I am trying to move files from a location to another one based on filename ONLY by using command line. The below works pretty fine but I do not want to specify both, the file name and the extension:
MOVE C:\firstlocation\file.sdltm "C:\secondlocation\file.sdltm"
The reason behind this is that I have the following files in the firstlocation:
file.sdltm
file.tmx
and my aim is to move both files named "file" to the secondlocation regardless the extension.
Sorry if this question has been asked before, - couldn't find any that matches my criteria.
Related
I have very little experience with the command line and I'm trying to do something very complicated (to me).
I have a directory with A LOT of subfolders and files in them. All file names contain the parent folder name, e.g.:
Folder1
data_Folder1.csv
other_file_Folder1.csv
Folder2
data_Folder2.csv
other_file_Folder2.csv
In another folder (all in one directory), I have a new version of all the data_FolderX.csv files and I need to replace them in the original folders. I cannot give them another name because of later analyses. Is there a way to replace the files in the original folders with the new version, in the command line?
I tried this Replacing a file into multiple folders/subdirectories but didn't work for me. Given that I have many .csv files in the derectories, I don't want to replace them all, so I don't think I should do it based on the file extension. I would also like to note that the name "FolderX" contains several other _, so in principal, I want to replace the .csv file starting with data in the FolderX.
Can anyone help?
Thanks in advance!
I have two folders with many files of the same naming format but differing extensions. I would like to loop through the files of the first folder and extract their filenames to check if a file of the same name exists in the other folder, then create one (with the other extension) if it doesn't. I can't get a grapple on bash for whatever reason, so I was wondering if there's a simple way to do this.
I need to rename 80k files in multiple folders & subfolders in the same directory. I have been trying to use ren but have been unsuccessful; I get an incorrect syntax error.
My old name looks like this:
c:/users/alice/BiDIR_DOCS_2017_Nov08020423\Company,LLC##NA##7967425.00##7967425.00\Company LLC A and A - Aug2017.pdf BiDIR_DOCS_2017_Nov08020423\Company, LLC##NA##7967425.00##7967425.00\document_# (x.y.z)-test~.pdf
and my new name looks like this:
c:/users/alice/BiDIR_DOCS_2017_Nov08020423\Company,LLC##NA##7967425.00##7967425.00\Company LLC A and A - Aug2017.pdf BiDIR_DOCS_2017_Nov08020423\Company, LLC##NA##7967425.00##7967425.00\system, a old name~ ` to # system b document (xyz)-test.pdf
I have the existing directory print in one column of Excel and in the next column what I want the directory print to be.
I'm not sure if I'm starting my ren command at the right hierarchy of my directory, or if I need quotation marks to keep the spaces and symbols in my new name.
I have tried improvising and testing on my own without success and I cannot find an article online on point.
Try FAR (find and replace) - it a free utility that works well.
http://findandreplace.sourceforge.net/
Oracle's sqlldr defaults to a .dat extension. That I want to override. I don't like to rename the file. When googled get to know few answers to use . like data='fileName.' which is not working. Share your ideas, please.
Error message is fileName.dat is not found.
Sqlloder has default extension for all input files data,log,control...
data= .dat
log= .log
control = .ctl
bad =.bad
PARFILE = .par
But you have to pass filename without apostrophe and dot
sqlloder pass/user#db control=control data=data
sqloader will add extension. control.ctl data.dat
Nevertheless i do not understand why you do not want to specify extension?
You can't, at least in Unix/Linux environments. In Windows you can use the trailing period trick, specifying either INFILE 'filename.' in the control file or DATA=filename. on the command line. WIndows file name handling allows that; you can for instance do DIR filename. at a command prompt and it will list the file with no extension (as will DIR filename). But you can't do that with *nix, from a shell prompt or anywhere else.
You said you don't want to copy or rename the file. Temporarily renaming it might be the simplest solution, but as you may have a reason not to do that even briefly you could instead create a hard or soft link to the file which does have an extension, and use that link as the target instead. You could wrap that in a shell script that takes the file name argument:
# set variable from correct positional parameter; if you pass in the control
# file name or other options, this might not be $1 so adjust as needed
# if the tmeproary file won't be int he same directory, need to be full path
filename=$1
# optionally check file exists, is readable, etc. but overkill for demo
# can also check temporary file does not already exist - stop or remove
# create soft link somewhere it won't impact any other processes
ln -s ${filename} /tmp/${filename##*/}.dat
# run SQL*Loader with soft link as target
sqlldr user/password#db control=file.ctl data=/tmp/${filename##*/}.dat
# clean up
rm -f /tmp/${filename##*/}.dat
You can then call that as:
./scriptfile.sh /path/to/filename
If you can create the link in the same directory then you only need to pass the file, but if it's somewhere else - which may be necessary depending on why renaming isn't an option, and desirable either way - then you need to pass the full path of the data file so the link works. (If the temporary file will be int he same filesystem you could use a hard link, and you wouldn't have to pass the full path then either, but it's still cleaner to do so).
As you haven't shown your current command line options you may have to adjust that to take into account anything else you currently specify there rather than in the control file, particularly which positional argument is actually the data file path.
I have the same issue. I get a monthly download of reference data used in medical application and the 485 downloaded files don't have file extensions (#2gb). Unless I can load without file extensions I have to copy the files with .dat and load from there.
I have a bunch of files I'm trying to organize quickly, and I had two questions about how to do that. I really appreciate any help! I tried searching but couldn't find anything on these specific commands for OSX.
First, I have about 100 folders in a directory - I'd like to place an folder in each one of those folders.
For example, I have
Cars/Mercedes/<br>
Cars/BMW/<br>
Cars/Audi/<br>
Cars/Jeep/<br>
Cars/Tesla/
Is there a way I can create a folder inside each of those named "Pricing" in one command, i.e. ->
Cars/Mercedes/Pricing <br>
Cars/BMW/Pricing<br>
Cars/Audi/Pricing<br>
Cars/Jeep/Pricing<br>
Cars/Tesla/Pricing
My second question is a little tougher to explain. In each of these folders, I'd like move certain files into these newly created folders (above) in the subdirectory.
Each file has a slightly different filename but contains the same string of letters - for example, in each of the above folders, I might have
Cars/Mercedes/payment123.html
Cars/BMW/payment432.html
Cars/Audi/payment999.html
Cars/Jeep/payment283.html
Is there a way to search each subdirectory for a file containing the string "payment" and move that file into a subfolder in that subdirecotry - i.e. into the hypothetical "Pricing" folders we just created above with one command for all the subdirectories in Cars?
Thanks so much~! help with either of these would be invaluable.
I will assume you are using bash, since it is the default shell in OS X. One way to do this uses a for loop over each directory to create the subdirectory and move the file. Wildcards are used to find all of the directories and the file.
for DIR in Cars/*/ ; do
mkdir "${DIR}Pricing"
mv "${DIR}payment*.html" "${DIR}Pricing/"
done
The first line finds every directory in Cars, and then runs the loop once for each, replacing ${DIR} with the current directory. The second line creates the subdirectory using the substitution. Note the double quotes, which are necessary only if the path could contain spaces. The third line moves any file in the directory whose name starts with "payment" and ends with ".html" to the subdirectory. If you have multiple files which match this, they will all be moved. The fourth line simply marks the end of the loop.
If you are typing this directly into the command line, you can combine it into a single line:
for DIR in Cars/*/ ; do mkdir "${DIR}Pricing"; mv "${DIR}payment*.html" "${DIR}Pricing/"; done