Shell Scripting sed Errors:
Cannot view /home/xx/htdocs/*/modules/forms/int.php
/bin/rm: cannot remove `/home/xx/htdocs/tmp.26758': No such file or directory
I am getting an error in my shell script. I am not sure if this for loop will work, it is intended to climb a large directory tree of PHP files and prepend a functions in every int.php file with a little validation. Don't ask me why this wasn't centralized/OO but it wasn't. I copied the script as best I could from here: http://www.cyberciti.biz/faq/unix-linux-replace-string-words-in-many-files/
#!/bin/bash
OLD="public function displayFunction(\$int)\n{"
NEW="public function displayFunction(\$int)\n{if(empty(\$int) || !is_numeric(\$int)){return '<p>Invalid ID.</p>';}"
DPATH="/home/xx/htdocs/*/modules/forms/int.php"
BPATH="/home/xx/htdocs/BAK/"
TFILE="/home/xx/htdocs/tmp.$$"
[ ! -d $BPATH ] && mkdir -p $BPATH || :
for f in $DPATH
do
if [ -f $f -a -r $f ]; then
/bin/cp -f $f $BPATH
sed "s/$OLD/$NEW/g" "$f" > $TFILE && mv $TFILE "$f"
else
echo "Error: Cannot view ${f}"
fi
done
/bin/rm $TFILE
Do wildcards like this even work? Can I check in every subdirectory across a tree like this? Do I need to precode an array and loop over that? How would I go about doing this?
Also is, the $ in the PHP code breaking the script at all?
I am terribly confused.
Problems in your code
You cannot use sed to replace multiple lines this way.
you are using / in OLD which is used in a s/// sed command. This won't work
[ ! -d $BPATH ] && mkdir -p $BPATH || : is horrible. use mkdir -p "$bpath" 2>/dev/null
Yes, wildcards like this will work but only because your string has no spaces
Doube-quote your variables, or your code will be very dangerous
Single quote your strings or you won't understand what you are escaping
Do not use capital variable names, you could accidentally replace a bash inner variable
do not rm a file that does not exist
Your backups will be overwritten as all files are named int.php
Assuming you are using GNU sed, I'm not used to other sed flavors.
If you are not using GNU sed, replacing the \n with a newline (inside the string) should work.
Fixed Code
#!/usr/bin/env bash
old='public function displayFunction(\$int)\n{'
old=${old//,/\\,} # escaping eventual commas
# the \$ is for escaping the sed-special meaning of $ in the search field
new='public function displayFunction($int)\n{if(empty($int) || !is_numeric($int)){return "<p>Invalid ID.</p>";}\n'
new=${new//,/\\,} # escaping eventual commas
dpath='/home/xx/htdocs/*/modules/forms/int.php'
for f in $dpath; do
[ -r "$f" ]; then
sed -i.bak ':a;N;$!ba;'"s,$old,$new,g" "$f"
else
echo "Error: Cannot view $f" >&2
fi
done
Links
Replace newline in sed
Inplace replace with sed with a backup
Using a different sed substitution separator
Existency not necessary if readable
Bash search and replace inside a variable
Bash guide
Related
I currently have a long list of files, which look somewhat like this:
Gmc_W_GCtl_E_Erz_Aue_Dl_281_heart_xerton
Gmc_W_GCtl_E_Erz_Aue_Dl_254_toe_taixwon
Gmc_W_GCtl_E_Erz_Homersdorf_Dl_201_head_xaubadan
Gmc_W_GCtl_E_Erz_Homersdorf_Dl_262_bone_bainan
Gmc_W_GCtl_E_Thur_Peuschen_Dl_261_blood_blodan
Gmc_W_GCtl_E_Thur_Peuschen_Dl_281_heart_xerton
The naming pattern all follow the same order, where I'm mainly seeking to group the files based on the part with "Aue", "Homersdorf", "Peuschen", and so forth (there are many others down the list), with the position of these keywords being always the same (e.g. they are all followed by Dl; they are all after the fifth underscore...etc.).
All the files are in the same folder, and I am trying to move these files into subfolders based on these keywords in bash, but I'm not quite certain how. Any help on this would be appreciated, thanks!
I am guessing you want something like this:
$ find . -type f | awk -F_ '{system("mkdir -p "$5"/"$6";mv "$0" "$5"/"$6)}'
This will move say Gmc_W_GCtl_E_Erz_Aue_Dl_281_heart_xerton into /Erz/Aue/Gmc_W_GCtl_E_Erz_Aue_Dl_281_heart_xerton.
Using the bash shell with a for loop.
#!/usr/bin/env bash
shopt -s nullglob
for file in Gmc*; do
[[ -d $file ]] && continue
IFS=_ read -ra dir <<< "$file"
echo mkdir -pv "${dir[4]}/${dir[5]}" || exit
echo mv -v "$file" "${dir[4]}/${dir[5]}" || exit
done
Place the script inside the directory in question make it executable and execute it.
Remove the echo's so it create the directories and move the files.
I am currently trying to list the size of all files in a directory which is passed as the first argument to the script, but the -f option in Linux is not working, or am I missing something.
