ms-access-97

How to connect to MS Access 97 (with workgroup mdw) database using pyodbc

由 我是研究僧i 提交于 2019-12-12 03:24:39
问题 In a Python script with pyodbc, I am trying to connect to a still used Access 97 database on our network but I have problems to get the connection (or the connection string) to work. I keep getting the following error: pyodbc.Error: ('IM002', '[IM002] [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified (0) (SQLDriverConnectW)') This is my connection string which fails with above mentioned error: conn = pyodbc.connect('Driver={Microsoft Access Driver (

Accessing another password protected database in an SQL query in Access 97

由 故事扮演 提交于 2019-12-10 15:17:06
问题 I am currently working on an SQL query for Access 97. Given are the following tables (simplified for demonstration purposes), each of which are located in separate mdb files: Table1 in C:\db\db1.mdb: PartyId (PK) Name ------------ -------- 1 A 2 B 3 C Table2 in C:\db\db2.mdb: PartyId (PK) Date (PK) Value ------------ --------- ----- 1 6/30/2014 4 1 7/1/2014 8 2 5/3/2014 3 3 5/5/2014 5 3 5/3/2014 1 3 5/2/2014 2 Here I would like to look for the most recent value of each party, based on a

Find the directory part (minus the filename) of a full path in access 97

由 半腔热情 提交于 2019-11-29 09:06:31
For various reasons, I'm stuck in Access 97 and need to get only the path part of a full pathname. For example, the name c:\whatever dir\another dir\stuff.mdb should become c:\whatever dir\another dir\ This site has some suggestions on how to do it: http://www.ammara.com/access_image_faq/parse_path_filename.html But they seem rather hideous. There must be a better way, right? That's about it. There is no magic built-in function... You can do something simple like: Left(path, InStrRev(path, "\")) Example: Function GetDirectory(path) GetDirectory = Left(path, InStrRev(path, "\")) End Function

Find the directory part (minus the filename) of a full path in access 97

由 一世执手 提交于 2019-11-27 06:34:58
问题 For various reasons, I'm stuck in Access 97 and need to get only the path part of a full pathname. For example, the name c:\whatever dir\another dir\stuff.mdb should become c:\whatever dir\another dir\ This site has some suggestions on how to do it: http://www.ammara.com/access_image_faq/parse_path_filename.html But they seem rather hideous. There must be a better way, right? 回答1: That's about it. There is no magic built-in function... 回答2: You can do something simple like: Left(path,
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