php中的MYSQL语法错误，但sql有效

Question

I'm trying to start a transaction is mysql and insert data into the database. The database source sql can be found on github here. Here is the error:

Error: START TRANSACTION; INSERT INTO Books(Title, PublicationDate, PurchaseDate, Description, LocationID, GenreID) VALUES('Simple Genius', '2008-4-1','2009-5-7','','Hardbook Library','Fiction'); SET @bookid = LAST_INSERT_ID(); INSERT INTO BookAuthors(FirstName, MiddleName, LastName) VALUES('David', '', 'Baldacci'); SET @authorid = LAST_INSERT_ID(); INSERT INTO AuthorsInBooks(AuthorID, BookID) VALUES(@authorid, @bookid); COMMIT; You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INSERT INTO Books(Title, PublicationDate, PurchaseDate, Description, LocationID,' at line 3

Near 'INSERT INTO Books(Title, PublicationDate, PurchaseDate, Description, LocationID,' doesn't make sense to me because it is missing GenreID after LocationID. Am i missing something? When I copy and paste this code into phpmyadmin it works fine. My php version is 5.4.

Here is php code:

$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() can only execute 1 query, if you want to execute multiple queries, you need:

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