Here is the code :
for tmp in "$1/*"
do
echo $tmp
if [ -f "$tmp" ]
then num=`ls -l $tmp | cut -d " " -f5`
echo $num
fi
done
How would I fix this problem?
I think the error is with your glob syntax which doesn't work in either single- or double-quotes,
for tmp in "$1"/*; do
..
Do the above to expand the glob outside the quotes.
There are couple more improvements possible in your script,
Double-quote your variables to prevent from word-splitting, e.g. echo "$temp"
Backtick command substitution `` is legacy syntax with several issues, use the $(..) syntax.
The [-f "filename"] condition check in linux is for checking the existence of a file and it is a regular file. For reference, use this text as reference,
-b FILE
FILE exists and is block special
-c FILE
FILE exists and is character special
-d FILE
FILE exists and is a directory
-e FILE
FILE exists
-f FILE
FILE exists and is a regular file
-g FILE
FILE exists and is set-group-ID
-G FILE
FILE exists and is owned by the effective group ID
I suggest you try with [-e "filename"] and see if it works.
Cheers!
At least on the command line, this piece of script does it:
for tmp in *; do echo $tmp; if [ -f $tmp ]; then num=$(ls -l $tmp | sed -e 's/ */ /g' | cut -d ' ' -f5); echo $num; fi; done;
If cut uses space as delimiter, it cuts at every space sign. Sometimes you have more than one space between columns and the count can easily go wrong. I'm guessing that in your case you just happened to echo a space, which looks like nothing. With the sed command I remove extra spaces.
deploy.sh
USERNAME="Tom"
PASSWORD="abc123"
FILE="config.conf"
sed -i "s/\PLACEHOLDER_USERNAME/$USERNAME/g" $FILE
sed -i "s/\PLACEHOLDER_PASSWORD/$PASSWORD/g" $FILE
config.conf
deloy="PLACEHOLDER_USERNAME"
pass="PLACEHOLDER_PASSWORD"
This file puts my variables defined in deploy into my config file. I can't source the file so I want put my variables in this way.
Question
I want a command that is generic to work for all placeholder variables using some sort of while loop rather than needing one command per variable. This means any term starting with placeholder_ in the file will try to be replaced with the value of the variable defined already in deploy.sh
All variables should be set and not empty. I guess if there is the ability to print a warning if it can't find the variable that would be good but it isn't mandatory for this.
Basically, use shell code to write a sed script and then use sed -i .bak -f sed.script config.conf to apply it:
trap "rm -f sed.script; exit 1" 0 1 2 3 13 15
for var in USERNAME PASSWORD
do
echo "s/PLACEHOLDER_$var/${!var}/"
done > sed.script
sed -i .bak -f sed.script config.conf
rm -f sed.script
trap 0
The main 'tricks' here are:
knowing that ${!var} expands to the value of the variable named by $var, and
knowing that sed will take a script full of commands via -f sed.script, and
knowing how to use trap to ensure temporary files are cleaned up.
You could also use sed -e "s/.../.../" -e "s/.../.../" -i .bak config.conf too, but the script file is easier, I think, especially if you have more than 2 values to substitute. If you want to go down this route, use a bash array to hold the arguments to sed. A more careful script would use at least $$ in the script file name, or use mktemp to create the temporary file.
Revised answer
The trouble is, although much closer to being generic, it is still not generic since I have to manually put in what variables I want to change. Can it not be more like "for each placeholder_, find the variable in deploy.sh and add that variable, so it can work for any number of variables.
So, find what the variables are in the configuration file, then apply the techniques of the previous answer to solve that problem:
trap "rm -f $tmp; exit 1" 0 1 2 3 13 15
for file in "$#"
do
for var in $(sed 's/.*PLACEHOLDER_\([A-Z0-9_]*\).*/\1/' "$file")
do
value="${!var}"
[ -z "$value" ] && { echo "$0: variable $var not set for $file" >&2; exit 1; }
echo "s/PLACEHOLDER_$var/$value/"
done > $tmp
sed -i .bak -f $tmp "$file"
rm -f $tmp
done
trap 0
This code still pulls the values from the environment. You need to clarify what is required if you want to extract the settings from the shell script, but it can be done — the script will have to be sufficiently self-aware to find its source so it can search it for the names. But the basics are in this answer; the rest is a question of tinkering until it does what you need.
#!/bin/ksh
TemplateFile=$1
SourceData=$2
(sed 's/.*/#V0r:PLACEHOLDER_&:r0V#/' ${SourceData}; cat ${TemplateFile}) | sed -n "
s/$/²/
H
$ {
x
s/^\(\n *\)*//
# also reset t flag
t varxs
:varxs
s/^#V0r:\([a-zA-Z0-9_]\{1,\}\)=\([^²]*\):r0V#²\(\n.*\)\"\1\"/#V0r:\1=\2:r0V#²\3\2/
t varxs
# clean the line when no more occurance in text
s/^[^²]*:r0V#²\n//
# and next
t varxs
# clean the marker
s/²\(\n\)/\1/g
s/²$//
# display the result
p
}
"
call like this: YourScript.ksh YourTemplateFile YourDataSourceFile where:
YourTemplateFile is the file that contain the structure with generic value like deloy="PLACEHOLDER_USERNAME"
YourDataSourceFile is the file that contain all the peer Generic value = specific value like USERNAME="Tom"
I am trying to replace the prefix of all files in the directory with another prefix (renaming).
