Recently, we had an issue in our application which is a single page application(SPA). We have all the functionality in a single page and most of the data and view changes happen through ajax requests and iframe page displays.We are using CodeIgniter framework for this application. So sometimes what happened was the user was logged out while accessing the application and this occurred once a while for 1 or 2 users, not a reproducible case one would say.Before starting on the case, I would like to specify the steps which CodeIgniter takes for checking session for any request.Steps:
1. Read session cookie sent from browser.
2. Read session cookie process proceeds with below step checks (If any one step fails below, it results in create one session cookie)
2a. Checks if session cookie is present
2b. Checks if it is a valid session cookie by using it's decryption algorithm check
2c. If  valid session, then check if …

Recently while writing a data loading application for a customer, we’ve come into a problematic situation with the Python MySQLdb module that can be installed with base RHEL repository or DVD. As a little background, this application uses an HA architecture where a Virtual IP can be assigned to different servers during a failover scenario. However, during failover, as long as the VIPs are not re-assigned, the application could hang waiting for query response. A quick strace reveals something like:

sendto(5, "W\0\0\0\3SELECT * FROM dataloader.bo"..., 91, 0, NULL, 0) = 91
recvfrom(5,

Further test reveals that this problem will not timeout until after about 15 minutes which matches the default tcp_retries2 value on the system (see man 7 tcp):

[root@node1 ~]# cat /proc/sys/net/ipv4/tcp_retries2
15

It turns out that this version of MySQLdb module has client net read and write timeouts set to 0 allowing the TCP setting …

This would have to be one of the most common MySQL error messages that is misleading to the end user developer. The MySQL Manual page confirms the broad range of possible conditions, but offers little to a PHP developer that does not speak MySQL Geek. I am commonly asked to help solve this issue from a developer.

The problem is that there are several conditions that can cause this error, and a more meaningful explanation to the end user would help in addressing the issue. In general terms, this actually means “Your SQL statement has failed because the connection to the database has been disconnected because of ???”.

Here are a few common situations and how to check for what “???” is.

1. Your MySQL server really did go away.

We can easily check this by looking at the server uptime and the server error log.

$ …

There are several parameters to set a timeout on MySQL :

But I would like to focus on wait_timeout (or interactive_timeout depending on how you connect)

This timeout allows MySQL to automatically close a connection in case of non-activity during the time defined by this parameter (default value is 28000 seconds).

The problem I wish to explain here may happen when this timeout is set to a low value (about 10 to 30 seconds).

Indeed, in this case, this timeout may have serious consequences, look at that :
[The wait_timeout parameter is set to 10 seconds in this case]

You are connected through the standard MySQL client and you need to run a delete query (bypass autocommit) :

• mysql> start transaction; …

There are several parameters to set a timeout on MySQL :

But I would like to focus on wait_timeout (or interactive_timeout depending on how you connect)

This timeout allows MySQL to automatically close a connection in case of non-activity during the time defined by this parameter (default value is 28000 seconds).

The problem I wish to explain here may happen when this timeout is set to a low value (about 10 to 30 seconds).

Indeed, in this case, this timeout may have serious consequences, look at that :
[The wait_timeout parameter is set to 10 seconds in this case]

You are connected through the standard MySQL client and you need to run a delete query (bypass autocommit) :

• mysql> start transaction;
• [ You …