PHP function return don't work

Question

i have a problem. I write this function tu recover a previous branch from my DB entity, and i want that the function return a string with I can see this information. this is the code:

function stampaOggettoCollezione($tipoOggetto, $id, $connection, $percorso="")
{
    $i=0;

    switch($tipoOggetto)
    {
        case 'casaeditrice':
            return $percorso;
            break;

        case 'storia':
            $dati = array("storia", "IDCasaEditrice", "casaeditrice");
            break;

        case 'testata':
            $dati = array("testata", "IDStoria", "storia");
            break;

        case 'albo':
            $dati = array("albo", "IDTestata", "testata");
            break;
    }

            $sql = "SELECT * FROM ".$dati[0]." WHERE ID=$id";
            $result=mysqli_query($connection,$sql);
            $IDSuperiore=mysqli_fetch_array($result)[$dati[1]];

            $sql = "SELECT * FROM ".$dati[2]." WHERE ID=$IDSuperiore";
            $result = mysqli_query($connection, $sql);
            $nomeSuperiore=mysqli_fetch_array($result)['nome'];

            $percorso = $percorso . $nomeSuperiore;

            stampaOggettoCollezione($dati[2], $IDSuperiore, $connection, $percorso);
}

i'm sure that code works, i try some echo in the code to check the final path. My problem is that i can't write the return like a string in my webpage. What do you think about? there's some problem with the declaration? Thanks

What do you mean by "write the return like a string in my webpage"? It's not really clear what you're asking. — David
– David, Commented Dec 18, 2015 at 1:52
sorry, is my problem with english. I use that function in a webpage, to write a path. if i try to use $var=stampaOggettoCollezione(...); echo $var; i can't view an output. — NicolaPez
– NicolaPez, Commented Dec 18, 2015 at 1:54
That should work, assuming a value is returned. Currently only one case in the function actually returns a value. So unless that case is invoked, nothing will be returned. — David
– David, Commented Dec 18, 2015 at 1:55
Replace stampaOggettoCollezione($dati[2], $IDSuperiore, $connection, $percorso); } with return stampaOggettoCollezione($dati[2], $IDSuperiore, $connection, $percorso); } — user5051310
– user5051310, Commented Dec 18, 2015 at 2:03

George · Accepted Answer · 2015-12-18 01:53:26Z

1

I'm not sure I understand because the english is difficult to understand, but it seems that all you need to do is write:

return $result;

at the end of the function, or whichever string variable you want to return. Then use echo stampaOggettoCollezione(arguments...) after the function to have this value echoed.

answered Dec 18, 2015 at 1:53

George

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2 Comments

NicolaPez Over a year ago

al the function, for the db structure, assuming the case 'casaeditrice' and go out. I don't understand why i don't have output. For the recursive function i must have the return only in the 'final' case, not at the end of function.

George Over a year ago

Then use return after recursive stampaOggettoCollezione(...) like return false; This means if nothing was returned it will be false. Otherwise return the value you want. What variable do you want to return as the 'final' case?

user5051310 · Accepted Answer · 2015-12-18 05:16:17Z

Replace

stampaOggettoCollezione($dati[2], $IDSuperiore, $connection, $percorso);

with

return stampaOggettoCollezione($dati[2], $IDSuperiore, $connection, $percorso);

You have to pass $percorso all the way back out of all the calls to stampaOggettoCollezione and back to the original call. Read more about recursion on the Internets.

Alternatively, I believe you could achieve the same effect if you change your function to build a query that joins on tables. It will remove the need for recursion, reduce the number of database calls, and should work without changing anything but the function itself.

// dynamically queries more tables based on what the inital table to query is
function stampaOggettoCollezione($tipoOggetto, $id, $connection, $percorso="")
{
    $selectQuery = "SELECT ";
    $columnsToSelect = "";
    $fromQuery = "";
    $whereClause = " WHERE ";

    switch($tipoOggetto) {
        case 'albo':
            $columnsToSelect .= ", a.nome AS aNome";
            $fromQuery = " LEFT JOIN albo a ON t.ID = a.IDTestata";
        case 'testata':
            $columnsToSelect .= ", t.nome AS tNome";
            $fromQuery = " LEFT JOIN testata t ON s.ID = t.IDStoria" . $fromQuery;
        case 'storia':
            $columnsToSelect .= ", s.nome AS sNome";
            $fromQuery = " LEFT JOIN storia s ON ce.ID = s.IDCasaEditrice" . $fromQuery;
        case 'casaeditrice':
            $columnsToSelect .= ", ce.nome AS ceNome";
            $fromQuery = " FROM casaeditrice ce" . $fromQuery;
            break;
    }
    // remove the first `, `
    $selectQuery .= substr($columnsToSelect, 2);

    switch($tipoOggetto)
    {
        case 'casaeditrice':
            $whereClause .= "c.ID = $id";
            break;
        case 'storia':
            $whereClause .= "s.ID = $id";
            break;
        case 'testata':
            $whereClause .= "t.ID = $id";
            break;
        case 'albo':
            $whereClause .= "a.ID = $id";
            break;
    }

    $qry = $selectQuery . $fromQuery . $whereClause;

    $result=mysqli_query($connection,$sql);
    $nomeSuperiore= "";
    $nomes = mysqli_fetch_array($result);
    // might take some work to get the output in the right order
    foreach($nomes as $nome)
        $nomeSuperiore .= $nome;

    return $nomeSuperiore;
}

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