Question

我正在尝试启动一个事务是mysql并将数据插入数据库。可以在github here上找到数据库源sql。这是错误：

错误：START TRANSACTION; INSERT INTO Books（Title，PublicationDate， PurchaseDate，Description，LocationID，GenreID）VALUES（'简单 Genius'，'2008-4-1'，'2009-5-7'，''，'Hardbook Library'，'Fiction'）;组 @bookid = LAST_INSERT_ID（）; INSERT INTO BookAuthors（FirstName， MiddleName，LastName）VALUES（'David'，''，'Baldacci'）; SET @authorid = LAST_INSERT_ID（）; INSERT INTO AuthorsInBooks（AuthorID，BookID）价值观（@authorid，@ bookid）;承诺;您的SQL中有错误句法;查看与MySQL服务器版本对应的手册正确的语法使用'INSERT INTO Books附近（标题， PublicationDate，PurchaseDate，Description，LocationID，'at line 3

靠近'INSERT INTO Books（Title，PublicationDate，PurchaseDate，Description，LocationID'）对我来说没有意义，因为它在LocationID之后缺少GenreID。我错过了什么？当我将这段代码复制并粘贴到phpmyadmin时工作正常。我的php版本是5.4。

这是php代码：

$sql = "
START TRANSACTION;

INSERT INTO Books(Title, PublicationDate, PurchaseDate, Description, LocationID, GenreID)
VALUES('".$Title."', '".$YearWritten."','".$YearPurchased."','".$Description."','".$Location."','".$Genre."');

SET @bookid =  LAST_INSERT_ID();

INSERT INTO BookAuthors(FirstName, MiddleName, LastName)
VALUES('".$AuthFirstName."', '".$AuthMiddleName."', '".$AuthLastName."');

SET @authorid =  LAST_INSERT_ID();

INSERT INTO AuthorsInBooks(AuthorID, BookID)
VALUES(@authorid, @bookid);

COMMIT;
";

if (mysqli_query($conn, $sql)) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);

Answer 1

mysqli_query()只能执行1个查询，如果要执行多个查询，则需要：

if (mysqli_multi_query($conn, $sql)) {

Answer 2

回复your comment“我能看到你的意思@eggyal的例子吗？”：

// mysqli provides API calls for managing transactions
mysqli_autocommit($conn, false);

// parameterise variables - NEVER concatenate them into dynamic SQL
$insert_book = mysqli_prepare($conn, '
  INSERT INTO Books
    (Title, PublicationDate, PurchaseDate, Description, LocationID, GenreID)
  VALUES
    (?, ?, ?, ?, ?, ?)
');

// bind the variables that (will) hold the actual values
mysqli_stmt_bind_param(
  $insert_book,
  'siisss', // string, integer, integer, string, string, string
  $Title, $YearWritten, $YearPurchased, $Description, $Location, $Genre
);

// execute the statement (you can change the values of some variables and
// execute repeatedly without repreparing, if so desired - much faster)
mysqli_stmt_execute($insert_book);

// mysqli provides API calls for obtaining generated ids of inserted records
$book_id = mysqli_insert_id($conn);

// ... etc ...

// use the API call to commit your transaction
mysqli_commit($conn);

// tidy up
mysqli_stmt_close($insert_book);

请注意，我之前没有包含任何错误检测/处理，您当然希望将其包含在任何实际代码中。

php中的MYSQL语法错误，但sql有效

2 个答案: