I want to search a line from file which contains "+", number and character. If any other character available in string then complete string should be discarded.
A1264
13255
1255+*
*6_54
54789+
Output should be
A1264
13255
54789+
Record number 3 and 4 should not come as it contains some other character also.
You can try something like:
grep -E '^[a-zA-Z0-9\+]+$'
This will accept only a to z chars (small and caps), digits and + sign
If you have other symbols you can edit the command line:
# grep -E '^[a-fA-F0-9©]+$' a1
A1264
13255
54789©
Related
TL;DR: grep for a string in a file and output the match that is closest before a given line number.
I'm trying to write a bash command line function to find a line within a file that contains a given string.
I don't know bash enough to write it; though I've been trying.
The input to the function is:
The string to search.
The file to search.
A line number; the match for the string will be somewhere before this line.
The match will definetly exist within the file.
There may be multiple matches for the string within the file, but the line that I'm interested in is the closest
to the line number, in reserve. For example, if the passed line number is 100, and there were matches on
lines 5, 30, 77, then I'm interested in the match at line 77.
I want the output of the function to be the value of the line (the whole string)
contains the searched string. Example
some foo value
some bar value
some zee value
and the search was for 'bar'. I want the output to be:
some bar value
head -n $num $file | tac | grep -m 1 $string
head -n to get first n lines
tac will reverse the order of lines returned by head to put the closest match at the top
grep -m 1 will return only the first line matching the string
Requirement is to find a string from txt file and store it to variable.
file look like this(rsa.txt)
Encrypting String
... Input string : Test_123
... Encrypted string : $ENC(JEVOQyhZbVpkQmM0L3ArT2c4M05TZks5TmxRPT1+KQ==)
Required output (variable name : encstring):
encstring = $ENC(JEVOQyhZbVpkQmM0L3ArT2c4M05TZks5TmxRPT1+KQ==)
I tried below code but showing no result
encstring=$(grep -oE '$ENC[^()]*==)' <<< rsa.txt)
With awk, could you please try following. Simply, search for string /Encrypted string along with a condition to check if last field of that line has $ENC in it then last field for that line by using $NF.
encstring=$(awk '/Encrypted string/ && $NF~/\$ENC/{print $NF}'
You can use
encstring=$(sed -n 's/.*\(\$ENC(.*)\).*/\1/p' rsa.txt)
# OR
encstring=$(grep -oP '\$ENC\(.*?\)' rsa.txt)
See an online demo:
s='Encrypting String
... Input string : Test_123
... Encrypted string : $ENC(JEVOQyhZbVpkQmM0L3ArT2c4M05TZks5TmxRPT1+KQ==)'
encstring=$(sed -n 's/.*\(\$ENC(.*)\).*/\1/p' <<< "$s")
echo "$encstring"
# => $ENC(JEVOQyhZbVpkQmM0L3ArT2c4M05TZks5TmxRPT1+KQ==)
The sed -n 's/.*\(\$ENC(.*)\).*/\1/p' command does the following:
-n suppresses the default line output
s/.*\(\$ENC(.*)\).*/\1/ - finds any text, then captures $ENC(...) into Group 1 and then matches the rest of the string, and replaces the match with the Group 1 value
p - prints the result of the substitution.
The grep -oP '\$ENC\(.*?\)' command extracts all $ENC(...) matches, with any chars, as few as possible, between ( and ).
You are searching for ENC which is followed by 0 or more occurances of something which is not an open or closed parenthesis. However, in your input file, there is an open parenthese after ENC. Therefore [^()]* matches the null string. After this you expect the string ==). This would match only for the input ENC==)`.
You need to escape $ as \$ as it means "end of string" with -E
So i have an alphanumeric string 10006cc2190ab011 i am trying to add a colon after every two letters in this alphanumeric string.
this is the string : 10006cc2190ab011
i want it be - 10:00:6c:c2:19:0a:b0:11
Thanks in advance.
A sed solution:
$ echo 10006cc2190ab011 | sed 's/../&:/g; s/:$//'
10:00:6c:c2:19:0a:b0:11
Replaces each non-overlapping pair of characters with the same pair plus :. In the end removes the trailing : (if input text had even length).
str=10006cc2190ab011; str="${str//??/${.sh.match}:}"; echo ${str%:}
is doing the same replacement without the use of an external command, just using ksh-internals.
Doing the same as in sed (the other answer). Replace in $str every // two charactes ?? with / the matched string and a : (every match is kept in the ksh-variable ${.sh.match}). Then print $str without the last % ':'.
I have a fixed length file in which some records have different special characters like Еӏєпа
I'm able to select those records containing special characters/.
