise you for security use mysqli prepare statements or PDO. I also advise you to read well How to avoid SQL injection in PHP ?