I am looking for any known tools or scripts which can convert my bcp files to csv files.
Input bcp file format:
Fields separated by 'XXXXXXX'
Rows separated by 'YYYYYYY'
Fields contains special characters like CRLF,CR,LF,", Tab, comma etc...
Output format I want:
Standard csv format file with comma delimited
Field values should contain original content including special characters (I mean no addition or deletion of special character(s), CR should also not be deleted)
The file able to cut by column index/name to select interested columns.
For this I did the following:
Transformed the bcp file to csv with few sed commands, with this I can open the file in MS excel program with proper alignment, and I could see content was not altered (as expected).
sed -i 's/\"/\"\"/g' $inFile
sed -i 's/XXXXXXX/","/g' $inFile
sed -i 's/YYYYYYY/"\n"/g' $inFile
sed -i '1s/^/\"/' $inFile
sed -i '$s/\"//' $inFile
sed -i -e '${/^$/d}' $inFile
sed -i '1s/^/"Header","added","here"\n/' $inFile
Tried csvkit tool: csvcut $infile
This tool is selecting the preferred columns but modifying the content like deleting the CR.
Any ideas in this kind conversion?
awk to the rescue!
awk -F='XXXXXXX' -v RS='YYYYYYY' -v OFS='","' -v ORS='\n'
'{gsub(/"/,"\""); $1=$1; print "\""$0"\""}' file
escape quotes, quote fields, change field delimiter to comma and record delimiter to newline. $1=$1 forces awk to apply the new delimiters.
Related
I have some data in the format:
-e, 's/,Chalk/,Cheese/g'
-e, 's/,Black/,White/g'
-e, 's/,Leave/,Remain/g'
in a file data.csv.
Using Gitbash, I use the file command to discover that this is ASCII text with CRLF terminators. If I also use the command cat -v , I see in Gitbash that each line ends ^M .
I want to remove those terminators, to leave a single line.
I've tried the following:
sed -e 's/'\r\n'//g' < data.csv > output.csv
taking care to put the \r\n in single quotes in order that the backslash is treated literally, but it does not work. No error, just no effect.
I'm using Gitbash for Windows.
Quotes within quotes cancel each other out, so you actually undo the quotes around the sed command for the newline characters. You could escape the quotes like 's|'\''\r\n'\''||g', but that would just include them in the string, which would not match anything in your case.
But that is not the only problem; sed by default only processes strings between newlines.
If you have the GNU version of sed, RAM to spare if the file is huge, and are sure the file does not contain data with null characters, try adding the -z argument, like:
sed -z -e 's|\r\n||g' < data.csv > output.csv
Though I guess you probably also want to replace it with a comma:
sed -z -e 's|\r\n|,|g' < data.csv > output.csv
For non-GNU versions of sed, you may have an easier time using tr instead, like:
tr '\r\n' ',' data.csv > output.csv
Have a file that has been created incorrectly. There are several space delimited fields in the file but one text field has some unwanted newlines. This is causing a big problem.
How can I remove these characters but not the wanted line ends?
file is:
'Number field' 'Text field' 'Number field'
1 Some text 999999
2 more
text 111111111
3 Even more text 8888888888
EOF
So there is a NL after the word "more".
I've tried sed:
sed 's/.$//g' test.txt > test.out
and
sed 's/\n//g' test.txt > test.out
But none of these work. The newlines do not get removed.
tr -d '\n' does too much - I need to remove ONLY the newlines that are preceded by a space.
How can I delete newlines that follow a space?
SunOS 5.10 Generic_144488-09 sun4u sparc SUNW,Sun-Fire-V440
A sed solution is
sed '/ $/{N;s/\n//}'
Explanation:
/ $/: whenever the line ends in space, then
N: append a newline and the next line of input, and
s/\n//: delete the newline.
It might be simplest with Perl:
perl -p0 -e 's/ \n/ /g'
The -0 flag makes Perl read the entire file as one line. Then we can substitute using s in the usual way. You can, of course, also add the -i option to edit the file in-place.
How can I delete newlines that follow a space?
If you want every occurrence of $' \n' in the original file to be replaced by a space ($' '), and if you know of a character (e.g. a control character) that does not appear in the file, then the task can be accomplished quite simply using sed and tr (as you requested). Let's suppose, for example, that control-A is a character that is not in the file. For the sake of simplicity, let's also assume we can use bash. Then the following script should do the job:
#!/bin/bash
A=$'\01'
tr '\n' "$A" | sed "s/ $A/ /g" | tr "$A" '\n'
I have tried to scan through the other posts in stack overflow for this, but couldn't get my code work, hence I am posting a new question.
Below is the content of file temp.
<?xml version="1.0" encoding="UTF-8"?>
<env:Envelope xmlns:env="http://schemas.xmlsoap.org/soap/envelope/<env:Body><dp:response xmlns:dp="http://www.datapower.com/schemas/management"><dp:timestamp>2015-01-
22T13:38:04Z</dp:timestamp><dp:file name="temporary://test.txt">XJzLXJlc3VsdHMtYWN0aW9uX18i</dp:file><dp:file name="temporary://test1.txt">lc3VsdHMtYWN0aW9uX18i</dp:file></dp:response></env:Body></env:Envelope>
This file contains the base64 encoded contents of two files names test.txt and test1.txt. I want to extract the base64 encoded content of each file to seperate files test.txt and text1.txt respectively.
