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How can I store the "find" command results as an array in Bash
(8 answers)
Closed 4 years ago.
How do I put the result of find $1 into an array?
In for loop:
for /f "delims=/" %%G in ('find $1') do %%G | cut -d\/ -f6-
I want to cry.
In bash:
file_list=()
while IFS= read -d $'\0' -r file ; do
file_list=("${file_list[#]}" "$file")
done < <(find "$1" -print0)
echo "${file_list[#]}"
file_list is now an array containing the results of find "$1
What's special about "field 6"? It's not clear what you were attempting to do with your cut command.
Do you want to cut each file after the 6th directory?
for file in "${file_list[#]}" ; do
echo "$file" | cut -d/ -f6-
done
But why "field 6"? Can I presume that you actually want to return just the last element of the path?
for file in "${file_list[#]}" ; do
echo "${file##*/}"
done
Or even
echo "${file_list[#]##*/}"
Which will give you the last path element for each path in the array. You could even do something with the result
for file in "${file_list[#]##*/}" ; do
echo "$file"
done
Explanation of the bash program elements:
(One should probably use the builtin readarray instead)
find "$1" -print0
Find stuff and 'print the full file name on the standard output, followed by a null character'. This is important as we will split that output by the null character later.
<(find "$1" -print0)
"Process Substitution" : The output of the find subprocess is read in via a FIFO (i.e. the output of the find subprocess behaves like a file here)
while ...
done < <(find "$1" -print0)
The output of the find subprocess is read by the while command via <
IFS= read -d $'\0' -r file
This is the while condition:
read
Read one line of input (from the find command). Returnvalue of read is 0 unless EOF is encountered, at which point while exits.
-d $'\0'
...taking as delimiter the null character (see QUOTING in bash manpage). Which is done because we used the null character using -print0 earlier.
-r
backslash is not considered an escape character as it may be part of the filename
file
Result (first word actually, which is unique here) is put into variable file
IFS=
The command is run with IFS, the special variable which contains the characters on which read splits input into words unset. Because we don't want to split.
And inside the loop:
file_list=("${file_list[#]}" "$file")
Inside the loop, the file_list array is just grown by $file, suitably quoted.
arrayname=( $(find $1) )
I don't understand your loop question? If you look how to work with that array then in bash you can loop through all array elements like this:
for element in $(seq 0 $((${#arrayname[#]} - 1)))
do
echo "${arrayname[$element]}"
done
This is probably not 100% foolproof, but it will probably work 99% of the time (I used the GNU utilities; the BSD utilities won't work without modifications; also, this was done using an ext4 filesystem):
declare -a BASH_ARRAY_VARIABLE=$(find <path> <other options> -print0 | sed -e 's/\x0$//' | awk -F'\0' 'BEGIN { printf "("; } { for (i = 1; i <= NF; i++) { printf "%c"gensub(/"/, "\\\\\"", "g", $i)"%c ", 34, 34; } } END { printf ")"; }')
Then you would iterate over it like so:
for FIND_PATH in "${BASH_ARRAY_VARIABLE[#]}"; do echo "$FIND_PATH"; done
Make sure to enclose $FIND_PATH inside double-quotes when working with the path.
Here's a simpler pipeless version, based on the version of user2618594
declare -a names=$(echo "("; find <path> <other options> -printf '"%p" '; echo ")")
for nm in "${names[#]}"
do
echo "$nm"
done
To loop through a find, you can simply use find:
for file in "`find "$1"`"; do
echo "$file" | cut -d/ -f6-
done
It was what I got from your question.
Related
Why isn't this bash array populating? I believe I've done them like this in the past. Echoing ${#XECOMMAND[#]} shows no data..
DIR=$1
TEMPFILE=/tmp/dir.tmp
ls -l $DIR | tail -n +2 | sed 's/\s\+/ /g' | cut -d" " -f5,9 > $TEMPFILE
i=0
cat $TEMPFILE | while read line ;do
if [[ $(echo $line | cut -d" " -f1) == 0 ]]; then
XECOMMAND[$i]="$(echo "$line" | cut -d" " -f2)"
(( i++ ))
fi
done
When you run the while loop like
somecommand | while read ...
then the while loop is executed in sub-shell, i.e. a different process than the main script. Thus, all variable assignments that happen in the loop, will not be reflected in the main process. The workaround is to use input redirection and/or command substitution, so that the loop executes in the current process. For example if you want to read from a file you do
while read ....
do
# do stuff
done < "$filename"
or if you wan't the output of a process you can do
while read ....
do
# do stuff
done < <(some command)
Finally, in bash 4.2 and above, you can set shopt -s lastpipe, which causes the last command in the pipeline to be executed in the current process.
