read Name?" Enter Full name "
First= echo $Name | cut -d ' ' -f1
I am using this command to cut Full name, but it automatically print first name. which I don't want.
I want an output like
James, This is prime number.
In the shell I use (bash on MacOS), this following works.
read -p "Enter full name " first last
echo Hi $first, you are the man
Please specify what shell you are using.
Another thing to note, is that in order to assign to the variable 'First' in your question, in bash you need to do surround the command with backticks `` or the $() construction, as follows.
First=$(echo $Name | cut -d ' ' -f1)
Related
I'm new to zsh scripting and I was wondering if it's possible to use the sha256sum function to encrypt every value in a list.
Here is what I have tried so far:
#!/bin/zsh
filenames=`cat filenames.txt`
output='shaNames.txt'
for name in $filenames
do
echo -n $name | sha256sum >> $output
done
What I'm trying to accomplish is to encrypt every name in the list and append it to a new text file.
Any suggestions on what am I doing wrong are appreciated.
You are assigning the output of cat filenames.txt to a multiline variable. The for loop will then only loop once over the content.
What you want to do instead is e.g.:
for name in $(cat filenames.txt)
do
echo -n "$name" | sha256sum >> "$output"
done
Note that while you can still use them, backticks are deprecated in favor of $(somecommand).
Also note that you should always put variables in double quotes, as they could contain spaces.
Your method would fail anyways if one line of your textfile would contain a space.
You could use the following instead:
while read name
do
echo -n "$name" | sha256sum >> "$output"
done < filenames.txt
To anyone who might need the same. What I was doing wrong was assigning the values in the file to a single string variable instead of a list.
To correct that one must use:
filenames=(`cat filenames.txt`)
The parenthesis indicates that a list or array is stored in the filenames variable.
I have a string in different ranges :
WATSON_AJAY_AB04_DOTHING.data
WATSON_NAVNEET_CK4_DOTHING.data
WATSON_PRASHANTH_KJ56_DOTHING.data
WATSON_ABHINAV_KD323_DOTHING.data
On these above string how can I extract
AB04,CK4,KJ56,KD323
in Unix?
echo "$string" | cut -d'_' -f3
You could use sed or grep for this task. But since the string is so simple, I dont think you will need to.
One method is to use the bash 'cut' command. Below is an example directly on the BASH shell/command line:
jimm#pi$ string='WATSON_AJAY_AB04_DOTHING.data'
jimm#pi$ cut -d '_' -f 3 <<< "$string"
AB04 <-- outputs the result directly
(edit: of course Lucas' answer above is also a quick 'one-liner' that does the same thing as above - he beat me to it) :)
The cut will take an _ character as the delimiter (the -d '_' part), then display the 3rd slice of the string (the -f 3 part).
Or, if you want to output that 3rd slice from a list of content (using your list above), you can write a simple BASH script.
First, save the lines above ('WATSON...etc') into something like text.txt. Then open up your favorite text editor and type:
#!/bin/sh
cut -d '_' -f 3 < $1
Save that script to some useful name like slice.sh, and make sure it is executable with something like chmod 775 slice.sh.
Then at the command line you can execute the script against your text file, and immediately get an output of those parts of the file you want (in this case the third set of text, separated by the _ character):
$ ./slice.sh text.txt
AB04
CK4
KJ56
KD323
Hope that helps! Bear in mind that the commands above may vary a bit, depending on the flavor of *nix you are using, but it should at least point you in the right direction.
The intended result of this program is to perform the deletion of a record from a temporary file by searching for the last name. If the name is in the file it will display a message that the record is deleted from the file. The message will have a last and first name of the person deleted. If there is no record for the name entered, display an error message indicating that the name is not in the file. The error message will have the last name of the person searched.
For the most part I have figured this code completely out. However, I am running into errors that are giving me a lot of trouble.
