I have tried different ways of doing this. I need to transform this:
5 dias, 3 meses, 4 anos
67 decimales
naranjas 45, limones 56
66 + 44 = 100.
7777a 7b 88c 777d
Into this:
0005 dias, 0003 meses, 0004 anos
0067 decimales
naranjas 0045, limones 0056
0066 + 0044 = 0100.
7777a 0007b 0088c 0777d
But I cannot manage to add zeros to the last line with the code I have so far, which is the following:
sed -e 's/\b[0-9]\{3\}\b/0&/g; s/\b[0-9]\{2\}\b/00&/g; s/\b[0-9]\b/000&/g' numbers.txt >output.txt
What am I missing?
You could use perl
perl -pe 's/\d+/sprintf "%04d",$&/ge' test
or if you really want to use sed
sed -r ':1;s/([^0-9]|^)([0-9]{1,3})([^0-9]|$)/\10\2\3/;t1' file
Related
I have tried to come up with this answer but everything I try does not work.
My code below is what I have come up with:
sort -k$field_number "$1".db > temp.txt && cp temp.txt "$1.db"
Shouldn't this line of code sort the .db file by ASCII value (the sort function should sort by ASCII by default?). In the code, field_number corresponds to the column I wish to sort the lines of the file by. When I use my code to format the file (where I am sorting by column 2), I get the output below.
Textfile (the .db file) format:
a 5 5 5
Green 72 72 72
Smith 84 72 93
Jones 85 73 94
z 9 9 9
Ford 92 64 93
Miller 93 73 87
bobua che Apple Xor
Maybe your problem is with your collection. Try this please:
LC_COLLATE=C sort -n --ignore-case -k$field_number "$1".db > temp.txt && cp temp.txt "$1.db"
I want to convert the output of command:
dmidecode -s system-serial-number
which is a string looking like this:
VMware-56 4d ad 01 22 5a 73 c2-89 ce 3f d8 ba d6 e4 0c
to:
564dad01-225a-73c2-89ce-3fd8bad6e40c
I suspect I need to first of all extract all letters and numbers after the "VMware-" part at that start and then insert "-" at the known positions after character 10, 14, 18, 22.
To try the first extraction I have tried:
$ echo `dmidecode -s system-serial-number | grep -oE '(VMware-)?[a0-Z9]'`
VMware-5 6 4 d a d 0 1 2 2 5 a 7 3 c 2 8 9 c e 3 f d 8 b a d 6 e 4 0 c
However this isn't going the right way.
EDIT:
This gets me to a single log string however it's not elegant:
$ echo `dmidecode -s system-serial-number | sed -s "s/VMware-//" | sed -s "s/-//" | sed -s "s/ //g"`
564dad01225a73c289ce3fd8bad6e40c
Like this :
dmidecode -s system-serial-number |
sed -E 's/VMware-//;
s/ +//g;
s/(.)/\1-/8;
s/(.)/\1-/13;
s/(.)/\1-/23'
You can use Bash sub string extraction:
$ s="VMware-56 4d ad 01 22 5a 73 c2-89 ce 3f d8 ba d6 e4 0c"
$ s1=$(echo "${s:7}" | tr -d '[:space:]')
$ echo "${s1:0:8}-${s1:8:4}-${s1:12:9}-${s1:21}"
564dad01-225a-73c2-89ce-3fd8bad6e40c
Or, built-ins only (ie, no tr):
$ s1=${s:7}
$ s1="${s1// /}"
$ echo "${s1:0:8}-${s1:8:4}-${s1:12:9}-${s1:21}"
$ cat grades.dat
santosh 65 65 65 65
john 85 92 78 94 88
andrea 89 90 75 90 86
jasper 84 88 80 92 84
santosh 99 99 99 99 99
Scripts:-
#!/usr/bin/bash
filename="$1"
while read line
do
a=`grep -w "santosh" $1 | awk '{print$1}' |wc -l`
echo "total is count of the file is $a";
done <"$filename"
O/p
total is count of the file is 2
total is count of the file is 2
total is count of the file is 2
total is count of the file is 2
total is count of the file is 2
Real O/P should be
total is count of the file is 2 like this right..please let me know,where i am missing in above scripts.
Whilst others have shown you better ways to solve your problem, the answer to your question is in the following line:
a=`grep -w "santosh" $1 | awk '{print$1}' |wc -l`
You are storing names in the variable "line" through the while loop, but it is never used. Instead your loop is always looking for "santosh" which does appear twice and because you run the same query for all 5 lines in the file being searched, you therefore get 5 lines of the exact same output.
You could alter your current script like so:
a=$(grep -w "$line" "$filename" | awk '{print$1}' | wc -l)
The above is not meant to be a solution as others have pointed out, but it does solve your issue.
Assume a text file with 40 lines of data. How can I remove lines 3 to 10, 13 to 20, 23 to 30, 33 to 40, in place using bash script?
I already know how to remove lines 3 to 10 with sed, but I wonder if there is a way to do all the removing, in place, with only one command line. I can use for loop but the problem is that with each iteration of loop the lines number will be changed and it needs some additional calculation of line numbers to be removed.
here is an awk oneliner, works for your needs no matter your file has 40 lines or 40k lines:
awk 'NR~/[12]$/' file
for example, with 50 lines:
kent$ seq 50|awk 'NR~/[12]$/'
1
2
11
12
21
22
31
32
41
42
sed -i '3,10d;13,20d;23,30d;33,40d' file
This might work for you (GNU sed):
sed '3~10,+7d' file
Deletes lines in the range of 3 and thereafter steps of 10 for the following 7 lines to be deleted.
If the file was longer than 40 lines and you were only interested in the first 40 lines:
sed '41,$b;3~10,+7d' file
The first instruction tells sed to ignore lines 41 to end-of-file.
Could also be written:
sed '1,40{3~10,+7d}' file
#Kent's answer is the way to go for this particular case, but in general:
$ seq 50 | awk '{idx=(NR%10)} idx>=1 && idx<=2'
1
2
11
12
21
22
31
32
41
The above will work even if you want to select the 4th through 7th lines out of every 13, for example:
$ seq 50 | awk '{idx=(NR%13)} idx>=4 && idx<=7'
4
5
6
7
17
18
19
20
30
31
32
33
43
44
45
46
its not constrained to N out of 10.
Or to select just the 3rd, 5th and 6th lines out of every 13:
$ seq 50 | awk 'BEGIN{split("3 5 6",tmp); for (i in tmp) tgt[tmp[i]]=1} tgt[NR%13]'
3
5
6
16
18
19
29
31
32
42
44
45
The point is - selecting ranges of lines is a job for awk, definitely not sed.
awk '{m=NR%10} !(m==0 || m>=3)' file > tmp && mv tmp file
For example, I want to count from 001 to 100. Meaning the zero buffer would start off with 2, 1, then eventually 0 when it reaches 100 or more.
ex:
001
002
...
010
011
...
098
099
100
I could do this if the numbers had a predefined number of zeroes with printf "%02d" $i. But that's static and not dynamic and would not work in my example.
If by static versus dynamic you mean that you'd like to be able to use a variable for the width, you can do this:
$ padtowidth=3
$ for i in 0 {8..11} {98..101}; do printf "%0*d\n" $padtowidth $i; done
000
008
009
010
011
098
099
100
101
The asterisk is replaced by the value of the variable it corresponds to in the argument list ($padtowidth in this case).
Otherwise, the only reason your example doesn't work is that you use "2" (perhaps as if it were the maximum padding to apply) when it should be "3" (as in my example) since that value is the resulting total width (not the pad-only width).
If your system has it, try seq with the -w (--equal-width) option:
$ seq -s, -w 1 10
01,02,03,04,05,06,07,08,09,10
$ for i in `seq -w 95 105` ; do echo -n " $i" ; done
095 096 097 098 099 100 101 102 103 104 105
In Bash version 4 (use bash -version) you can use brace expansion. Putting a 0 before either limit forces the numbers to be padded by zeros
echo {01..100} # 001 002 003 ...
echo {03..100..3} # 003 006 009 ...
#!/bin/bash
max=100;
for ((i=1;i<=$max;i++)); do
printf "%0*d\n" ${#max} $i
done
The code above will auto-pad your numbers with the correct number of 0's based upon how many digits the max/terminal value contains. All you need to do is change the max variable and it will handle the rest.
Examples:
max=10
01
02
03
04
05
06
07
08
09
10
max=100
001
002
003
004
005
006
...
097
098
099
100
max=1000
0001
0002
0003
0004
0005
0006
...
0997
0998
0999
1000
# jot is available on FreeBSD, Mac OS X, ...
jot -s " " -w '%03d' 5
jot -s " " -w '%03d' 10
jot -s " " -w '%03d' 50
jot -s " " -w '%03d' 100
If you need to pad values up to a variable number with variable padding:
$values_count=514;
$padding_width=5;
for i in 0 `seq 1 $(($values_count - 1))`; do printf "%0*d\n" $padding_width $i; done;
This would print out 00000, 00001, ... 00513.
(I didn't find any of the current answers meeting my need)