echo中的函数为什么会出错？

Question

I read some topics here and in other forums and find a way to set a function in a echo. But it doesnt show my stuff. My Code

echo "<tr>
        <td>
           Choose people:<br>
            ",choose_kunde(),"
        </td>
    </tr>";

my function:

   function choose_kunde()
{
    require './config/config.inc.php';
    if($result2 = mysqli_query($db, "SELECT kunden_id, vname, nname FROM kunden"))
    {
        echo "<select name='kunde_a'>";
        while($adr = mysqli_fetch_assoc($result2))
        {
            echo "<option value='".$adr['vname']."|".$adr['nname']."|".$adr['kunden_id']."'>".$adr['vname']." ".$adr['nname']."</option>";
        }
        echo "</select>";
    }
    else
    {
        echo "Daten konnten nicht aus der Datenbank gelesen werden.<br>";
        echo mysqli_error($db);
    }
}

Answer 1

You need to change your function to return string instead of echoing it.

example:

function choose_kunde()
{
    $output = '';
    require './config/config.inc.php';

    if($result2 = mysqli_query($db, "SELECT kunden_id, vname, nname FROM kunden"))
    {
        $output .= "<select name='kunde_a'>";

        while($adr = mysqli_fetch_assoc($result2))
        {
            $output .= "<option value='".$adr['vname']."|".$adr['nname']."|".$adr['kunden_id']."'>".$adr['vname']." ".$adr['nname']."</option>";
        }

        $output .= "</select>";
    }
    else
    {
        $output .= "Daten konnten nicht aus der Datenbank gelesen werden.<br>";    
        $output .= mysqli_error($db);
    }

    return $output;
}

and then concatenate it

echo "<tr>
    <td>
       Choose people:<br>
        " . choose_kunde() ."
    </td>
</tr>";

Still you need work on the code to make it more clean.