我有這個queries
是JOINS 2個表和我計劃增加更多的表,但是,我已經有兩個tables
如何在mysqli中連接多個表?
$query = "SELECT * FROM info WHERE JOIN crew_rank ON info.id = crew_rank.crew_rank_id WHERE info.id = ?";
$stmt = mysqli_prepare($conn, $query);
mysqli_stmt_bind_param($stmt, 'i', $_GET['id']);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $id, $full_name, $phone_number, $crew_rank,$date_of_birth,$age,$telephone_number,$vessel,$place_of_birth,$religion,$joining_date);
麻煩你能不能幫我傢伙這有什麼query
問題?我有一個錯誤這樣
警告:mysqli_stmt_bind_param()預計參數1被mysqli_stmt,布爾在C中給出:\ XAMPP \ htdocs中\ practice1 \管理員在第9行 \ edit_info_docs.php(這是
mysqli_stmt_bind_param($stmt, 'i', $_GET['id']);
)警告:mysqli_stmt_execute()預計參數1被mysqli_stmt,布爾在C中給出:\ XAMPP \ htdocs中\ practice1 \管理員\ edit_info_docs.php第10行 (這是
mysqli_stmt_execute($stmt);
)警告: mysqli_stmt_bind_result()需要參數1爲mysqli_stmt,布爾值在C:\ xampp \ htdocs \ pr中給出actice1 \ ADMIN \ edit_info_docs.php上線11 (這是
mysqli_stmt_bind_result($stmt, $id, $full_name, $phone_number, $crew_rank,$date_of_birth,$age,$telephone_number,$vessel,$place_of_birth,$religion,$joining_date);
)
預先感謝您
你的查詢是不正確的閱讀文檔。 http://dev.mysql.com/doc/refman/5.7/en/join.html –