Querying Data From One Table Against Another Table For All Results

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Querying Data From One Table Against Another Table For All Results

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michaelhamptonwork

Posted Monday, August 13, 2012 3:03 PM

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Hello,
I am new to the forums and overall new to SQL and SRSS. I have two 2008 SQL databases that I am trying to generate a report from two different data sets.

Essentially, I have one database that contains all of our supported buildings with their subnets:

Table - Building_Information

Building_Name Subnet
Building 1 192.168.1.%
Building 2 192.168.2.%
Building 3 192.168.3.%

And then in another database, I have a list of all of our devices:

Table - Device_Information

DeviceNum IP_Address
Device1 192.168.1.2
Device2 192.168.1.3
Device3 192.168.2.2
Device4 192.168.3.1
Device5 192.168.3.2
Device6 192.168.3.3

What I am trying to achieve is a report by Building of the total number of devices in each Building.

I have a Dataset for Building_Information:

SELECT *
FROM Building_Information

and a second Dataset for Device_Information:

SELECT count(IP_Address)
FROM Device_Information
WHERE (IP_Address LIKE (@Subnet))

matching the parameter of @Subnet on the queried values of the Subnet from the Building_Information query.

However, on the multiple value pass it gives an error because of the like statement (Incorrect syntax ","). When I try to substitute the LIKE for IN, it will only match the value if it is an exact match.

Any ideas would be great.

Thank you.

Post #1344441

IgorMi

Posted Monday, August 13, 2012 3:44 PM

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Hi,

You must miss something simple. I don't know your exact code, however put attention on data types in tables of both databases.

select col1,col2,...,colN
from db1.shcema1.table1 tb1
join db2.schema2.table2 tb2 on convert(nvarchar(100), tb1.id) = convert(nvarchar(100),tb2.id)
where ... 'your conditions'

also check if the collations on both databases are same.
:)

Post #1344459

rVadim

Posted Monday, August 13, 2012 3:56 PM

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Please next time provide DDL and INSERTs like this:


CREATE TABLE #Building_Information (
  Building_Name VARCHAR(50)
 ,Subnet VARCHAR(15)
);

INSERT INTO #Building_Information (Building_Name, Subnet)
VALUES ('Building 1', '192.168.1.%'),
       ('Building 2', '192.168.2.%'),
       ('Building 3', '192.168.3.%')

SELECT * FROM #Building_Information

CREATE TABLE #Device_Information (
  DeviceNum VARCHAR(50)
 ,IP_Address VARCHAR(15)
);

INSERT INTO #Device_Information (DeviceNum, IP_Address)
VALUES ('Device1', '192.168.1.2'),
       ('Device2', '192.168.1.3'),
       ('Device3', '192.168.2.2'),
       ('Device4', '192.168.3.1'),
       ('Device5', '192.168.3.2'),
       ('Device6', '192.168.3.3')

SELECT * FROM #Device_Information

Now what do you mean by "multiple value pass"? What is the value of @Subnet when error occures?

Post #1344463

Lynn Pettis

Posted Monday, August 13, 2012 4:11 PM

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Do you mean something like this (based on test setup provided by rVadim):


CREATE TABLE #Building_Information (
  Building_Name VARCHAR(50)
 ,Subnet VARCHAR(15)
);

INSERT INTO #Building_Information (Building_Name, Subnet)
VALUES ('Building 1', '192.168.1.%'),
       ('Building 2', '192.168.2.%'),
       ('Building 3', '192.168.3.%')

SELECT * FROM #Building_Information

CREATE TABLE #Device_Information (
  DeviceNum VARCHAR(50)
 ,IP_Address VARCHAR(15)
);

INSERT INTO #Device_Information (DeviceNum, IP_Address)
VALUES ('Device1', '192.168.1.2'),
       ('Device2', '192.168.1.3'),
       ('Device3', '192.168.2.2'),
       ('Device4', '192.168.3.1'),
       ('Device5', '192.168.3.2'),
       ('Device6', '192.168.3.3')

SELECT * FROM #Device_Information;

SELECT
    *
FROM
    #Building_Information bi
    INNER JOIN #Device_Information di
        ON (di.IP_Address LIKE bi.Subnet);

