Bash – Checking if a ruby gem is installed from bash script

bashgem

I need to, from a bash script, check to see if certain Ruby gems are installed .

I thought I could do something like

if ! gem list <name>; then do_stuff; fi

but testing on the command line using echo $? shows that gem list <name> returns 0 regardless of if name is actually found.

Does this mean I have to use grep to filter the output of gem list, or is there a better way I can check to see if a gem is installed?

Best Answer

gem list <name> -i will return the string true if the gem is installed and false otherwise. Also, the return codes are what you would expect.

For more informations, see gem help list.

Edit: @Riateche correctly observed that this might give false positives if you search for a gem name that is a substring of an otherwise installed gem. To avoid this, use a regex syntax:

gem list '^<name>$' -i

(Example: gem list '^mini$' -i).

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use scriptable ftp client like lftp.

you use it like this:

lftp -u login,pass host.com -e "get file.txt;exit "

or even simpler - use wget:

wget -O file ftp://login:pass@host.com/file.txt

Bash avoide intrpretate special characters

I don't understand what particular problem you are stuck on. Your description is really unclear. So I'll give some general advice on simplifying the script; if it's not enough to solve your problem, try to come up with a clearer explanation.

I'm pretty sure this script is much more complicated than it needs to be. Spending a few minutes browsing the documentation of a command to see if one of its options could help you can save hours of debugging. Spending a few minutes thinking about the general structure of the script can save hours of debugging.

Here are several ways in which you could have made your script simpler.

All variable substitutions should be inside double quotes, i.e., always write "$foo" and not just ~~$foo~~. You've done it sometimes, but not systematically. Always use double quotes unless you know why you do not want them in a particular case.

Here's a simpler way of writing your help function; it's called a “here document”.

function help()
{
    cat <<EOF
This script is responsible for simply recursively searching patterns
given in [INPUT FILE] in [PATH]. If pattern is not found then it is
written to [OPTIONAL OUTPUT FILE]. If [OPTIONAL OUTPUT FILE] is not
given the default [OUTPUT FILE] name is: out
Usage: $0 [INPUT FILE] [PATH TO DIRECTORY] [OPTIONAL OUTPUT FILE]
e.g. : $0 input_file /var/www/html/ output_file
or   : $0 help -> this help
EOF
}

Give your script a non-zero exit code to indicate failure:

if [ $# -lt 2 ]; then help; exit 2; fi
if [ ! -e "$in" ]; then echo "Input file: $1 does not exist"; exit 2; fi
if [ ! -d "$path" ]; then  echo "Path: $path does not exist"; exit 2; fi

Modifying the input file in is surprising, and you can combine the whitespace-only line removal with the multiple sed expressions that add a backslash before some characters.
```
<"$in" sed -e '/^ *$/d' -e 's,[-\\".],\\&,g' > "$tmp"
```
However the quoting you perform here is strange. Why are you quoting - and ", which are not special to grep, but not * and [ which are special? What is the syntax of the patterns supposed to be?

If you meant the patterns to be literal strings to look for, all this work is unnecessary (except for removing whitespace-only lines): call grep -F.
In the part that reads the lines from $tmp, you don't need the counter variable, you can just append to the array. You also need to pass the -r argument to the read built-in, so that it doesn't strip some backslashes.
```
while read -r line; do
    linesTable+=("$line")
done <"$tmp"
```
In the loop over the patterns, you store the output of grep in a variable, but all you're doing is testing if grep found a match. It would be a lot easier (and faster) to use the return code of grep for that. (I've also removed what is presumably debugging output from the loop; you don't need tee to append to a file, just use the redirection operator >>.)
```
for line in "${linesTable[@]}"; do
    grep -q -r -- "$line" "$path"
    if [ $? -ne 0 ]; then echo "$line" >>"$tmp"; fi
done
```
You don't need to free memory at the end of the script. If this was really a function that was part of a bigger script, you should declare them with the local builtin.

Here's a restructured version of the part of your script after the command line parsing. I've incorporated the local changes outlined above and used functions to make the structure clearer. Note that the clearer structure means I don't need to use a temporary file. I don't know if the resulting script does what you want, since you don't explain precisely what you want.

function extract_patterns () {
    <"$in" sed -e '/^ *$/d' -e 's,[-\\".],\\&,g'
}

function report_missing_patterns () {
  local pattern
  for pattern in "$@"; do
    grep -q -r -- "$pattern" "$path"
    if [ $? -ne 0 ]; then printf "%s\n" "$pattern"; fi
  done
}

process_patterns () {
  local patterns line
  patterns=()
  while read -r line; do
      patterns+=("$line")
  done
  report_missing_patterns "${patterns[@]}" | sort -u >"$out"
}

extract_patterns | process_patterns

Best Answer

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