How to get a password from a shell script without echoing
I have a script that automates a process that needs access to a password protected system. The system is accessed via a command-line program that accepts the user password as an argument.
I would like to prompt the user to type in their password, assign it to a shell variable, and then use that variable to construct the command line of the accessing program (which will of course produce stream output that I will process).
I am a reasonably competent shell programmer in Bourne/Bash, but I don't know how to accept the user input without having it echo to the terminal (or maybe having it echoed using '*' characters).
Can anyone help with this?
Here is another way to do it:
#!/bin/bash
# Read Password
echo -n Password:
read -s password
echo
# Run Command
echo $password
The read -s will turn off echo for you. Just replace the echo on the last line with the command you want to run.
#!/bin/bash
stty -echo
printf "Password: "
read PASSWORD
stty echo
printf "n"
One liner:
read -s -p "Password: " password
Under Linux (and cygwin) this form works in bash and sh. It may not be standard Unix sh, though.
For more info and options, in bash, type "help read".
$ help read
read: read [-ers] [-a array] [-d delim] [-i text] [-n nchars] [-N nchars] [-p prompt] [-t timeout] [-u fd] [name ...]
Read a line from the standard input and split it into fields.
...
-p prompt output the string PROMPT without a trailing newline before
attempting to read
...
-s do not echo input coming from a terminal
上一篇: Ruby中的访问器？
下一篇: 如何从shell脚本获取密码而不回显