Database Administrators Stack Exchange is a question and answer site for database professionals who wish to improve their database skills and learn from others in the community. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a Java application that configured with a connection pool. I have ensured that the application terminates the connections properly.

However when I run this command on mysql command line -

show status like '%onn%';

| Variable_name            | Value |
| Aborted_connects         | 2     |
| Connections              | 72    |
| Max_used_connections     | 10    |
| Ssl_client_connects      | 0     |
| Ssl_connect_renegotiates | 0     |
| Ssl_finished_connects    | 0     |
| Threads_connected        | 2     |

Why does mysql keep the connections open when they have been terminated from the client?

The contents of my.cnf file are (comments and unimportant sections omitted) -

port        = 3306
socket      = /var/run/mysqld/mysqld.sock

user        = mysql
pid-file            = /var/run/mysqld/
socket      = /var/run/mysqld/mysqld.sock
port        = 3306
basedir     = /usr
datadir     = /var/lib/mysql
tmpdir      = /tmp
lc-messages-dir = /usr/share/mysql
lower_case_table_names = 1

key_buffer      = 16M
max_allowed_packet  = 16M
thread_stack    = 192K
thread_cache_size   = 8

myisam-recover      = BACKUP

query_cache_limit   = 1M
query_cache_size    = 16M

expire_logs_days    = 10
max_binlog_size     = 100M
share|improve this question
up vote 2 down vote accepted

According to MySQL documentation the Connections status variable shows "The number of connection attempts (successful or not) to the MySQL server." This means that it does not represent current open connections, but all connection attempts since the server has been started.

The variable showing number of current open connections, again according to MySQ docs, is Threads_connected.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.