Does starting a new transaction forces the current transaction to commit? - mysql

Question

I have server that uses transactions to write data to database and if all the queries are successful it will commit otherwise it will rollback. Now I want to have two instances of the server work at the same time on the same database and tables.
When I was reading mysql's transaction documentation I noticed this sentence: "Beginning a transaction causes any pending transaction to be committed". Does this mean that if server A start transaction A and while transaction A is not completed yet, the server B start transaction B, transaction A is forced to commit? This doesn't make sense to me. If this is the case how I can insure that transaction B is not executed until transaction A is completed normally?
Would SET autocommit = 0 be an alternative to this problem?

Answer 1

Assuming when you say you want two instances of the server to work at the same time you mean two separate sessions running on the same server.
The sentence "Beginning a transaction causes any pending transaction to be committed" refers to any pending transaction within the same session only. From http://dev.mysql.com/doc/refman/5.7/en/implicit-commit.html
The statements listed in this section (and any synonyms for them) implicitly end any transaction active in the current session, as if you had done a COMMIT before executing the statement.
So, if Session B starts a transaction before Session A is committed it will not force Session A to commit.

Answer 2

mysql> CREATE TEMPORARY TABLE super(id int);
Query OK, 0 rows affected (0.04 sec)
mysql> START TRANSACTION;
Query OK, 0 rows affected (0.00 sec)
mysql> INSERT INTO super VALUE(1);
Query OK, 1 row affected (0.00 sec)
mysql> START TRANSACTION;
Query OK, 0 rows affected (0.00 sec)
mysql> SELECT * FROM super;
+------+
| id |
+------+
| 1 |
+------+
1 row in set (0.00 sec)

Answer 3

Starting the next transaction implicitly commits the previous pending transaction only in the same session on MySQL.
For example, there is person table with id and name on MySQL as shown below.
person table:
id
name
1
John
2
Tom
First, with only one command prompt(with the same session), you take these steps with the default isolation level REPEATABLE READ on MySQL as shown below:
Flow
Command Prompt 1(CP1)
Explanation
Step 1
mysql -u root -p
CP1 logs in MySQL.
Step 2
USE test
CP1 uses test database.
Step 3
BEGIN;
CP1 starts a transaction.
Step 4
SELECT * FROM person;1 John2 Tom
CP1 reads person table.
Step 5
UPDATE person set name = "Lisa" where id = 2;
CP1 updates Tom to Lisa.
Step 6
BEGIN;
Now, CP1 starts the next transaction so the previous pending transaction is implicitly committed in the same session.
Step 7
ROLLBACK;
CP1 rollbacks.
Step 8
ROLLBACK;
CP1 rollbacks again.
Step 9
SELECT * FROM person;1 John2 Lisa
CP1 reads person table.Lisa is not rollbacked to Tom so starting the next transaction implicitly commits the previous pending transaction in the same session.
Second, with two command prompts(with two different sessions), you take these steps with the default isolation level REPEATABLE READ on MySQL as shown below:
Flow
Command Prompt 1(CP1)
Command Prompt 2(CP2)
Explanation
Step 1
mysql -u root -p
CP1 logs in MySQL.
Step 2
mysql -u root -p
CP2 logs in MySQL.
Step 3
USE test
CP1 uses test database.
Step 4
USE test
CP2 uses test database.
Step 5
BEGIN;
CP1 starts a transaction.
Step 6
SELECT * FROM person;1 John2 Tom
CP1 reads person table.
Step 7
UPDATE person set name = "Lisa" where id = 2;
CP1 updates Tom to Lisa.
Step 8
BEGIN;
Now, CP2 starts the next transaction but the previous pending transaction in CP1's different session is not implicitly committed.
Step 7
ROLLBACK;
CP1 rollbacks.
Step 8
ROLLBACK;
CP2 rollbacks.
Step 9
SELECT * FROM person;1 John2 Tom
CP1 reads person table.Lisa is rollbacked to Tom so starting the next transaction in a different session doesn't implicitly commits the previous pending transaction in a different session.