slj90 Posted December 6, 2017 Share Posted December 6, 2017 $username = "user1"; $result = single_row_prepared($username); echo $username . " - " . $country . " - " . $about . " - " . $age; It says the variables are undefined. How do I echo them out? function single_row_prepared($username) { global $connect; $sql = "SELECT profile_username, profile_country, profile_about, profile_age "; $sql .= "FROM profiles "; $sql .= "WHERE profile_username = ?"; $stmt = mysqli_prepare($connect, $sql); mysqli_stmt_bind_param($stmt, 's', $username); mysqli_stmt_execute($stmt); mysqli_stmt_bind_result($stmt, $username, $country, $about, $age); mysqli_stmt_fetch($stmt); mysqli_stmt_close($stmt); } [06-Dec-2017 08:19:54 UTC] PHP Notice: Undefined variable: country in /home/buysnapchats/public_html/examples/prepared_statement_single_row.php on line 13 [06-Dec-2017 08:19:54 UTC] PHP Notice: Undefined variable: about in /home/buysnapchats/public_html/examples/prepared_statement_single_row.php on line 13 [06-Dec-2017 08:19:54 UTC] PHP Notice: Undefined variable: age in /home/buysnapchats/public_html/examples/prepared_statement_single_row.php on line 13 Quote Link to comment https://forums.phpfreaks.com/topic/305847-echo-result-from-prepared-statement-function/ Share on other sites More sharing options...
Solution requinix Posted December 6, 2017 Solution Share Posted December 6, 2017 Variables defined inside of a function are not available outside of a function. Return everything you need using an array. "single_row_prepared" sounds like it can handle any query, but it can't. It's only about profiles. So name it in a way that shows it gets a profile. Like "get_profile". function get_profile($connect, $username) { $sql = "SELECT profile_username, profile_country, profile_about, profile_age "; $sql .= "FROM profiles "; $sql .= "WHERE profile_username = ?"; $stmt = mysqli_prepare($connect, $sql); mysqli_stmt_bind_param($stmt, 's', $username); mysqli_stmt_execute($stmt); mysqli_stmt_bind_result($stmt, $username, $country, $about, $age); mysqli_stmt_fetch($stmt); mysqli_stmt_close($stmt); return ["username" => $username, "country" => $country, "about" => $about, "age" => $age]; } $username = "user1"; $result = get_profile($username); echo $result["username"] . " - " . $result["country"] . " - " . $result["about"] . " - " . $result["age"]; Quote Link to comment https://forums.phpfreaks.com/topic/305847-echo-result-from-prepared-statement-function/#findComment-1554502 Share on other sites More sharing options...
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