I want to read 2 columns from those records and update it with '*' padded with blanks
Sample Data :
1234562013-09-01 01:05:30Еӏєпа Нцвѡі A other
5657812011-05-05 02:34:56abu jaya B other
Specifically, the 3rd and 4th column containing special characters, should be replaced with a single '*' padded with blanks to fill the length
I need result like below
1234562013-09-01 01:05:30* * A2013-09-01 02:03:40other
5657812011-05-05 02:34:56abu jaya B2013-09-01 07:06:10other
Tried the following commands :
sed -r "s/^(.{56}).{510}/\1$PAD/g;s/^(.{511}).{1023}/\1$PAD/g" errorline.txt
cut -c 57-568
Could someone help me out with this?
I would go with awk, something like:
awk '/[LIST__OF_SPECIAL_CHARS]/ {
l=$0
# for 3rd col
# NOTE the * must be padded if you have a fixed length file
# This can be done with spaces and/or (s)printf, read the docs
if (substr($0,FROM,NUM_OF_CHARS) ~ /[LIST__OF_SPECIAL_CHARS]/) {
l=substr(l,1,START_OF_3RD_COL_MINUS_1) "*" substr(l,START_OF_4TH_COL)
}
# for 4th col
# NOTE the * must be padded if you have a fixed length file
# This can be done with spaces and/or (s)printf, read the docs
if (substr($0,START_OF_4TH_COL,NUM_OF_CHARS) ~ /[LIST__OF_SPECIAL_CHARS]/) {
l=substr(l,1,START_OF_4TH_COL_MINUS_1) "*" substr(l,END_OF_4TH_COL_PLUS_1)
}
# after printing this line, skip to next record.
print l
next
}
{ # prints every other record
print }' INPUTFILE
sed "/.\{56\}.*[^a-zA-Z0-9 ].*.\{7\}/ s/\(.\{56\}\).\{20\}\(.\{7\}\)/\1* * \2/"errorline.txt
where:
56 is the first part of your line that don't contain special char
20 is the second part taht contain maybe special char
7 is the last part, end of your string.
"* * " is the string that will replace your special char section.
Adapt those values to your string structure
This sed read all the file and replace only the lines with special char.
I have a text file in which I want to find all of ID:= "abc123" when it finds that I want it to take that value of abc123 and create a new line and have a set string, newId:= "abc123 How can I do this within terminal?
I'd like to use bash, below are some examples, find the string '"ID": ", copy the value (abc123) and make a new line with this data.
"ID": "abc123"
"newID": "abc123"
You can do this:
sed -e 's/^"ID": "\(.*\)"/&\
"newID": "\1"/' myfile.txt
First, I'll try to explain the regular expression that searches for matches:
^ Matches the start of the line
"ID": " Matches that exact string
\(.*\) Matches a sequence of zero or more (*) of any character (.). Placing this expression between backslashed parenthesis creates a "capture", which allows us to store the resulting part of the match into an auxiliary variable \1.
" Matches the double-quote character
When it finds a match, it replaces it with:
& the match itself. This operator is an auxiliary variable that represents what was matched.
\<new-line> the backslash followed by an actual new line character escapes a new line, ie. it allows us to print a new line character into the replacement
"newId": " prints that exact string
\1 prints the contents of our capture, so it prints the ID we found
" prints a double quote character.
Hope this helps =)
Try doing this :
sed -r 's#^"ID": "([a-Z0-9]+)"#"newID": "\1"#' file.txt
sed : the executable
-r : extented mode (no need to backslash parenthesis)
s : we perform a substitution, skeleton is s#origin#replacement# (the separator can be anything)
^ : means start of line in regex
( ) : parenthesis is a capture
"newID": is the start of the new string
\1 : is the end of the substituted string (the captured string)
Considering your question is very vague I made some assumptions which will become apparent in my implementation.
INPUT FILE -- call it t
ID="one"
dkkd
ID="two"
ffkjf
ID="three"
ldl
Command ran on input file
for line in `cat t`; do newID=`echo $line | grep ID | cut -d= -f2`; if [[ "$newID" != "" ]]; then echo $line >> t2; echo newID=$newID >> t2; else echo $line >> t2; fi; done
OUTPUT FILE -- Name is t2 (apparent from the command)
ID="one"
newID="one"
dkkd
ID="two"
newID="two"
ffkjf
ID="three"
newID="three"
ldl
Basically this command goes line by line in the file (in this case called t) looks for an ID line. If it finds one it gets its value, prints the original line with the ID and then prints another one with a newID following right after. If the line in question does not have and ID then it just prints the line it self.
Things to note:
If you have any other line in the file that contains "ID" in it but is not the normal ID that you requested, this will not work.