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Home/ Questions/Q 7584795
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T19:00:32+00:00 2026-05-30T19:00:32+00:00

Nested function inside of fetch (which is inside of another function) does not perform.

  • 0

Nested function inside of fetch (which is inside of another function) does not perform.

fn_smth1 is nested inside of fn_smth2 and should output result via fn_smth2

Example below is a simplified version.

function fn_smth1 ($id){
    global $mysqli;
    $stmt = $mysqli->stmt_init();
    if ($stmt->prepare("SELECT code FROM at WHERE id = ?")){
        $stmt->bind_param("i",$id);
        $stmt->execute();
        $stmt->bind_result($code);
        if ($stmt->fetch()){
            $code_displ = $code; 
        }
    }
    $stmt->close;
    return $code_displ;
}


function fn_smth2($id){
    global $mysqli;
    $stmt = $mysqli->stmt_init();
    if ($stmt->prepare("SELECT idx, name FROM at WHERE id = ?")){
        $stmt->bind_param("i",$id);
        $stmt->execute();
        $stmt->bind_result($idx, $name);
        if ($stmt->fetch()){
            $code_displ = $name.' === '.fn_smth1($idx); 
        }
    }
    $stmt->close;
    return $code_displ;
}


echo fn_smth2(1);

//expected

some name here === some code here

//received

some name here === null (function fn_smth1 does not give a value)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T19:00:34+00:00Added an answer on May 30, 2026 at 7:00 pm

    You’re trying to execute second prepared statement, while the resultset from the first one has not been stored yet. Use mysqli_stmt::store_result() before trying to execute second statement.

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