Aggregate NodeID totals in report
Aggregate NodeID totals in report
I've been having trouble with editing my current query to get the total of all interfaces and interfaceID
's associated with a single nodeID
to show on one row.
interfaceID
nodeID
More specifically, I want to get the sum of the MAXIMUM INPUT BPS (90)
, MAXIMUM INPUT BPS (95)
, MAXIMUM INPUT BPS (99)
, and other relevant columns. For instance, for nodeID = 113
, there are multiple interfaceID
's associated to that nodeID
and therefore will output the results to have multiple rows for nodeID = 113
.
MAXIMUM INPUT BPS (90)
MAXIMUM INPUT BPS (95)
MAXIMUM INPUT BPS (99)
nodeID = 113
interfaceID
nodeID
nodeID = 113
What I want to do is get the sum of the MAXIMUM INPUT BPS (90)
and other numerical columns to be displayed in one singular row for nodeID = 113
and show the numerical outputs (I'm fine with removing the interfaceID
's and interface name columns).
MAXIMUM INPUT BPS (90)
nodeID = 113
interfaceID
I'll be attaching an image of the current report to help clear things up for what I have now based off of the query I have written:
Any suggestions would be great. Thank you!
I'm not sure if I am allowed to do that as I have just added onto the query given by SolarWinds, the application I'm using for this sort of reporting.
– Nilay
Aug 28 at 17:27
1 Answer
1
SolarWinds Query Language (SWQL) supports aggregate functions such as SUM that you can use here to obtain the result that you are looking for.
SELECT NodeID, SUM([MAX INPUT BPS(90)])
FROM
<TableName>
GROUP BY NodeID
Please let me know if this helps answer your question.
To see what's supported by SWQL, you can visit this page
I tried "SELECT Interfaces.InterfaceId, Nodes.NodeID, NodeName, SUM(Interface_Caption), SUM(Maxbps_In90), SUM(Maxbps_Out90), SUM(Maxbps_In95), SUM(Maxbps_Out95), SUM(Maxbps_In99), SUM(Maxbps_Out99)" and ended with GROUP BY Nodes.NodeID and had no budge. This is just a code snippet but I can direct message you the rest if you need it.
– Nilay
Sep 5 at 13:20
Can I get an update please?
– Nilay
Sep 12 at 13:16
By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.
Hi. Welcome to SO. If you can post your current query that you are having problems with, we can help you better.
– Saravana Kannadasan Gandhi
Aug 28 at 15:57