Question

我有一个问题，您看到我借了一本书来获取book_id，并且借阅日期设置为今天的日期，但是每本书都有一个days_id值，需要将其添加到borrow_date的一天，现在我很难从我选择的date_id中获取book_id并将其添加到borrow_date中，以便生成{自动{1}}

这是我的PHP代码

due_date

这是我的HTML代码

<?php

    include 'connect.php';


     $librarian_id  =   $_POST['librarian_id'];
     $member_id     =   $_POST['member_id'];
     $book_id       =   $_POST['book_id'];
     $date_borrow   =   $_POST['date_borrow'];
     $status_id     =   $_POST['status_id'];   


    $days = "SELECT days_id, book_id FROM book_setup 
            WHERE book_id  = '$book_id' AND days_id = " . $row['book_id'] . " ";
    $result = mysqli_query($conn, $days);
    $due_date = date('Y-m-d', strtotime($date_borrow. ' +  $result days'));


if      ($librarian_id == '') {
    echo "librarian is empty";
}
else if     ($member_id == '') {
    echo "member id is empty";
}
else if     ($book_id == '') {
    echo "book is empty";
}
else if   ($date_borrow == '') {
    echo "date borrow is empty";
}

else if ($status_id == '') {
    echo "status is empty";
}
else if ($due_date == '') {
    echo "due date is empty";
}



else {



    $sql = " INSERT INTO borrow_book (librarian_id, member_id, 
            book_id, date_borrow, due_date, status_id) 
            VALUES 
            ('$librarian_id', '$member_id', '$book_id', 
             '$date_borrow', '$due_date',  '$status_id' ) ";

    if ($conn->query($sql) === TRUE) 
        {
            $message = "Borrowing book successful";
            echo "<script type='text/javascript'>alert('$message');</script>";

        } 

    else 

        {
            $message = "Borrowing book failed Failed";
            echo "<script type='text/javascript'>alert('$message');</script>";
        } 

    }

$conn->close();
?>

this is my database for borrowing book

and this is where I get the book that is available to borrow

Answer 1

执行以下操作：

$res = mysqli_query($con, " SELECT days_id FROM book_setup WHERE book_id  = '$book_id' ");

$days = mysqli_fetch_assoc($res)['days_id'];

如何查询值并将其添加到日期

1 个答案: