I have a file address.txt containing
`0x0003FFB0'
at line 1 and column 2
i want to paste it in another file 'linker.txt' at line 20 and column 58
How can i do that using bash scripting?
Note that the content of the input file can be random , doesn't have to be same each time.
But the length of the word to be copied will always be the same
you can use SED
sed -i.old '20s/^.{58}/&0x0003FFB0/' file
it will produce a file.old with the original content and file will be updated with this address. Quickly explain
sed '20command' file # do command in line 20
sed '20s/RE/&xxx/' file # search for regular expression, replace by the original text (&) + xxx
to read the address and put in this sed, cut it possible
ADDRESS=$(head -1 address.txt | cut -f2)
sed -i.old "20s/^.{58}/&${ADDRESS}/" file
You can use a combination of head and tail to get any line number. To get line 2, get the last line (using tail) of the first two lines (using head):
ADDRESS=$(head -2 address.txt | tail -1 | cut -f2)
For the third line:
ADDRESS=$(head -3 address.txt | tail -1 | cut -f2)
And so on.
Related
I am trying to simply count the lines in the .CSV per column, while at the same time ignoring empty lines.
I use below and it works for the 1st column:
cat /path/test.csv | cut -d, -f1 | grep . | wc -l` >> ~/Desktop/Output.csv
#Outputs: 8
And below for the 2nd column:
cat /path/test.csv | cut -d, -f2 | grep . | wc -l` >> ~/Desktop/Output.csv
#Outputs: 6
But when I try to count 3rd column, it simply Outputs the Total number of lines in the whole .CSV.
cat /path/test.csv | cut -d, -f3 | grep . | wc -l` >> ~/Desktop/Output.csv
#Outputs: 33
#Should be: 19?
I've also tried to use awk instead of cut, but get the same issue.
I have tried creating new file thinking maybe it had some spaces in the lines, still the same.
Can someone clarify what is the difference? Betwen reading 1-2 column and the rest?
20355570_01.tif,,
20355570_02.tif,,
21377804_01.tif,,
21377804_02.tif,,
21404518_01.tif,,
21404518_02.tif,,
21404521_01.tif,,
21404521_02.tif,,
,22043764_01.tif,
,22043764_02.tif,
,22095060_01.tif,
,22095060_02.tif,
,23507574_01.tif,
,23507574_02.tif,
,,23507574_03.tif
,,23507804_01.tif
,,23507804_02.tif
,,23507804_03.tif
,,23509247_01.tif
,,23509247_02.tif
,,23509247_03.tif
,,23527663_01.tif
,,23527663_02.tif
,,23527663_03.tif
,,23527908_01.tif
,,23527908_02.tif
,,23527908_03.tif
,,23535506_01.tif
,,23535506_02.tif
,,23535562_01.tif
,,23535562_02.tif
,,23535636_01.tif
,,23535636_02.tif
That happens when input file has DOS line endings (\r\n). Fix your file using dos2unix and your command will work for 3rd column too.
dos2unix /path/test.csv
Or, you can remove the \r at the end while counting non-empty columns using awk:
awk -F, '{sub(/\r/,"")} $3!=""{n++} END{print n}' /path/test.csv
The problem is in the grep command: the way you wrote it will return 33 lines when you count the 3rd column.
It's better instead to use the following command to count number of lines in .CSV for each column (example below is for the 3rd column):
cat /path/test.csv | cut -d , -f3 | grep -cve '^\s*$'
This will return the exact number of lines for each column and avoid of piping into wc.
See previous post here:
count (non-blank) lines-of-code in bash
edit: I think oguz ismail found the actual reason in their answer. If they are right and your file has windows line endings you can use one of the following commands without having to convert the file.
cut -d, -f3 yourFile.csv cut | tr -d \\r | grep -c .
cut -d, -f3 yourFile.csv | grep -c $'[^\r]' # bash only
old answer: Since I cannot reproduce your problem with the provided input I take a wild guess:
The "empty" fields in the last column contain spaces. A field containing a space is not empty altough it looks like it is empty as you cannot see spaces.
