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Say that you have a data file called c:\dissertation\salary6.sd2. Because the extension of the file is .sd2 we know it is a Windows SAS 6.xx file. You can read a file in version 8 much like you would have in version 6, except that you need to explicitly tell SAS that the file is a version 6 file, as shown in the example below. Note the v6 in the example below -- this tells SAS that the libname diss6 will read a version 6.xx file from the directory c:\dissertation.
We see the output from this program below. This shows us that we read the file successfully.libname diss6 v6 'c:\dissertation\'; proc contents data=diss6.salary6; run; proc print data=diss6.salary6; run;
The CONTENTS Procedure
Data Set Name: DISS6.SALARY6 Observations: 2
Member Type: DATA Variables: 5
Engine: V6 Indexes: 0
Created: 16:53 Thursday, November 16, 2000 Observation Length: 40
Last Modified: 16:53 Thursday, November 16, 2000 Deleted Observations: 0
Protection: Compressed: NO
Data Set Type: Sorted: NO
Label:
-----Engine/Host Dependent Information-----
<output edited to save space>
File Name: c:\dissertation\salary6.sd2
Release Created: 6.08.00
Host Created: WIN
-----Alphabetic List of Variables and Attributes-----
# Variable Type Len Pos
-----------------------------------
1 SAL1996 Num 8 0
2 SAL1997 Num 8 8
3 SAL1998 Num 8 16
4 SAL1999 Num 8 24
5 SAL2000 Num 8 32
Obs SAL1996 SAL1997 SAL1998 SAL1999 SAL2000
1 10000 10500 11000 12000 12700
2 14000 16500 18000 22000 29000
We wee the output from this program below. You can see that the file is now called c:\dissertation\salary8.sas7bdat and you can see that SAS says that the release that created it is version 8 (actually 8.0101M0, i.e., version 8). You can now read and use this file as a version 8 SAS data file.libname diss6 v6 'c:\dissertation\'; data 'c:\dissertation\salary8' ; set diss6.salary6; run; proc contents data='c:\dissertation\salary8'; run; proc print data='c:\dissertation\salary8'; run;
The CONTENTS Procedure
Data Set Name: c:\dissertation\salary8 Observations: 2
Member Type: DATA Variables: 5
Engine: V8 Indexes: 0
Created: 16:53 Thursday, November 16, 2000 Observation Length: 40
Last Modified: 16:53 Thursday, November 16, 2000 Deleted Observations: 0
Protection: Compressed: NO
Data Set Type: Sorted: NO
Label:
-----Engine/Host Dependent Information-----
<output edited to save space>
File Name: c:\dissertation\salary8.sas7bdat
Release Created: 8.0101M0
Host Created: WIN_NT
-----Alphabetic List of Variables and Attributes-----
# Variable Type Len Pos
-----------------------------------
1 SAL1996 Num 8 0
2 SAL1997 Num 8 8
3 SAL1998 Num 8 16
4 SAL1999 Num 8 24
5 SAL2000 Num 8 32
Obs SAL1996 SAL1997 SAL1998 SAL1999 SAL2000
1 10000 10500 11000 12000 12700
2 14000 16500 18000 22000 29000
Say that you had numerous SAS version 6 files in c:\dissertation\ that you wanted to convert to version 8. For simplicity say that the files were called file1 file2 and file3, but you could have many such files. The example below shows how you could do the conversion using PROC COPY. Note that the files are read from a directory called c:\dissertation\ and then copied to a directory called c:\dissertation8\ . It is recommended that you use this kind of strategy to copy the files from one location to another.
libname diss6 v6 'c:\dissertation\'; libname diss8 v8 'c:\dissertation8\';proc copy in=diss6 out=diss8; select file1 file2 file3; run;
libname first v6 "c:\diss6"; libname second v8 "c:\diss8"; proc catalog cat=first.FORMATS; copy out=second.FORMATS; run;proc format library=second fmtlib; run;
We omit the output from this, but the output would show the formats associated with the new (version 8) format library that was created.
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