I think you're trying to construct an array consisting of the names of all zero-length files and directories in $DIR. If so, you can do it like this:
mapfile -t ZERO_LENGTH < <(find "$DIR" -maxdepth 1 -size 0)
(Add -type f to the find command if you're only interested in regular files.)
This sort of solution is almost always better than trying to parse ls output.
The use of process substitution (< <(...)) rather than piping (... |) is important, because it means that the shell variable will be set in the current shell, not in an ephimeral subshell.
I am very new to Unix shell script and trying to get some knowledge in shell scripting. Please check my requirement and my approach.
I have a input file having data
ABC = A:3 E:3 PS:6
PQR = B:5 S:5 AS:2 N:2
I am trying to parse the data and get the result as
ABC
A=3
E=3
PS=6
PQR
B=5
S=5
AS=2
N=2
The values can be added horizontally and vertically so I am trying to use an array. I am trying something like this:
myarr=(main.conf | awk -F"=" 'NR!=1 {print $1}'))
echo ${myarr[1]}
# Or loop through every element in the array
for i in "${myarr[#]}"
do
:
echo $i
done
or
awk -F"=" 'NR!=1 {
print $1"\n"
STR=$2
IFS=':' read -r -a array <<< "$STR"
for i in "${!array[#]}"
do
echo "$i=>${array[i]}"
done
}' main.conf
But when I add this code to a .sh file and try to run it, I get syntax errors as
$ awk -F"=" 'NR!=1 {
> print $1"\n"
> STR=$2
> FS= read -r -a array <<< "$STR"
> for i in "${!array[#]}"
> do
> echo "$i=>${array[i]}"
> done
>
> }' main.conf
awk: cmd. line:4: FS= read -r -a array <<< "$STR"
awk: cmd. line:4: ^ syntax error
awk: cmd. line:5: for i in "${!array[#]}"
awk: cmd. line:5: ^ syntax error
awk: cmd. line:8: done
awk: cmd. line:8: ^ syntax error
How can I complete the above expectations?
This is the awk code to do what you want:
$ cat tst.awk
BEGIN { FS="[ =:]+"; OFS="=" }
{
print $1
for (i=2;i<NF;i+=2) {
print $i, $(i+1)
}
print ""
}
and this is the shell script (yes, all a shell script does to manipulate text is call awk):
$ awk -f tst.awk file
ABC
A=3
E=3
PS=6
PQR
B=5
S=5
AS=2
N=2
A UNIX shell is an environment from which to call UNIX tools (find, sort, sed, grep, awk, tr, cut, etc.). It has its own language for manipulating (e.g. creating/destroying) files and processes and sequencing calls to tools but it is NOT intended to be used to manipulate text. The guys who invented shell also invented awk for shell to call to manipulate text.
Read https://unix.stackexchange.com/questions/169716/why-is-using-a-shell-loop-to-process-text-considered-bad-practice and the book Effective Awk Programming, 4th Edition, by Arnold Robbins.
First off, a command that does what you want:
$ sed 's/ = /\n/;y/: /=\n/' main.conf
ABC
A=3
E=3
PS=6
PQR
B=5
S=5
AS=2
N=2
This replaces, on each line, the first (and only) occurrence of = with a newline (the s command), then turns all : into = and all spaces into newlines (the y command). Notice that
this works only because there is a space at the end of the first line (otherwise it would be a bit more involved to get the empty line between the blocks) and
this works only with GNU sed because it substitutes newlines; see this fantastic answer for all the details and how to get it to work with BSD sed.
As for what you tried, there is almost too much wrong with it to try and fix it piece by piece: from the wild mixing of awk and Bash to syntax errors all over the place. I recommend you read good tutorials for both, for example:
The BashGuide
Effective AWK Programming
A Bash solution
Here is a way to solve the same in Bash; I didn't use any arrays.
#!/bin/bash
# Read line by line into the 'line' variable. Setting 'IFS' to the empty string
# preserves leading and trailing whitespace; '-r' prevents interpretation of
# backslash escapes
while IFS= read -r line; do
# Three parameter expansions:
# Replace ' = ' by newline (escape backslash)
line="${line/ = /\\n}"
# Replace ':' by '='
line="${line//:/=}"
# Replace spaces by newlines (escape backslash)
line="${line// /\\n}"
# Print the modified input line; '%b' expands backslash escapes
printf "%b" "$line"
done < "$1"
Output:
$ ./SO.sh main.conf
ABC
A=3
E=3
PS=6
PQR
B=5
S=5
AS=2
N=2
On the web I found answers to find if an element of array is present in the string. But I want to find if each element in the array is present in the string.
eg. str1 = "This_is_a_big_sentence"
Initially str2 was like
str2 = "Sentence_This_big"
Now I wanted to search if string str1 contains "sentence"&"this"&"big" (All 3, ignore alphabetic order and case)
So I used arr=(${str2//_/ })
How do i proceed now, I know comm command finds intersection, but it needs a sorted list, also I need to ignore _ underscores.