Code is as follows:
#While loop
delete_choice="y"
until [[ $delete_choice == "n" ]]
do
#Create Delete Message
clear
echo " Delete Record "
echo -en '\n'
echo -en '\n'
echo "What is the last name of the person you want to delete:"
read last_name
if line=$( grep -Fi "$last_name" "$2")
then
IFS=: read c1 c2 rest <<< "$line"
echo -e "Last Name: $c1\nFirst Name: $c2\nSTUDENT RECORD HAS BEEN DELETED FROM FILE"
sed "/$line/d" $2
else
echo "ERROR: $last_name is not in database"
echo "Would you like to search again (y/n)"
read delete_choice
fi
done
;;
So what happens when I execute this code is it brings up the delete message and asks me to input a last name. I put in a last name "smith", when I do this it skips the whole if statement and goes right to the end of the loop and then brings me right up to asking me what the last name is of the person I want to delete. So obviously it is concentrated in the grep statement somewhere. Another odd thing is if I put a name that I know is not in there it will take me to the else statement and give me the error message and ask me if I want to search again.
Any help would be appreciated, I been searching for hours with the grep statement and cannot figure it out.
Additionally: On a side note does anybody know how to make it so I can input "n" or "N" in the Until...Do statement to keep the loop going?
EDIT:
Ok I fixed all the other problems in this code but there is just one problem I cannot fix. Every time I delete an entry from the file it doesn't allow me to execute the echo command.
Code is as follows:
d|D)
#While loop
delete_choice="y"
while true
do
#Create Delete Message
clear
echo " Delete Record "
echo -en '\n'
echo -en '\n'
echo "What is the last name of the person you want to delete:"
read last_name
if line=$(grep -i "^${last_name}:" "$2")
then
echo "$line" |
while IFS=: read c1 c2 rest; do
last=$c1
first=$c2
sed -i "/^$c1:$c2:/d" "$2"
done
echo -e "Last Name: $last\nFirst Name: $first\nSTUDENT RECORD HAS BEEN DELETED FROM FILE"
else
echo "ERROR: $last_name is not in database"
echo "Would you like to search again (y/n):"
read delete_choice
case $delete_choice in [Nn]) break;; esac
fi
done
;;
As you can see I execute the echo command but it never displays. However, I can tell that it deletes the entry from the file when I exit the program and check it. Does anybody know how to make it display the echo command correctly?
You have used <<< which is a Bash-only feature.
The proper way to debug your problem is to run your script with ksh -x script arg (or possibly sh -x script arg if your script was a pure Bourne shell script; but this one isn't).
However, you have some peculiarities in your code. Allow me to offer a refactoring.
delete_choice="y"
until [[ "$delete_choice" == "n" ]]
do # Indent your control structures
clear
echo " Delete Record"
echo # Massive trailing whitespace removed ----------^^^
echo # Just echo to output a new line
echo "What is the last name of the person you want to delete:"
read last_name
case $last_name in [Nn]) break;; esac # Exit on "n"
if line=$(grep -Fi "$last_name" "$2")
then # Indent your control structures
file=$2
oldIFS=$IFS
IFS=:
set -- "$line" # break input into $1, $2, etc
IFS=$oldIFS
echo -e "Last Name: $1\nFirst Name: $2\nI SCREAM IN UPPER CASE"
sed -i "/^$1:$2:/d" "$file" # note quoting and -i and changed variable
else
echo "ERROR: $last_name is not in database"
echo "Would you like to search again (y/n)"
read delete_choice
fi
done
Another problem here was unrelated to your question. Directly executing the user's input in the regex would be extremely careless -- what if somebody input just a for a last name? The changed code will only delete one user, regardless of how short the input; but perhaps the grep should also be anchored to use a tighter search expression.
There is still a possibility that a peculiar name in the database would not match itself, or match more than itself. For example, I use my second given name, so where my full name is required, I often input my name as "First *Middle Last" with an asterisk next to the name I use, which is the convention where I live; but the string *Middle will not match itself in a regular expression.
Also, without -i, the sed script would simply print a copy of the database to standard output.
The delete_choice variable is not getting updated anywhere within your code so you could change the exterior loop to just while true and handle exiting by breaking out of the loop, like I have done. I didn't change the until on the vague speculation that this might be used in other parts of your code which you aren't showing.
Goal
In ZSH script, for a given args, I want to obtain the first string and the rest.
For instance, when the script is named test
sh test hello
supposed to extract h and ello.
ZSH manual
http://zsh.sourceforge.net/Doc/zsh_a4.pdf
says:
Subscripting may also be performed on non-array values, in which case the subscripts specify a
substring to be extracted. For example, if FOO is set to ‘foobar’, then ‘echo $FOO[2,5]’ prints
‘ooba’.
