What is the location of the SQL Server Fallback Certificate?

Question

My dataframe consists of the following table:

Time X Y
0100 5 9
0200 7 10
0300 11 12
0400 3 13
0500 4 14

My goal is to find the index of the value of Y which corresponds to a certain number (e.g.: 9) and return the corresponding X value from the table.

My idea previously was for a for-loop (as I have a number of Ys) to loop through and find all the values which match and then to create an empty array to store the values of X as such:

for i in (list of Ys):
    empty_storing_array.append(df[index_of_X].loc[df[Y] == i])

Problem is (if my newbie understanding of Pandas holds true), the values that loc gives is no number, but rather something else. How should I do it so that empty_storing_array then lists the numbers of X which corresponds to the values in array Y?

Solution

By default, the certificate is located in the registry, at:

HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Microsoft SQL Server\MSSQL.x\MSSQLServer\SuperSocketNetLib

(Where the MSSQL.x is MSSQL.x is a placeholder for the corresponding value of the instance of SQL Server.)

If SQL server cannot find a certificate at that location, it will fall back to searching the certificate store for a certificate with the FQDN of the computer it is installed on. By default, that certificate would be the machine certificate, found in Local Computer -> Personal -> Certificates through the MMC Certificates snap-in.

You would check the key length (and other properties) of this (or any) certificate through the MMC Certificates snap-in, by browsing to it, double clicking on it, and then hitting the Details tab in the ensuing dialogue box that comes up for the certificate.

If this cert is also unavailable, then SQL Server will generate a self-signed fallback certificate and use this for encryption. You can view the thumbprint or certificate being used when viewing the error logs around when SQL service restarts. The location of this fallback certificate is not officially known and other forums have suggested that this likely resides in memory. Please refer here for more info

Comments

[user3483203]

Is it always a 1-1 relationship? If it is, just set_index to Y and use loc: df.set_index('Y').loc[9, 'X']

[BENY]

check .loc and bool slice

[Joker]

@user3483203 No, there is no 1-1 relationship, the sizes of both columns are different