I get my str2 by finding the extension of a particular type of file using the command
for i in `ls snooze.*`; do echo $i | cut -d "." -f2
# Till here i get str2 and need to check as mentioned above. Not sure how to do this, i tried putting str2 as array and now just need to check if all elements of my array occur in str1 (ignore case,order)
Any help would be highly appreciated. I did try to use This link
Now I wanted to search if string a contains "sentence"&"this"&"big"
(All 3, ignore alphabatic order and case)
Here is one approach:
#!/bin/bash
str1="This_is_a_big_sentence"
str2="Sentence_This_big"
if ! grep -qvwFf <(sed 's/_/\n/g' <<<${str1,,}) <(sed 's/_/\n/g' <<<${str2,,})
then
echo "All words present"
else
echo "Some words missing"
fi
How it works
${str1,,} returns the string str1 with all capitals replaced by lower case.
sed 's/_/\n/g' <<<${str1,,} returns the string str1, all converted to lower case and with underlines replaced by new lines so that each word is on a new line.
<(sed 's/_/\n/g' <<<${str1,,}) returns a file-like object containing all the words in str1, each word lower case and on a separate line.
The creation of file-like objects is called process substitution. It allows us, in this case, to treat the output of a shell command as if it were a file to read.
<(sed 's/_/\n/g' <<<${str2,,}) does the same for str2.
Assuming that file1 and file2 each have one word per line, grep -vwFf file1 file2 removes from file2 every occurrence of a word in file2. If there are no words left, that means that every word in file2 appears in file1.
By adding the option -q, grep will return no output but will set an exit code that we can use in our if statement.
In the actual command, file1 and file2 are replaced by our file-like objects.
The remaining grep options can be understood as follows:
-w tells grep to look for whole words only.
-F tells grep to look for fixed strings, not regular expressions.
-f tells grep to look for the patterns to match in the file (or file-like object) which follows.
-v tells grep to remove (the default is to keep) the words which match.
Here is an awk solution to check existence of all the words from a string in another string:
str1="This_is_a_big_sentence"
str2="Sentence_This_big"
awk -v RS=_ 'FNR==NR{a[tolower($1)]; next} {delete a[tolower($1)]} END{print (length(a)) ? "Not all words" : "All words"}' <(echo "$str2") <(echo "$str1")
With indentation:
awk -v RS=_ 'FNR==NR {
a[tolower($1)];
next
}
{ delete a[tolower($1)] }
END {
print (length(a)) ? "Not all words" : "All words"
}' <(echo "$str2") <(echo "$str1")
Explanation:
-v RS=_ We use record separator as _
FNR==NR - Execute this block for str2
a[tolower($1)]; next - Populate an array a with each lowercase word as key
{delete a[tolower($1)]} - For each word in str1 delete key in array a
END - If length of array a is still not 0 then there are some words left.
Here's another solution:
#!/bin/bash
str1="This_is_a_big_sentence"
str2="sentence_This_big"
var=0
var2=0
while read in
do
if [ $(echo $str1 | grep -ioE $in) ]
then
var=$((var+1))
fi
var2=$((var2+1))
done < <(echo $str2 | sed -e 's/\(.*\)/\L\1/' -e 's/_/\n/g')
if [[ $var -eq $var2 && $var -ne 0 ]]
then
echo "matched"
else
echo "not matched"
What this script does make str2 all lower case with sed -e 's/\(.*\)/\L\1/' which is a substitution of any character with its lower case, then replace underscores _ with return lines \n with the following sed expression: sed -e 's/_/\n/g', which is another substitution.
Now the individual words are fed into a while loop that compares str1 with the word that was fed in. Every time there's a match, increment var and every time we iterate though the while, we increment var2. If var == var2, then all the words of str2 were found in str1. Hope that helps.
Here's an approach.
if [ "$(echo "This_BIG_senTence" | grep -ioE 'this|big|sentence' | wc -l)" == "3" ]; then echo "matched"; fi
How it works.
grep options -i makes the grep case insensitive, -E for extended regular expressions, and -o separates the matches by line. Now that it is separated by line use wc with -l for line count. Since we had 3 conditions we check if it equals 3. Grep will return the lines where the match occurred, so if you are only working with a string, the example above will return the string for each condition, in this case 3, so there won't be any problems.