Q1
So, I wrote a shell script in a file named test
echo $1
echo $1[1,1]
terminal:
$ sh test hello
hello
hello[1,1]
the result fails. What's wrong with the code?
Q2
Also I don't know how to extract subString from n to the last. Perhaps do I have to use Array split by regex?
EDIT: Q3
This may be another question, so if it's proper to start new Thread, I will do so.
Thanks to #skishore Here is the further code
#! /bin/zsh
echo $1
ARG_FIRST=`echo $1 | cut -c1`
ARG_REST=`echo $1 | cut -c2-`
echo ARG_FIRST=$ARG_FIRST
echo ARG_REST=$ARG_REST
if $ARG_FIRST = ""; then
echo nullArgs
else
if $ARG_FIRST = "#"; then
echo #Args
else
echo regularArgs
fi
fi
I'm not sure how to compare string valuables to string, but for a given args hello
result:
command not found: h
What's wrong with the code?
EDIT2:
What I've found right
#! /bin/zsh
echo $1
ARG_FIRST=`echo $1 | cut -c1`
ARG_REST=`echo $1 | cut -c2-`
echo ARG_FIRST=$ARG_FIRST
echo ARG_REST=$ARG_REST
if [ $ARG_FIRST ]; then
if [ $ARG_FIRST = "#" ]; then
echo #Args
else
echo regularArgs
fi
else
echo nullArgs
fi
EDIT3:
As the result of whole, this is what I've done with this question.
https://github.com/kenokabe/GitSnapShot
GitSnapShot is a ZSH thin wrapper for Git commands for easier and simpler usage
A1
As others have said, you need to wrap it in curly braces. Also, use a command interpreter (#!...), mark the file as executable, and call it directly.
#!/bin/zsh
echo $1
echo ${1[1,1]}
A2
The easiest way to extract a substring from a parameter (zsh calls variables parameters) is to use parameter expansion. Using the square brackets tells zsh to treat the scalar (i.e. string) parameter as an array. For a single character, this makes sense. For the rest of the string, you can use the simpler ${parameter:start:length} notation instead. If you omit the :length part (as we will here), then it will give you the rest of the scalar.
File test:
#!/bin/zsh
echo ${1[1]}
echo ${1:1}
Terminal:
$ ./test Hello
H
ello
A3
As others have said, you need (preferably double) square brackets to test. Also, to test if a string is NULL use -z, and to test if it is not NULL use -n. You can just put a string in double brackets ([[ ... ]]), but it is preferable to make your intentions clear with -n.
if [[ -z "${ARG_FIRST}" ]]; then
...
fi
Also remove the space between #! and /bin/zsh.
And if you are checking for equality, use ==; if you are assigning a value, use =.
RE:EDIT2:
Declare all parameters to set the scope. If you do not, you may clobber or use a parameter inherited from the shell, which may cause unexpected behavior. Google's shell style guide is a good resource for stuff like this.
Use builtins over external commands.
Avoid backticks. Use $(...) instead.
Use single quotes when quoting a literal string. This prevents pattern matching.
Make use of elif or case to avoid nested ifs. case will be easier to read in your example here, but elif will probably be better for your actual code.
Using case:
#!/bin/zsh
typeset ARG_FIRST="${1[1]}"
typeset ARG_REST="${1:1}"
echo $1
echo 'ARG_FIRST='"${ARG_FIRST}"
echo 'ARG_REST='"${ARG_REST}"
case "${ARG_FIRST}" in
('') echo 'nullArgs' ;;
('#') echo '#Args' ;;
(*)
# Recommended formatting example with more than 1 sloc
echo 'regularArgs'
;;
esac
using elif:
#!/bin/zsh
typeset ARG_FIRST="${1[1]}"
typeset ARG_REST="${1:1}"
echo $1
echo 'ARG_FIRST='"${ARG_FIRST}"
echo 'ARG_REST='"${ARG_REST}"
if [[ -z "${ARG_FIRST}" ]]; then
echo nullArgs
elif [[ '#' == "${ARG_FIRST}" ]]; then
echo #Args
else
echo regularArgs
fi
RE:EDIT3
Use "$#" unless you really know what you are doing. Explanation.