This is my script
# Script to rename the files
#!/bin/bash
for file in $1*;
do
mv $file `echo $file | sed -e 's/^$1/$2/'`;
done
Upon executing the script with
rename.sh BIT SIT
I get the following errors
mv: `BITfile.h' and `BITFile.h' are the same file
mv: `BITDefs.cpp' and `BITDefs.cpp' are the same file
mv: `BITDefs.h' and `BITDefs.h' are the same file
Seems like sed is treating $1 and $2 as the same value, but when I print those variables on another line it shows that they are different.
As Roman Newaza says, you can use " instead of ' to tell Bash that you want variables to be expanded. However, in your case, it would be safest to write:
for file in "$1"* ; do
mv -- "$file" "$2${file#$1}"
done
so that weird characters in filenames, or in your script parameters, cannot cause any problems.
You can also use parameter expansion to replace the prefix of all files in a directory.
for file in "$1"*;
do
mv ${file} ${file/#$1/$2}
done
Use double quotes instead:
# ...
mv "$file" `echo $file | sed -e "s/^$1/$2/"`
# ...
And learn Quotes and escaping in Bash.
When you don't use double quotes, the variables won't expand.
I will rather use this
#!/bin/bash
for file in $1*;
do
mv "$file" "$1${file:${#2}}"
done
where
${file:${#2}
means substring, from length of the argument 2 to the end
I have a directory called "images" filled with about one million images. Yep.
I want to write a shell command to rename all of those images into the following format:
original: filename.jpg
new: /f/i/l/filename.jpg
Any suggestions?
Thanks,
Dan
for i in *.*; do mkdir -p ${i:0:1}/${i:1:1}/${i:2:1}/; mv $i ${i:0:1}/${i:1:1}/${i:2:1}/; done;
The ${i:0:1}/${i:1:1}/${i:2:1} part could probably be a variable, or shorter or different, but the command above gets the job done. You'll probably face performance issues but if you really want to use it, narrow the *.* to fewer options (a*.*, b*.* or what fits you)
edit: added a $ before i for mv, as noted by Dan
You can generate the new file name using, e.g., sed:
$ echo "test.jpg" | sed -e 's/^\(\(.\)\(.\)\(.\).*\)$/\2\/\3\/\4\/\1/'
t/e/s/test.jpg
So, you can do something like this (assuming all the directories are already created):
for f in *; do
mv -i "$f" "$(echo "$f" | sed -e 's/^\(\(.\)\(.\)\(.\).*\)$/\2\/\3\/\4\/\1/')"
done
or, if you can't use the bash $( syntax:
for f in *; do
mv -i "$f" "`echo "$f" | sed -e 's/^\(\(.\)\(.\)\(.\).*\)$/\2\/\3\/\4\/\1/'`"
done
However, considering the number of files, you may just want to use perl as that's a lot of sed and mv processes to spawn:
#!/usr/bin/perl -w
use strict;
# warning: untested
opendir DIR, "." or die "opendir: $!";
my #files = readdir(DIR); # can't change dir while reading: read in advance
closedir DIR;
foreach my $f (#files) {
(my $new_name = $f) =~ s!^((.)(.)(.).*)$!$2/$3/$4/$1/;
-e $new_name and die "$new_name already exists";
rename($f, $new_name);
}
That perl is surely limited to same-filesystem only, though you can use File::Copy::move to get around that.
You can do it as a bash script:
#!/bin/bash
base=base
mkdir -p $base/shorts
for n in *
do
if [ ${#n} -lt 3 ]
then
mv $n $base/shorts
else
dir=$base/${n:0:1}/${n:1:1}/${n:2:1}
mkdir -p $dir
mv $n $dir
fi
done
Needless to say, you might need to worry about spaces and the files with short names.
I suggest a short python script. Most shell tools will balk at that much input (though xargs may do the trick). Will update with example in a sec.
#!/usr/bin/python
import os, shutil
src_dir = '/src/dir'
dest_dir = '/dest/dir'
for fn in os.listdir(src_dir):
os.makedirs(dest_dir+'/'+fn[0]+'/'+fn[1]+'/'+fn[2]+'/')
shutil.copyfile(src_dir+'/'+fn, dest_dir+'/'+fn[0]+'/'+fn[1]+'/'+fn[2]+'/'+fn)
Any of the proposed solutions which use a wildcard syntax in the shell will likely fail due to the sheer number of files you have. Of the current proposed solutions, the perl one is probably the best.
However, you can easily adapt any of the shell script methods to deal with any number of files thus:
ls -1 | \
while read filename
do
# insert the loop body of your preference here, operating on "filename"
done
I would still use perl, but if you're limited to only having simple unix tools around, then combining one of the above shell solutions with a loop like I've shown should get you there. It'll be slow, though.