PHP / MySQL：查询正确执行，受影响的行返回-1

Question

I'm trying to delete a row from a table, and check if it worked / if a row has been found and removed. I use the following query to do so: DELETE FROM TableName WHERE val1=1 AND val2=2 AND val3=3 . I send that code via the connection $conn using a prepare and execute function, and it works just fine. The row I'm looking for is being found and deleted. But if I want to check if it worked using $conn->affected_rows it returns -1, which means that an error occured. I don't understand, where the error could be, that affected_rows returns -1, even though the query has been executed properly...

If you do not understand the issue, feel free to ask in a comment.

code:

$query = "DELETE FROM ComTasks WHERE comid=? AND obj=? AND type=0";
$response = paramQuery($conn, $query, array($communityid, $userid), "ss");
echo $conn->affected_rows; //=> -1

function paramQuery($conn, $query, $params, $str){
    if(!is_array($params)) $params = array($params);
    $stmt = $conn->prepare($query);
    if($stmt === false) return -1;
    for($i=0; $i<count($params); $i++){
        $params[$i] = &$params[$i];
    }
    array_unshift($params, $str);
    call_user_func_array(array($stmt, 'bind_param'), $params);
    $stmt->execute();
    $result = $stmt->get_result();
    $stmt->close();
    return $result;
}

Answer 1

Since you are using a prepared statement and $stmt, the affected_rows should also be called from $stmt:

$stmt->affected_rows;

However since you wrapped it in a function, and do a $stmt-close(), this will not work outside of the function call. You may have to adjust your function to store that value before close, and before the return of the function.

Some more info here on php manual: http://php.net/manual/en/mysqli-stmt.affected-rows.php