You can use the cut command:
echo $1 | cut -c1
echo $1 | cut -c2-
Use $() to assign these values to variables:
ARG_FIRST=$(echo $1 | cut -c1)
ARG_REST=$(echo $1 | cut -c2-)
echo ARG_FIRST=$ARG_FIRST
echo ARG_REST=$ARG_REST
You can also replace $() with backticks, but the former is recommended and the latter is somewhat deprecated due to nesting issues.
So, I wrote a shell script in a file named test
$ sh test hello
This isn't a zsh script: you're calling it with sh, which is (almost certainly) bash. If you've got the shebang (#!/bin/zsh), you can make it executable (chmod +x <script>) and run it: ./script. Alternatively, you can run it with zsh <script>.
the result fails. What's wrong with the code?
You can wrap in braces:
echo ${1} # This'll work with or without the braces.
echo ${1[3,5]} # This works in the braces.
echo $1[3,5] # This doesn't work.
Running this: ./test-script hello gives:
./test-script.zsh hello
hello
llo
./test-script.zsh:5: no matches found: hello[3,5]
Also I don't know how to extract subString from n to the last. Perhaps do I have to use Array split by regex?
Use the [n,last] notation, but wrap in braces. We can determine how long our variable is with, then use the length:
# Store the length of $1 in LENGTH.
LENGTH=${#1}
echo ${1[2,${LENGTH}]} # Display from `2` to `LENGTH`.
This'll produce ello (prints from the 2nd to the last character of hello).
Script to play with:
#!/usr/local/bin/zsh
echo ${1} # Print the input
echo ${1[3,5]} # Print from 3rd->5th characters of input
LENGTH=${#1}
echo ${1[2,${LENGTH}]} # Print from 2nd -> last characters of input.
You can use the cut command:
But that would be using extra baggage - zsh is quite capable of doing all this on it's own without spawning multiple sub-shells for simplistic operations.
I am using the below code for replacing a string
inside a shell script.
echo $LINE | sed -e 's/12345678/"$replace"/g'
but it's getting replaced with $replace instead of the value of that variable.
Could anybody tell what went wrong?
If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.
Try:
echo $LINE | sed -e "s/12345678/${replace}/g"
Transcript:
pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _
Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.
you can use the shell (bash/ksh).
$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc
Not specific to the question, but for folks who need the same kind of functionality expanded for clarity from previous answers:
# create some variables
str="someFileName.foo"
find=".foo"
replace=".bar"
# notice the the str isn't prefixed with $
# this is just how this feature works :/
result=${str//$find/$replace}
echo $result
# result is: someFileName.bar
str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo $result
# result is: someFileName.sally because ".foo" was not found
Found a graceful solution.
echo ${LINE//12345678/$replace}
Single quotes are very strong. Once inside, there's nothing you can do to invoke variable substitution, until you leave. Use double quotes instead:
echo $LINE | sed -e "s/12345678/$replace/g"
Let me give you two examples.
Using sed:
#!/bin/bash
LINE="12345678HI"
replace="Hello"
echo $LINE | sed -e "s/12345678/$replace/g"
Without Using sed:
LINE="12345678HI"
str_to_replace="12345678"
replace_str="Hello"
result=${str//$str_to_replace/$replace_str}
echo $result
Hope you will find it helpful!
echo $LINE | sed -e 's/12345678/'$replace'/g'
you can still use single quotes, but you have to "open" them when you want the variable expanded at the right place. otherwise the string is taken "literally" (as #paxdiablo correctly stated, his answer is correct as well)
To let your shell expand the variable, you need to use double-quotes like
sed -i "s#12345678#$replace#g" file.txt
This will break if $replace contain special sed characters (#, \). But you can preprocess $replace to quote them:
replace_quoted=$(printf '%s' "$replace" | sed 's/[#\]/\\\0/g')
sed -i "s#12345678#$replace_quoted#g" file.txt
I had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:
replace="salt & pepper"
echo "pass the salt" | sed "s/salt/${replace/&/\&}/g"
use # if you want to replace things like /. $ etc.
result=$(echo $str | sed "s#$oldstr#$newstr#g")
the above code will replace all occurrences of the specified replacement term
if you want, remove the ending g which means that the only first occurrence will be replaced.
Use this instead
echo $LINE | sed -e 's/12345678/$replace/g'
this works for me just simply remove the quotes
I prefer to use double quotes , as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion .
so use double quotes instaed.
echo $LINE | sed -e "s/12345678/$replace/g"