Convert Unix Shell Script to Windows Batch File

Question

I have a shell script (.sh) that works in Unix but I'd like to convert it into a Windows batch file (.bat):

cat >flog.ctl <<_EOF
LOAD DATA
INFILE '$1.log' "str ';'"
APPEND INTO TABLE flog
fields terminated by '=' TRAILING NULLCOLS
(
filename constant '$1'
,num  char
,data char
)
_EOF

sqlldr <username>/<password>@<instance> control=flog.ctl data=$1.log

I'm not too knowledgable on batch files but if an answer is some hints rather than a complete solution then I'm sure I'll muddle through. This is to help with an answer to another question here.

JamieSee · Accepted Answer · 2012-02-03 00:16:45Z

3

Windows batch files do not support inlining data this way. You'll have to

ECHO firstLine > flog.ctl
ECHO additionalLine >> flog.ctl

Windows batch files denote variables in ECHO statements using % signs, i.e. %1%.

So you're resulting file would be something like:

@ECHO OFF
ECHO LOAD DATA > flog.ctl
ECHO INFILE '%1%.log' "str ';'" >> flog.ctl
ECHO APPEND INTO TABLE flog >> flog.ctl
ECHO fields terminated by '=' TRAILING NULLCOLS >> flog.ctl
ECHO ( >> flog.ctl
ECHO filename constant '%1%' >> flog.ctl
ECHO ,num  char >> flog.ctl
ECHO ,data char >> flog.ctl
ECHO ) >> flog.ctl

sqlldr <username>/<password>@<instance> control=flog.ctl data=%1%.log

edited Feb 3, 2012 at 0:16

answered Feb 3, 2012 at 0:07

JamieSee

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6 Comments

John Doyle Over a year ago

Is there a difference between %1% and %1 as used in @voithos's code? I've tested both and both work.

voithos Over a year ago

@JohnDoyle: The second % is not necessary. I think James might be used to using two signs to delimit variables in batch files, like shown here. But the "correct" way to reference command line arguments, as far as I know, is to use a single %.

JamieSee Over a year ago

Hi, @JohnDoyle. Neither way is more correct than the other. It's true that you can always reference the argument variables 0-9 with a single % because they get special treatment when expanded. I make always using the second % a habit because when using named variables, ECHO %Asometext%B will have the wrong output and ECHO %A%sometext%B% will not. Give it a try some time and you'll see what I mean.

JamieSee Over a year ago

@voithos See my comment above.

voithos Over a year ago

@James: I still prefer the single %. The double % works because command-line arguments get mapped to environment variables of the same name. I like differentiating between arguments and environment variables, and I know of at least one reason to do so: %* can be used a shorthand for "all the arguments", whereas %*% is an error, and will just give *. In short, even though you can use either one most of the time, they're not equivalent.

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voithos · Accepted Answer · 2012-02-03 00:14:57Z

2

There are some complicated solutions as far as the inlining goes, but as was already mentioned, a simple solution would be to use echo.

@echo off

echo LOAD DATA > flog.ctl
echo INFILE '%1.log' "str ';'" >> flog.ctl
echo APPEND INTO TABLE flog >> flog.ctl
echo fields terminated by '=' TRAILING NULLCOLS >> flog.ctl
echo ( >> flog.ctl
echo filename constant '%1' >> flog.ctl
echo ,num  char >> flog.ctl
echo ,data char >> flog.ctl
echo ) >> flog.ctl

sqlldr <username>/<password>@<instance> control=flog.ctl data=%1.log

answered Feb 3, 2012 at 0:14

voithos

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