sqlwatch | check the sql syntax and find the tables | SQL Database library

by zjjxxlgb C++ Version: Current License: GPL-2.0

X-Ray Key Features Code Snippets Community Discussions(1)Vulnerabilities Install Support

kandi X-RAY | sqlwatch Summary

sqlwatch is a C++ library typically used in Database, SQL Database, MariaDB, JPA, Oracle applications. sqlwatch has no bugs, it has no vulnerabilities, it has a Strong Copyleft License and it has low support. You can download it from GitHub.

A tool can check the sql syntax of mysql and find the tables which been modified.

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Support

sqlwatch has a low active ecosystem.

It has 13 star(s) with 9 fork(s). There are 4 watchers for this library.

It had no major release in the last 6 months.

sqlwatch has no issues reported. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of sqlwatch is current.

Quality

sqlwatch has no bugs reported.

Security

sqlwatch has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

sqlwatch is licensed under the GPL-2.0 License. This license is Strong Copyleft.

Strong Copyleft licenses enforce sharing, and you can use them when creating open source projects.

Reuse

sqlwatch releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

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sqlwatch Key Features

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sqlwatch Examples and Code Snippets

No Code Snippets are available at this moment for sqlwatch.

Community Discussions

Trending Discussions on sqlwatch

SQLWatch - notifications not being sent

QUESTION

SQLWatch - notifications not being sent

Asked 2020-Mar-26 at 23:08

I’m wondering if someone with knowledge/experience of SQLWatch could help me out with something. We have SQLWatch set up on 2 DEV servers and 1 Central monitoring server, its working fine and the data from the 2 DEV servers is coming over to the central server, I can see alerts are being recorded in the table - [dbo].[sqlwatch_logger_check].

However, our issue that we are not being notified by any means (email, Powershell script running). What’s interesting is that if we drop a row into the table [dbo].[sqlwatch_meta_action_queue] then alert notification does happen.

So our issue seems to be for some reason alerts are being raised but the record is not being inserted into the queue table. I suspect some sort of mapping issue but as it stands now it all looks ok, I use the following to check

...

ANSWER

Answered 2020-Mar-26 at 23:08

I am the creator of SQLWATCH.

Firstly, just to clarify, default notifications that come with SQLWATCH only work in a local scope i.e. they will happen on each monitored instance where @@SERVERNAME = sql_instance. If you are expecting the default notifications to fire from the central server for a remote instance this will not happen. The default notifications on the central server will only fire for the central server itself and not for data imported from the remote instances. This is done to avoid a situation where pull into the central repository is rare and thus notifications could be well delayed.

However, there is nothing stopping you from creating Check Rules or Reports to fire on the back of the imported data.

Secondly, the checks are not alerts per se. Checks are just... well, checks... that run periodically and make sure everything is in order. Checks can trigger an action to send an email. For this, as you have worked out, there is an association table that links together checks and actions.

As for your problem, is the actual action enabled? All actions that are not associated with report are disabled by default as they need to be configured first:

Add a column to your query to bring action_enabled column:

Source https://stackoverflow.com/questions/60870432

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install sqlwatch

You can download it from GitHub.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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