Bash “command not found” error

问题

I am trying to edit my .bashrc file with a custom function to launch xwin. I want it to be able to open in multiple windows, so I decided to make a function that accepts 1 parameter: the display number. Here is my code:

function test(){
    a=$(($1-0))
    "xinit -- :$a -multiwindow -clipboard &"
}

The reason why I created a variable "a" to hold the input is because I suspected that the input was being read in as a string and not a number. I was hoping that taking the step where I subtract the input by 0 would convert the string into an integer, but I'm not actually sure if it will or not. Now, when I call

test 0

I am given the error

-bash: xinit -- :0 -multiwindow -clipboard &: command not found

How can I fix this? Thanks!

回答1:

Because the entire quoted command is acting as the command itself:

$ "ls"
a b c
$ "ls -1"
-bash: ls -1: command not found

Get rid of the double quotation marks surrounding your xinit:

xinit -- ":$a" -multiwindow -clipboard &

回答2:

In addition to the double-quotes bishop pointed out, there are several other problems with this function:

• test is a standard, and very important, command. Do not redefine it! If you do, you risk having some script (or sourced file, or whatever) run:

  if test $num -eq 5; then ...
  

  Which will fire off xinit on some random window number, then continue the script as if $num was equal to 5 (whether or not it actually is). This way lies madness.

• As chepner pointed out in a comment, bash doesn't really have an integer type. To it, an integer is just a string that happens to contain only digits (and maybe a "-" at the front), so converting to integer is a non-opertation. But what you might want to do is check whether the parameter got left off. You can either check whether $1 is empty (e.g. if [[ -z "$1" ]]; then echo "Usage: ..." >&2 etc), or supply a default value with e.g. ${1:-0} (in this case, "0" is used as the default).

• Finally, don't use the function keyword. bash tolerates it, but it's nonstandard and doesn't do anything useful.

So, here's what I get as the cleaned-up version of the function:

launchxwin() {
    xinit -- ":${1:-0}" -multiwindow -clipboard &
}

回答3:

That happens because bash interprets everything inside quotes as a String. A command is an array of strings which the first element is a binary file or a internal shell command. Subsequent strings in the array are taken as argument.

When you type:

"xinit -- :$a -multiwindow -clipboard &"

the shell thinks that everything you wrote is a command. Depending on the command/program you ran all the rest of the arguments can be a single string. But mostly you use quotes only if you are passing a argument that has spaces inside like:

mkdir "My Documents"

That creates a single directory named My Documents. Also, you could escape spaces like this.

mkdir My\ Documents

But remember, "$" is a special character like "\". It gets interpreted by the shell as a variable. "$a" will be substituted by its value before executing. If you use a simple quote ('$a') it will not be interpreted by the shell.

Also, "&" is a special character that executes the command in background. You should probably pass it outside the quotes also.

来源：https://stackoverflow.com/questions/32745607/bash-command-not-found-error