sending variable to function to execute odbc SELECT

I am trying to execute the function below but does not display anything.

function displayNr($x){
    $sql="SELECT D4741 FROM table_x WHERE D4711='".$x."';       
    if (!$result = odbc_exec($pconn, $sql)) {
        echo "Query error! ODBC: ", odbc_error();
    } else {
        while ($row = odbc_fetch_array($result)) {
        echo $row["D4741"] . "\n";
    }
    }
}

displayNr('name');

However, if I remove the function it works correctly:

x='name';

$sql="SELECT D4741 FROM table_x WHERE D4711='".$开发者_JAVA技巧x."';       
if (!$result = odbc_exec($pconn, $sql)) {
    echo "Query error! ODBC: ", odbc_error();
} else {
    while ($row = odbc_fetch_array($result)) {
    echo $row["D4741"] . "\n";
}
}

What could be the problem?

$pconn is not set in the function.

in the function, $pconn is a new local variable (with a scope of the function) and is not the same $pconn defined outside the function. pass it in as a parameter:

function displayNr($x,$pconn){
    $sql="SELECT D4741 FROM table_x WHERE D4711='".$x."';       
    if (!$result = odbc_exec($pconn, $sql)) {
        echo "Query error! ODBC: ", odbc_error();
    } else {
        while ($row = odbc_fetch_array($result)) {
        echo $row["D4741"] . "\n";
    }
    }
}

displayNr('name');

watch out for SQL injection!!! your code is a perfect example of what not to do!!:

$sql="SELECT D4741 FROM table_x WHERE D4711='".$x."';

see this: SQL Injection or just google it

function displayNr($x, $pconn)
{
    $sql = "SELECT D4741 FROM table_x WHERE D4711='" . $x . "'";        //here " is remaining
    if (!$result = odbc_exec($pconn, $sql)) {
        echo "Query error! ODBC: ", odbc_error();
    } else {
        while ($row = odbc_fetch_array($result)) {
            echo $row["D4741"] . "\n";
        }
    }
}

displayNr('name');