To count only fields that contain something other than a space adapt your regex from . (any symbol) to [^ ] (any symbol other than space).
cut -d, -f3 yourFile.csv | grep -c '[^ ]'
I want to delete one or more specific line numbers from a file. How would I do this using sed?
If you want to delete lines from 5 through 10 and line 12th:
sed -e '5,10d;12d' file
This will print the results to the screen. If you want to save the results to the same file:
sed -i.bak -e '5,10d;12d' file
This will store the unmodified file as file.bak, and delete the given lines.
Note: Line numbers start at 1. The first line of the file is 1, not 0.
You can delete a particular single line with its line number by
sed -i '33d' file
This will delete the line on 33 line number and save the updated file.
and awk as well
awk 'NR!~/^(5|10|25)$/' file
$ cat foo
1
2
3
4
5
$ sed -e '2d;4d' foo
1
3
5
$
This is very often a symptom of an antipattern. The tool which produced the line numbers may well be replaced with one which deletes the lines right away. For example;
grep -nh error logfile | cut -d: -f1 | deletelines logfile
(where deletelines is the utility you are imagining you need) is the same as
grep -v error logfile
Having said that, if you are in a situation where you genuinely need to perform this task, you can generate a simple sed script from the file of line numbers. Humorously (but perhaps slightly confusingly) you can do this with sed.
sed 's%$%d%' linenumbers
This accepts a file of line numbers, one per line, and produces, on standard output, the same line numbers with d appended after each. This is a valid sed script, which we can save to a file, or (on some platforms) pipe to another sed instance:
sed 's%$%d%' linenumbers | sed -f - logfile
On some platforms, sed -f does not understand the option argument - to mean standard input, so you have to redirect the script to a temporary file, and clean it up when you are done, or maybe replace the lone dash with /dev/stdin or /proc/$pid/fd/1 if your OS (or shell) has that.
As always, you can add -i before the -f option to have sed edit the target file in place, instead of producing the result on standard output. On *BSDish platforms (including OSX) you need to supply an explicit argument to -i as well; a common idiom is to supply an empty argument; -i ''.
The shortest, deleting the first line in sed
sed -i '1d' file
As Brian states here, <address><command> is used, <address> is <1> and <command> <d>.
I would like to propose a generalization with awk.
When the file is made by blocks of a fixed size
and the lines to delete are repeated for each block,
awk can work fine in such a way
awk '{nl=((NR-1)%2000)+1; if ( (nl<714) || ((nl>1025)&&(nl<1029)) ) print $0}'
OriginFile.dat > MyOutputCuttedFile.dat
In this example the size for the block is 2000 and I want to print the lines [1..713] and [1026..1029].
NR is the variable used by awk to store the current line number.
% gives the remainder (or modulus) of the division of two integers;
nl=((NR-1)%BLOCKSIZE)+1 Here we write in the variable nl the line number inside the current block. (see below)
|| and && are the logical operator OR and AND.
print $0 writes the full line
Why ((NR-1)%BLOCKSIZE)+1:
(NR-1) We need a shift of one because 1%3=1, 2%3=2, but 3%3=0.
+1 We add again 1 because we want to restore the desired order.
+-----+------+----------+------------+
| NR | NR%3 | (NR-1)%3 | (NR-1)%3+1 |
+-----+------+----------+------------+
| 1 | 1 | 0 | 1 |
| 2 | 2 | 1 | 2 |
| 3 | 0 | 2 | 3 |
| 4 | 1 | 0 | 1 |
+-----+------+----------+------------+
cat -b /etc/passwd | sed -E 's/^( )+(<line_number>)(\t)(.*)/--removed---/g;s/^( )+([0-9]+)(\t)//g'
cat -b -> print lines with numbers
s/^( )+(<line_number>)(\t)(.*)//g -> replace line number to null (remove line)
s/^( )+([0-9]+)(\t)//g #remove numbers the cat printed
I'm attempting to remove a certain pattern from a line, but not the entire line itself. An example would be:
Original:
user=dannyBoy
Desired:
dannyBoy
I have a file that is full of lines like that, so I was wondering how I would be able to cut a specific part of the text off, whether that be just removing the first five characters from the list or searching for the pattern "user=" and removing it.
