PHP Add Element Array

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PHP Add Element Array

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#1 by (4 votes)

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$conexion2 = mysqli_connect('localhost', 'root', '', 'ProbandoMysql');

$consulta = "SELECT * FROM persona";


  if ($resultado = mysqli_query($conexion2, $consulta)) {

        /* obtener el array por índices */
        while ($fila = mysqli_fetch_assoc($resultado)) {
            $array [] = $fila; 
        }

print_r ($ array);

        foreach ($array as $k=>$v){
        echo $k." - ".$v;
    }

He shows me the html:

Array ([0] => Array ([dni] = > 1a [name] = & edu [age] => 22) [1] => Array ([dni] = > 1b [name] = > diego [age] = > 30) [2] = > Array ([dni] = > 3f [name] = > carlos [age] = > 11))

Notice: Array to string conversion in C:\xampp\htdocs\PHP\MysqlProce\verarray.php on line 55
0 - Array
Notice: Array to string conversion in C:\xampp\htdocs\PHP\MysqlProce\verarray.php on line 55
1 - Array
Notice: Array to string conversion in C:\xampp\htdocs\PHP\MysqlProce\verarray.php on line 55
2 - Array

php

asked by EduBw 28.10.2017 в 22:42

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1 answer

redirect page if it is entered from url what is the equivalent of lblconteo.Text = Len (txtMessage.Text); in c #?

score 4 · Accepted Answer

Is that by doing so:

       $array= array("Dni" => $fila[0],"Nombre"=> $fila[1],"edad"=> $fila[2]."\n");

You are replacing the array value in each pass of the loop.

You should do it like this:

       $array []= array("Dni" => $fila[0],"Nombre"=> $fila[1],"edad"=> $fila[2]);

The \n\ does not make any sense in the array.

It is also not necessary to declare the variable $array at the beginning.

Although the best way to do this is with fetch_assoc that creates a associative array whose keys are the column names. That way you pass each row directly to the array without having to type the key names manually.

Example:

if ($resultado = mysqli_query($conexion2, $consulta)) {

    /* obtener el array por índices */
    while ($fila = mysqli_fetch_assoc($resultado)) {
       $array []= $fila; //La declaración de $array arriba no es necesaria
    }
    /*Aquí puedes ver tu resultado*/
    print_r ($array); //La salida aquí es sólo para depurar.
}

If you want to show some data on the screen within the same while . Making $fila["nombre-columna"] access each value:

if ($resultado = mysqli_query($conexion2, $consulta)) {

    /* obtener el array por índices */
    while ($fila = mysqli_fetch_assoc($resultado)) {
       echo "DNI: ".$fila["dni"]."\n";
       echo "Nombre: ".$fila["nombre"]."\n";
       echo "Edad: ".$fila["edad"]."\n";

    }
}

If you want to read it outside the while :

VERY IMPORTANT: Remove the statement above: $array=array() ; If you have not removed it, you will be creating an array of arrays.

if ($resultado = mysqli_query($conexion2, $consulta)) {

    /* obtener el array por índices */
    while ($fila = mysqli_fetch_assoc($resultado)) {
       $array []= $fila; //La declaración de $array arriba no es necesaria
    }

    /*Aquí puedes ver tu resultado*/
    foreach ($array as $k=>$v){
        if (is_array($v)){
            foreach ($v as $kk=>$vv){
                echo $kk." - ".$vv."\n";
            }

         }else{

            echo $k." - ".$v;
         }
    }

}

The problem is that mysqli, I do not know why, unlike PDO, it creates an array of array, so you have to read the values within two foreach .

Or, as you have said @ D.Bulten, you can read it with a% co-count%:

for ($i=0; $i<count($array); $i++) { 
    echo $array[$i]['dni']."\n"; 
    echo $array[$i]['nombre']."\n"; 
}