To achieve this, I have to remove the xml tags around the base64 contents. I am trying below commands to achieve this. However, it is not working as expected.
sed -n '/test.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test.txt">##g'|perl -p -e 's#</dp:file>##g' > test.txt
sed -n '/test1.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test1.txt">##g'|perl -p -e 's#</dp:file></dp:response></env:Body></env:Envelope>##g' > test1.txt
Below command:
sed -n '/test.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test.txt">##g'|perl -p -e 's#</dp:file>##g'
produces output:
XJzLXJlc3VsdHMtYWN0aW9uX18i
<dp:file name="temporary://test1.txt">lc3VsdHMtYWN0aW9uX18i</dp:response> </env:Body></env:Envelope>`
Howeveer, in the output I am expecting only first line XJzLXJlc3VsdHMtYWN0aW9uX18i. Where I am commiting mistake?
When i run below command, I am getting expected output:
sed -n '/test1.txt"\>/,/\<\/dp:file\>/p' temp | perl -p -e 's#<dp:file name="temporary://test1.txt">##g'|perl -p -e 's#</dp:file></dp:response></env:Body></env:Envelope>##g'
It produces below string
lc3VsdHMtYWN0aW9uX18i
I can then easily route this to test1.txt file.
UPDATE
I have edited the question by updating the source file content. The source file doesn't contain any newline character. The current solution will not work in that case, I have tried it and failed. wc -l temp must output to 1.
OS: solaris 10
Shell: bash
sed -n 's_<dp:file name="\([^"]*\)">\([^<]*\).*_\1 -> \2_p' temp
I add \1 -> to show link from file name to content but for content only, just remove this part
posix version so on GNU sed use --posix
assuming that base64 encoded contents is on the same line as the tag around (and not spread on several lines, that need some modification in this case)
Thanks to JID for full explaination below
How it works
sed -n
The -n means no printing so unless explicitly told to print, then there will be no output from sed
's_
This is to substitute the following regex using _ to separate regex from the replacement.
<dp:file name=
Regular text
"\([^"]*\)"
The brackets are a capture group and must be escaped unless the -r option is used( -r is not available on posix). Everything inside the brackets is captured. [^"]* means 0 or more occurrences of any character that is not a quote. So really this just captures anything between the two quotes.
>\([^<]*\)<
Again uses the capture group this time to capture everything between the > and <
.*
Everything else on the line
_\1 -> \2
This is the replacement, so replace everything in the regex before with the first capture group then a -> and then the second capture group.
_p
Means print the line
Resources
http://unixhelp.ed.ac.uk/CGI/man-cgi?sed
http://www.grymoire.com/Unix/Sed.html
/usr/xpg4/bin/sed works well here.
/usr/bin/sed is not working as expected in case if the file contains just 1 line.
below command works for a file containing only single line.
/usr/xpg4/bin/sed -n 's_<env:Envelope\(.*\)<dp:file name="temporary://BackUpDir/backupmanifest.xml">\([^>]*\)</dp:file>\(.*\)_\2_p' securebackup.xml 2>/dev/null
Without 2>/dev/null this sed command outputs the warning sed: Missing newline at end of file.
This because of the below reason:
Solaris default sed ignores the last line not to break existing scripts because a line was required to be terminated by a new line in the original Unix implementation.
GNU sed has a more relaxed behavior and the POSIX implementation accept the fact but outputs a warning.
I'm trying to change the text in a file from this:
file.txt,c:\path\to\file
file.txt,c:\path with spaces\to\file
file.txt,c:\path\to\file with spaces
file.txt,c:\path\to\file with spaces
file.txt,c:\path\to\file with spaces
To this kind of output (one path to the file):
c:\path\to\file\file.txt
c:\path with spaces\to\file\file.txt
c:\path\to\file with spaces\file.txt
c:\path\to\file with spaces\file.txt
c:\path\to\file with spaces\file.txt
This ALMOST works but requires a ',' on the end of the line :
sed 's#\(.*\),\(.*\),\(.*\)#\2,\1,\3#g' file
Any help would be appreciated, I don't know sed all that well...
EDIT
This worked for me but I would still like to add a "\" in there:
sed 's#\(.*\),\(.*\)#\2,\1,\3#g' file
Escape the backslash, \\.
sed 's/^\(.*\),\(.*\)$/\2\\\1/g' file
--^^--
Also, you only need two capturing groups.
But since I'm more an awk guy.
awk -F, '{ print $2 "\\" $1 }' file
How do I split a file to N files using as a filename the first 2 chars on the line.
Ex input file:
AA23409234TEXT
BA23201202Other Text
AA23509234YADA
BA23202202More Text.
C1000000000000000000
Should generate 3 files:
AA.txt
AA23409234TEXT
AA23509234YADA
BA.txt
BA23201202Other Text
BA23202202More Text.
C1.txt
C1000000000000000000
I'm thinking of using a sed script similar to this
/^(..)/w \1
But what that really does is create a file named '\1' instead of the capture group.
Any ideas?
$ awk '{fname=substr($0, 0, 2); print >>fname}' input.txt
Or
$ while read line; do echo "$line" >>"${line:0:2}"; done <input.txt
The first thing you need to do is determine all of your file names:
filenames=$(sed 's/\(..\).*/\1/' listOfStrings.txt | sort | uniq)
Then, loop through those filenames
for filename in $filenames
do
sed -n '/^$filename/ p' listOfStrings.txt > $filename.txt
done
I have not tested this, but I think it should work.
This might work for you:
sed 's/\(..\).*/echo "&" >>\1.txt/' file | sh
or if you have GNU sed:
sed 's/\(..\).*/echo "&" >>\1.txt/e' file