Can anyone explain this behavior? Running:
#!/bin/sh
echo "hello world" | read var1 var2
echo $var1
echo $var2
results in nothing being ouput, while:
#!/bin/sh
echo "hello world" > test.file
read var1 var2 < test.file
echo $var1
echo $var2
produces the expected output:
hello
world
Shouldn't the pipe do in one step what the redirection to test.file did in the second example? I tried the same code with both the dash and bash shells and got the same behavior from both of them.
produces no output because pipelines run each of their components inside a subshell. Subshells inherit copies of the parent shell's variables, rather than sharing them. Try this:
The parentheses define a region of code that gets run in a subshell, and $foo retains its original value after being modified inside them.
Now try this:
The braces are purely for grouping, no subshell is created, and the $foo modified inside the braces is the same $foo modified outside them.
Now try this:
Inside the braces, the read builtin creates $var1 and $var2 properly and you can see that they get echoed. Outside the braces, they don't exist any more. All the code within the braces has been run in a subshell because it's one component of a pipeline.
You can put arbitrary amounts of code between braces, so you can use this piping-into-a-block construction whenever you need to run a block of shell script that parses the output of something else.
A recent addition to
bash
is thelastpipe
option, which allows the last command in a pipeline to run in the current shell, not a subshell, when job control is deactivated.will indeed output
My take on this issue (using Bash):
Try:
The problem, as multiple people have stated, is that var1 and var2 are created in a subshell environment that is destroyed when that subshell exits. The above avoids destroying the subshell until the result has been echo'd. Another solution is:
This has already been answered correctly, but the solution has not been stated yet. Use ksh, not bash. Compare:
To:
ksh is a superior programming shell because of little niceties like this. (bash is the better interactive shell, in my opinion.)
Allright, I figured it out!
This is a hard bug to catch, but results from the way pipes are handled by the shell. Every element of a pipeline runs in a separate process. When the read command sets var1 and var2, is sets them it its own subshell, not the parent shell. So when the subshell exits, the values of var1 and var2 are lost. You can, however, try doing
which returns the expected answer. Unfortunately this only works for single variables, you can't set many at a time. In order to set multiple variables at a time you must either read into one variable and chop it up into multiple variables or use something like this:
While I admit it's not as elegant as using a pipe, it works. Of course you should keep in mind that read was meant to read from files into variables, so making it read from standard input should be a little harder.