I want to give a file as input while executing the script
For example i have the code:
while read line
do
done<xyz
while executing the script
$sh file.sh input.txt
the file input.txt should go as input in the place of xyz for the while loop.
Use $1 to get the parameter:
while read line
do
...
done < "$1"
$1 will contain the first parameter you give to the script.
Example
$ cat a
#!/bin/bash
echo "I have been given this parameter --> $1"
$ ./a hello
I have been given this parameter --> hello
Related
For example,there is an interactive script that read user input his name.And then write the name in a file.
#! /bin/bash
read name
echo $name>>name.txt
If I can't change the interactive script , how can I use anothor bash to quote the interactive script?
For example,I want to write a bash that extracts the name from a text and then calls this interactive script.
So can I achieve my idea?
Use a pipe to combine output of one program with the input of the next program:
echo 'name from other script' | ./script1.sh
echo can be replaced with any other executable file or script:
$ cat >script2.sh
#!/bin/sh
echo 'first input'
echo 'second input'
^D
$ chmod u+x script2.sh
$ ./script2.sh | ./script1.sh
If your intended input name is already in a file, use IO redirection instead:
$ ./script2.sh < file_containing_name.txt
echo -n "*.xcodeproj directory: ";
read fileDirectory;
echo -n $fileDirectory;
fileExtension="pbxproj";
find $fileDirectory -name "*.${fileExtension}";
It shows "find: XXXX"(fileDirectory) no such file or directory
However if I replace read fileDirectory by
fileDirectory=$1
It works.
So what's the difference?
$1 is the first argument passed to bash script or to a function inside the script
for example:
mybashfunction /dirtofind
inside the function if you write:
echo "$1"
It should print:
/dirtofind
Edit 1:
You must place the shebang in the beginning of you file
~$ cat a.sh
#!/bin/bash
echo -n "*.xcodeproj directory: ";
read fileDirectory;
echo -n $fileDirectory;
fileExtension="pbxproj";
find "$fileDirectory" -name "*.${fileExtension}";
~$ chmod +x a.sh
~$ ./a.sh
*.xcodeproj directory: /home
/home/home/leonardo/Qt/Tools/QtCreator/share/qtcreator/qbs/share/qbs/examples/cocoa-touch-application/CocoaTouchApplication.xcodeproj/project.pbxproj
/home/leonardo/Qt/Tools/QtCreator/share/qtcreator/qbs/share/qbs/examples/cocoa-application/CocoaApplication.xcodeproj/project.pbxproj
:~$
Works like charm here. Place the shebang
#!/bin/bash
Edit 2
Yes you can use eval. Your script will be like this:
#!/bin/bash
echo -n "*.xcodeproj directory: ";
read fileDirectory;
echo -n $fileDirectory;
fileExtension="pbxproj";
eval fileDirectory=$fileDirectory
find "$fileDirectory" -name "*.${fileExtension}";
read reads data from STDIN (by default), not from positional parameters (arguments).
As you are passing the data as first argument ($1) to the script, read would not catch it; it would catch the input you are providing interactively.
Just to note, you should quote your variable expansions to avoid word splitting and pathname expansion; these are unwanted in most cases.
I have written shell script which reads commands from input file and execute commands. I have command like:
cat linux_unit_test_commands | grep "dmesg"
in the input file. I am getting below error message while executing shell script:
cat: |: No such file or directory
cat: grep: No such file or directory
cat: "dmesg": No such file or directory
Script:
#!/bin/bash
while read line
do
output=`$line`
echo $output >> logs
done < $1
Below is input file(example_commands):
ls
date
cat linux_unit_test_commands | grep "dmesg"
Execute: ./linux_unit_tests.sh example_commands
Please help me to resolve this issue.
Special characters like | and " are not parsed after expanding variables; the only processing done after variable expansion is word splitting and wildcard expansions. If you want the line to be parsed fully, you need to use eval:
while read line
do
output=`eval "$line"`
echo "$output" >> logs
done < $1
You might be wondering why its not working with cat command.
Then here is the answer for your question.
output=`$line` i.e. output=`cat linux_unit_test_commands | grep "dmesg"`
here the cat command will take (linux_unit_test_commands | grep "dmesg") all these as arguments i.e. fileNames.
From Man page:
SYNTAX : cat [OPTION]... [FILE]...
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Script is OK!
#!/bin/bash
while read line;
do
output=`$line`
echo $output >> logs
done < $1
To make it work you need to change 'cat: "dmesg": No such file or directory' to 'grep "dmesg" linux_unit_test_commands'. It will work!
cat linux_unit_test_commands
ls
date
grep "dmesg" linux_unit_test_commands
I am trying to learn bash commands, and some very basic commands are not working as I expect...http://www.tutorialspoint.com/unix/unix-special-variables.htm
http://i.stack.imgur.com/F5VGK.png
Script:
#!/bin/bash
name="john"
other="shawn"
echo $name
echo $other
echo $1
echo $2
echo $#
echo $#
Output:
$ new
john
shawn
0
$
$1, $2, etc and $# have special meaning in bash scripts. They refer to the arguments passed to the bash script, so if you have a script in a file called foo.sh like:
#!/bin/bash
echo "Number of arguments: $#";
echo "First argument: $1";
echo "Second argument: $2";
If you chmod +x foo.sh and then run:
./foo.sh first second
You will see:
Number of arguments: 2
First argument: first
Second argument: second
$1 refers to the first command line argument passed to the script. The script is foo.sh, so anything after the script name will become a command line argument.
The default command line argument separator is the "space", so when you type ./foo.sh first second, bash stores first into $1 and second into $2.
If you typed:
./foo.sh first second third FOURTH fifth
bash would store third in the variable $3, FOURTH in the variable $4, and so on.
Is your script named 'new' ? In that case run it as follows one by one and you will get an idea how this works:
./new
./new a
./new a b
when you ran your script you did not pass any arguments. the number of arguments passed to the scripts are show by echoing "echo $#". and your output clearly shows that the "echo $#" command returned "0" count. pass the argument when you call your script like below
./new argument1 argument2
I am making a bash script and I want it to be just one line, meaning it will not have any interaction with the users and the parameters will be on the same line as the command. Once the user clicks return, it will output the result.
Right now, I have something that looks like this:
#! \bin\bash
read $1 $2
do something with $1 and $2
However, if I name my script "test" when I type in test at the beginning of the command line, I will have to type enter for the rest of the script to be executed. How should I modify it so that I can run the entire thing on just one line?
The standard way to pass parameters to a script is not with read (which actively waits for input from stdin), but just to call your script with the parameters on the same line:
./my_script.sh param1 param2
Then inside the script, you can access these parameters using $1, $2, etc. Example (note also the first line - this describes what shell should be used to run the script, and should be a valid path - ie /bin/bash, not backslashes):
#!/bin/bash
echo "First: $1 Second: $2"
Then call the script:
$ ./my_script.sh Hello There
First: Hello Second: There
What you probably need is this :
You script name test.sh contains the following:
#!/bin/bash
echo "$1 $2"
Then, change permission so that you can execute the script on the command line :
chmod u+x test.sh
and run the script with arguments (two in this case) :
./test.sh tik tak
will return
tik tak