SELECT
    bi.Building_Name,
    COUNT(di.DeviceNum) NumberOfDevices
FROM
    #Building_Information bi
    INNER JOIN #Device_Information di
        ON (di.IP_Address LIKE bi.Subnet)
GROUP BY
    Building_Name;

GO

DROP TABLE #Building_Information;
DROP TABLE #Device_Information;
GO

Lynn Pettis

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Post #1344472

andersg98

Posted Monday, August 13, 2012 4:16 PM

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you will probably need to split the IP addresses to deal with the third octet reaching double digits. But with the data you supplied this works:

SELECT B.Building_Name, D.Device_cnt
FROM Building_Information AS B
LEFT JOIN (SELECT LEFT(IP_Address,9) AS Subnet, COUNT(DeviceNum) AS Device_cnt
FROM Device_Information
GROUP BY LEFT(IP_Address,9)) AS D
ON LEFT(B.Subnet,9) = D.Subnet

Post #1344475

michaelhamptonwork

Posted Tuesday, August 14, 2012 6:43 AM

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Thank you all for your help. I think I will try going the route that Lynn suggested and possibly try to import the table from my second database into my first as to eliminate that issue.

Thank you all.

Post #1344663

ChrisM@Work

Posted Tuesday, August 14, 2012 6:57 AM

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Here's an alternative using PARSENAME()

SELECT
    bi.Building_Name,
    COUNT(di.DeviceNum) NumberOfDevices
FROM #Building_Information bi
INNER JOIN #Device_Information di
	ON PARSENAME(bi.Subnet,4) = PARSENAME(di.IP_Address,4)
	AND PARSENAME(bi.Subnet,3) = PARSENAME(di.IP_Address,3)
	AND PARSENAME(bi.Subnet,2) = PARSENAME(di.IP_Address,2)
GROUP BY Building_Name;

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

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Post #1344680

Lynn Pettis

Posted Tuesday, August 14, 2012 8:12 AM

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ChrisM@Work (8/14/2012)

Here's an alternative using PARSENAME()

SELECT
    bi.Building_Name,
    COUNT(di.DeviceNum) NumberOfDevices
FROM #Building_Information bi
INNER JOIN #Device_Information di
	ON PARSENAME(bi.Subnet,4) = PARSENAME(di.IP_Address,4)
	AND PARSENAME(bi.Subnet,3) = PARSENAME(di.IP_Address,3)
	AND PARSENAME(bi.Subnet,2) = PARSENAME(di.IP_Address,2)
GROUP BY Building_Name;

Only issue with using parsename() is that it is slow and will disallow the use of indexes on the joining columns if they exist.

Cool

Lynn Pettis

SQL Musings from the Desert Fountain Valley SQL (My Mirror Blog)

Post #1344740

ChrisM@Work

Posted Tuesday, August 14, 2012 8:17 AM

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Lynn Pettis (8/14/2012)

ChrisM@Work (8/14/2012)

Here's an alternative using PARSENAME()

SELECT
    bi.Building_Name,
    COUNT(di.DeviceNum) NumberOfDevices
FROM #Building_Information bi
INNER JOIN #Device_Information di
	ON PARSENAME(bi.Subnet,4) = PARSENAME(di.IP_Address,4)
	AND PARSENAME(bi.Subnet,3) = PARSENAME(di.IP_Address,3)
	AND PARSENAME(bi.Subnet,2) = PARSENAME(di.IP_Address,2)
GROUP BY Building_Name;

Only issue with using parsename() is that it is slow and will disallow the use of indexes on the joining columns if they exist.

Absolutely. Your alternative, using LIKE with a wildcard to the right of the test string, is SARGable. Thanks for the reminder, Lynn.

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

Post #1344746

SQLKnowItAll

Posted Tuesday, August 14, 2012 8:35 AM

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I think we are missing a key aspect of what the OP is asking. This is closer to an SSRS question, I believe. There are 2 datasets in his report; 1 to provide the selected buildings as a drop-down list and another to provide the data based on the selection. Then, in SSRS you can allow "multiple selection" in the drop-down (report parameter). Check this out if you are trying to understand the multi-value parameter in SSRS:
http://social.msdn.microsoft.com/Forums/en-US/sqlreportingservices/thread/c02370b5-aeda-47ec-a3a8-43b2ec1e6c26/

Thanks,

Jared
SQL Know-It-All

How to post data/code on a forum to get the best help - Jeff Moden

Post #1344759

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