I am wondering if it is possible to have a comment within a batch file command. Specifically, I have a long SED command like the following:
#SED -r -e "s/.../.../"^
-e "s/.../.../"^
-e "s/.../.../"^
fileName >outFileName
I would like to add a comment to each of the "-e" options, as indicated in the following examples:
:: Option #1: At the end of the line
#SED -r -e "s/.../.../"^ // First comment
-e "s/.../.../"^ // Second comment
-e "s/.../.../"^ // Third comment
fileName >outFileName
:: Option #2: Between lines
#SED -r
#REM First comment
-e "s/.../.../"^
#REM Second comment
-e "s/.../.../"^
#REM Third comment
-e "s/.../.../"^
fileName >outFileName
Is there any way to accomplish this?
Give this a try. I don't have sed so I just tested with echo.
#echo off
:: Option #1: At the end of the line
echo SED -r -e "s/.../.../" %= First comment =%^
-e "s/.../.../" %= second comment =%^
-e "s/.../.../" %= third comment =%
:: Option #2: Between lines
echo SED -r^
%= First comment =%^
-e "s/.../.../"^
%= second comment =%^
-e "s/.../.../"^
%= third comment =%^
-e "s/.../.../"
pause
Output
SED -r -e "s/.../.../" -e "s/.../.../" -e "s/.../.../"
SED -r -e "s/.../.../" -e "s/.../.../" -e "s/.../.../"
Press any key to continue . . .
Related
I need to run about 50 scripts in a folder using sqlcmd from a batch file. Each script's query results need to be sent to its own output file. I have a working batch file that just runs each from a separate line:
sqlcmd -S %INSTANCE% -d %DATABASE% -U %USERNAME% -P "%PASSWORD%" -i "%SCRIPTFOLDER%\master_departments.sql" -s "|" -o "%OUTPUTFOLDER%\master_departments.csv" -W
sqlcmd -S %INSTANCE% -d %DATABASE% -U %USERNAME% -P "%PASSWORD%" -i "%SCRIPTFOLDER%\master_companies.sql" -s "|" -o "%OUTPUTFOLDER%\master_companies.csv" -W
sqlcmd -S %INSTANCE% -d %DATABASE% -U %USERNAME% -P %PASSWORD% -i "%SCRIPTFOLDER%\bill_history.sql" -s "|" -o "%OUTPUTFOLDER%\bill_history.csv" -W
sqlcmd -S %INSTANCE% -d %DATABASE% -U %USERNAME% -P %PASSWORD% -i "%SCRIPTFOLDER%\episodes.sql" -s "|" -o "%OUTPUTFOLDER%\episodes.csv" -W
Is there any way to run this in some kind of loop? I've seen examples that run a loop of all SQL scripts in a folder, but nothing that I've seen does it with an output file set.
Per #LotPings' suggestion I used the below code:
set INSTANCE=<someinstance>
set DATABASE=<somedb>
set USERNAME=<someuser>
set PASSWORD=<somepassword>
set "SCRIPTFOLDER=D:\<pathToScripts>\"
set "OUTPUTFOLDER=D:\<pathForOutput>\"
#Echo off
For /F "tokens=*" %%S in ('Dir /B "%SCRIPTFOLDER%*.sql" '
) do echo sqlcmd -S %INSTANCE% -d %DATABASE% -U %USERNAME% -P "%PASSWORD%" -i "%%~fS" -s "|" -o "%%~dpnS.csv" -W
#pause
I ran that in a batch file and when it paused, the last line said, "The system cannot find the file specified."
Thinking it was perhaps the backslashes in my paths, I removed them and put a slash before the .sql in the for line, but I got the same results.
Removing the backslash altogether resulted in a "File not found" message when I ran it like that.
In case your output file name matches the script name (without extension)
and your parameters are the same for all scripts
#Echo off
For /F "tokens=*" %%S in ('Dir /B "%SCRIPTFOLDER%*.sql" '
) do echo sqlcmd -S %INSTANCE% -d %DATABASE% -U %USERNAME% -P "%PASSWORD%" -i "%%~fS" -s "|" -o "%%~dpnS.csv" -W
The echo in front of sqlcmd prevents execution and allows to review the output. If all looks OK, remove the echo.
The for variable behaviour can be changed with ~ modifiers, see For /? or visit ss64.com/nt/for.html / syntax-args
To pass a folder to the batch you can input via set /P or hand over via command line arguments.
