我有这样的表格:
提交表格后,个人详细信息将被插入到个人表格和书籍详细信息中。我想通过一个唯一的ID将一个个人表行链接到book表中的多行(因为它使用相同的查询插入),因此我可以确定这些书在以后与这个人相关。
这是我的PHP和MySQL脚本:
$mysqli = new mysqli($host,$user,$password,$database);
$users_firstname = $_POST['firstname'];
$users_middlename = $_POST['middlename'];
$users_lastname = $_POST['lastname'];
$users_gender= $_POST['gender'];
$users_location= $_POST['location'];
$users_email= $_POST['email'];
$users_mobile= $_POST['mobile'];
$query = "INSERT INTO personaldetails(FirstName ,MiddleName,LastName,
Gender,Location,Email,Mobile) VALUES ('$users_firstname',
'$users_middlename', '$users_lastname', '$users_gender','$users_location','$users_email','$users_mobile');";
foreach($_POST['booktitle'] as $key => $bookTitle) {
$bookTitle = mysqli_real_escape_string($mysqli, $bookTitle);
$bookGenre = mysqli_real_escape_string($mysqli, $_POST['bookgenre'][$key]);
$bookWriter = mysqli_real_escape_string($mysqli, $_POST['bookwriter'][$key]);
$bookDescription = mysqli_real_escape_string($mysqli, $_POST['bookdescription'][$key]);
$query .= "INSERT INTO bookdetails(BookTitle ,BookGenre,BookWriter,
BookDescription) VALUES('$bookTitle',
'$bookGenre', '$bookWriter', '$bookDescription');";
}
$result = mysqli_multi_query($mysqli, $query);
答案 0 :(得分:1)
一个可能的解决方案是使用mysqli_insert_id。
但您还必须在图书表中创建一个额外的列,用于存储用户的ID以将图书与用户相关联,让它为 user_id (将存储我们将通过mysqli_insert_id()
)
并且您还必须单独执行查询以获取新插入的用户ID。
所以代码就像 -
$mysqli = new mysqli($host,$user,$password,$database);
$users_firstname = $_POST['firstname'];
$users_middlename = $_POST['middlename'];
$users_lastname = $_POST['lastname'];
$users_gender= $_POST['gender'];
$users_location= $_POST['location'];
$users_email= $_POST['email'];
$users_mobile= $_POST['mobile'];
$user_query = "INSERT INTO personaldetails(FirstName ,MiddleName,LastName,
Gender,Location,Email,Mobile) VALUES ('$users_firstname',
'$users_middlename', '$users_lastname', '$users_gender','$users_location','$users_email','$users_mobile');";
//execute the user query
$result = mysqli_query($mysqli, $user_query);
//get the user id of newly inserted user
$user_id = mysqli_insert_id($mysqli);
foreach($_POST['booktitle'] as $key => $bookTitle) {
$bookTitle = mysqli_real_escape_string($mysqli, $bookTitle);
$bookGenre = mysqli_real_escape_string($mysqli, $_POST['bookgenre'][$key]);
$bookWriter = mysqli_real_escape_string($mysqli, $_POST['bookwriter'][$key]);
$bookDescription = mysqli_real_escape_string($mysqli, $_POST['bookdescription'][$key]);
//use the user id here to relate it with the book
$book_query = "INSERT INTO bookdetails(BookTitle ,BookGenre,BookWriter,
BookDescription, user_id) VALUES('$bookTitle',
'$bookGenre', '$bookWriter', '$bookDescription', '$user_id');";
//execute the query for book
$result = mysqli_query($mysqli, $book_query);
}