There are many ways to do this:
cut -d'=' -f2- file
sed 's/^[^=]*//' file
awk -F= '{print $2}' file #if just one = is present
cut sets a delimiter (-d'=) and then prints all the fields starting from the 2nd one (-f2-).
sed looks for all the content from the beginning up to the first = and removes it.
awk sets = as field separator and prints the second field.
Using ex:
echo user=dannyBoy | ex -s +"norm df=" +%p -cq! /dev/stdin
where ex is equivalent to vi -e/vim -e which basically executes vi command: df= (delete until finds =), then print the buffer (%p).
If you've multiple lines like that, then it would be simpler by using substitution:
ex -s +"%s/^.*=//g" +%p -cq! foo.txt
To edit file in place, change -cq! to -cwq.
The command below deletes the first 5 characters:
$ echo "user=dannyboy" | cut -c 6-
You can use it on a file with cut -c 6- inputfilename as well.
Do you have any bash solution to remove N lines from stdout?
like a 'head' command, print all lines, only except last N
Simple solition on bash:
find ./test_dir/ | sed '$d' | sed '$d' | sed '$d' | ...
but i need to copy sed command N times
Any better solution?
except awk, python etc...
Use head with a negative number. In my example it will print all lines but last 3:
head -n -3 infile
if head -n -3 filename doesn't work on your system (like mine), you could also try the following approach (and maybe alias it or create a function in your .bashrc)
head -`echo "$(wc -l filename)" | awk '{ print $1 - 3; }'` filename
Where filename and 3 above are your file and number of lines respectively.
The tail command can skip from the end of a file on Mac OS / BSD. tail accepts +/- prefix, which facilitates expression below, which will show 3 lines from the start
tail -n +3 filename.ext
Or, to skip lines from the end of file, use - prefixed, instead.
tail -n -3 filenme.ext
Typically, the default for tail is the - prefix, thus counting from the end of the file. See a similar answer to a different question here: Print a file skipping first X lines in Bash
I just want to delete the line which contain the number of selected rows in a query. I mean the one in the last line. please help.
[root#machine-test scripts]# ./hit_ratio.sh
193830 432
185260 125
2 rows selected.
If you know you want to delete the last line, but not other lines which contain similar text, or you don't know what text it will contain, sed is uniquely suitable.
./hit_ratio.sh | sed '$d'
You don't need the power of sed or the super-powers of awk if all you want is to delete a line based on a pattern. You can use:
./hit_ratio.sh | grep -v ' rows selected.'
You can do it with awk and sed but it's a bit like trying to kill a fly with a thermo-nuclear warhead:
pax> ./hit_ratio.sh | sed '/ rows selected./d'
193830 432
185260 125
pax> ./hit_ratio.sh | awk '$2!="rows"{print}'
193830 432
185260 125
Alternatively, do something with your SQL script. Sometimes, turning on the set nocount on statement eliminates the "rows affected" line.
My first recommendation is not to have that line outputted, list hit_ratio.sh here maybe it can be modified not to output that line
Anyway if you have to remove only the last line the easiest would be to use head:
./hit_ratio.sh | head -n -1
Using -n and a negative number, makes head print all but the last N lines of the input
Use head to get the first N - 1 lines of your file, where N is the length of the file (calculated with wc -l)
head -n $(($(cat lipsum.log | wc -l) - 1)) lipsum.log works
Pipe through
sed -e '/\w*[0-9]\+ rows\? selected/d'