My script is designed to get files (and folders, eventually) and then list them with a while iteration.
format should be:
1) file.txt
2) newfile.txt
3) new folder
and it works! but then the while cycle wouldn't stop. given that i'm a newbie in bash can you tell me what am i doing wrong?
#!/bin/bash
shopt -s nullglob
FILES=(*)
var2="0"
while [ ${FILES[var2] -n } ] ; do
echo "$var2 ${FILES[var2]}"
((var2++))
done
The syntax is [ -n "string to check" ]:
#!/bin/bash
shopt -s nullglob
FILES=(*)
var2="0"
while [ -n "${FILES[var2]}" ] ; do
echo "$var2 ${FILES[var2]}"
((var2++))
done
You can also just do:
#!/bin/bash
shopt -s nullglob
num=0
for file in *
do
echo "$num) $file"
(( num++ ))
done
If this is the basis for a selection menu, you can instead use a select loop which would be much simpler.
I am using widows XP OS and have batch file where I set few variable values.
when i echo those varibales i can see the values, but when i use it in some commands i get empty sting as its value.
Sample Batch file
#ECHO OFF
SET "output=select * from employee where empid='160'"
CALL SET output=%%output:'=''%%
ECHO "%output%"
sqlcmd -b -h-1 -m-1 -V1 -S testsvr -E -Q "%output%' " -d tesdb
o/p select * from employee where empid=''160''
and value of variable in sqlcmd command is empty space.
I'm not sure, what you want to do.
#ECHO OFF
SET output=select * from employee where empid="160"
ECHO %output%
ECHO sqlcmd -b -h-1 -m-1 -V1 -S testsvr -E -Q '%output%' -d tesdb
..output is:
select * from employee where empid="160"
sqlcmd -b -h-1 -m-1 -V1 -S testsvr -E -Q 'select * from employee where empid="160"' -d tesdb
Try to place the quotes as follows:
SET output="select * from employee where empid='160'"
sqlcmd -b -h-1 -m-1 -V1 -S testsvr -E -Q %output% -d tesdb
I have the following code for finding a string in a file then delete the line which contains that string.
echo `sed /$string/d file.txt` > file.txt
the problem is that if initially file.txt contains:
a
b
c
after deleting "a" (string=a) file.txt will become
b c
instead of
b
c
can any one help me?
This is because of the backticks. Do this instead:
sed -i /$string/d file.txt
Note, if you want to do this in-place, you need to use -i to sed as > will destroy the file before sed can read it.
You do not need the echo wrap, simply try:
sed -i '/a/d' file.txt
You need to quote the command's output:
echo -n "sed /$string/d file.txt" > file.txt
sed has an in-place editing option. It's more proper to use that in your senario.
e.g.
sed -i /$string/d file.txt
For the problem of your case, as the output of `` is not enclosed in double quotes, word splitting is done, by bash. And the newlines are removed.
To use echo in this case, do it like this:
echo "`sed /$string/d file.txt`" > file.txt
I'm trying to remove all lines of text that contain a double quote, and I have tried this:
sed -ne '/\"/!p' theinput > theproduct
It left the lines untouched. What do I do? Here is my script:
`touch tmp.txt
open tmp.txt
read -sn 1 -p "Paste in data and press any key to convert"
echo
touch tmp.txt
open tmp.txt
read -sn 1 -p "Paste in data and press any key to convert"
echo
sed -e 's/-/ /g' tmp.txt > tmp2.txt
grep -v '"' tmp2.txt > final.txt
open final.txt
echo Study Conversion Successful
The first sed command works. It replaces a hyphen with a bunch of spaces (don't ask why I need that). The grep command, which I added from a response, does not work. It leaves the lines with quotes untouched.
Its not necessary to escape the double quote:
sed -ne '/"/!p' theinput > theproduct
Very strange. It "works for me"
$ cat data.txt
dsklfljs
sdjflk"Sdgsd"
sdfj sldkfj "Sdfsd"
sdfj
sdf
sdjflks
$ sed -ne '/\"/!p' data.txt
dsklfljs
sdfj
sdf
sdjflks
Perhaps it is a version issue with sed?
However, you can also consider using grep -v for this.
$ grep -v '"' data.txt
dsklfljs
sdfj
sdf
sdjflks