Note you can also create a grep chain and see if its empty.
if [ $(echo "This_BIG_SenTence" | grep -i this | grep -i big | grep -i sentence) ]; then echo matched; else echo not_matched; fi
Now I know what you mean. Try this:
#!/bin/bash
# add 4 non-matching examples
> snooze.foo_bar
> snooze.bar_go
> snooze.go_foo
> snooze.no_match
# add 3 matching examples
> snooze.foo_bar_go
> snooze.goXX_XXfoo_XXbarXX
> snooze.bar_go_foo_Ok
str1=("foo" "bar" "go")
for i in `ls snooze.*`; do
str2=${i#snooze.}
j=0
found=1
while [[ $j -lt ${#str1[#]} ]]; do
if ! echo $str2 | eval grep \${str1[$j]} >& /dev/null; then
found=0
break
fi
((j++))
done
if [[ $found -ne 0 ]]; then
echo Match found: $str2
fi
done
Resulting print of this script:
Match found: bar_go_foo_Ok
Match found: foo_bar_go
Match found: goXX_XXfoo_XXbarXX
alternatively, the if..grep line above can be replaced by
if [[ ! $str2 =~ `eval echo \${str1[$j]}` ]]; then
utilizing bash's regular expression match.
Note: I am not too careful about special characters in the search string, such as "\" or " " (space), which may cause problem.
--- Some explanations ---
In the if .. grep line, $j is first evaluated to the running index, from 0 to the number of elements in $str1 minus 1. Then, eval will re-evaluate the whole grep command again, causing ${str1[jjj]} to be re-evaluated (Here, jjj is the already evaluated index)
The strategy is to set found=1 (found by default), and then when any grep fails, we set found to 0 and break the inner j-loop.
Everything else should be straightforward.
I need to search a pattern in a directory and save the names of the files which contain it in an array.
Searching for pattern:
grep -HR "pattern" . | cut -d: -f1
This prints me all filenames that contain "pattern".
If I try:
targets=$(grep -HR "pattern" . | cut -d: -f1)
length=${#targets[#]}
for ((i = 0; i != length; i++)); do
echo "target $i: '${targets[i]}'"
done
This prints only one element that contains a string with all filnames.
output: target 0: 'file0 file1 .. fileN'
But I need:
output: target 0: 'file0'
output: target 1: 'file1'
.....
output: target N: 'fileN'
How can I achieve the result without doing a boring split operation on targets?
You can use:
targets=($(grep -HRl "pattern" .))
Note use of (...) for array creation in BASH.
Also you can use grep -l to get only file names in grep's output (as shown in my command).
Above answer (written 7 years ago) made an assumption that output filenames won't contain special characters like whitespaces or globs. Here is a safe way to read those special filenames into an array: (will work with older bash versions)
while IFS= read -rd ''; do
targets+=("$REPLY")
done < <(grep --null -HRl "pattern" .)
# check content of array
declare -p targets
On BASH 4+ you can use readarray instead of a loop:
readarray -d '' -t targets < <(grep --null -HRl "pattern" .)
Problem:
Need to sort array before operating on them with function.
First, array is loaded with files:
unset a i
counter=1
while IFS= read -r -d $'\0' file; do
a[i++]="$file"
done < <(find $DIR -type f -print0)
Next, each member of array is sent to function
for f in "${a[#]}"
do
func_hash "$f"
[ $(expr $counter % 20) -eq 0 ] && printf "="
counter=$((counter + 1))
done
Somehow a sort needs to be thrown into the above for loop. Have looked
through the SO posts on sorting arrays but somehow my crazy file names
cause issues when I try to tack on a sort.
Ideas?
Thanks!
Bubnoff
UPDATE: Here's code with sort:
while IFS= read -r -d $'\0' file; do
func_hash "$file"
[ $(expr $counter % 20) -eq 0 ] && printf "="
counter=$((counter + 1))
done < <(find $DIR -type f -print0 | sort -z +1 -1)
It's sorting by full path rather than file name. Any ideas on how to
sort by file name given that the path is needed for the function?
UPDATE 2: Decided to compromise.
My main goal was to avoid temp files using sort. GNU sort can write back to the original
file with its '-o' option so now I can:
sort -o $OUT -t',' -k 1 $OUT
Anyone have a more 'elegant' solution ( whatever that means ).
SOLVED See jw013's answer below. Thanks man!
EDIT
while IFS= read -r -d/ && read -r -d '' file; do
a[i++]="$file"
done < <(find "$DIR" -type f -printf '%f/%p\0' | sort -z -t/ -k1 )
Rationale:
I make the assumption that / is never a legal character within a file name (which seems reasonable on most *nix filesystems since it is the path separator).
The -printf is used to print the file name without leading directories, then then full file name with path, separated by /. The sort takes place on the first field separated by / which should be the full file name without path.
The read is modified to first use / as a delimiter to throw out the pathless file name.
side note
Any POSIX shell should support the modulo operator as part of its arithmetic expansion. You can replace line with the call to external command expr in the second loop with
[ $(( counter % 20 )) -eq 0 ] ...