Re: Help with String Extraction

[Eduardo A . Bustamante López]

There are a few issues with your script.

On Mon, Sep 16, 2024 at 02:06:22PM -0400, Steve Matzura wrote:
(...)
> #!/bin/sh
> for f in $(ls *.xml)

Do not parse the output of ls [1]. Instead, just use:

  for f in *.xml

> do
>   echo $f # Display the filename

The use of (the right kind of) quotes is crucial in shell scripts. As a general
rule, wrap parameter expansions [2] in double quotes:

    echo "$f"

>   stringZ = $(grep -i svrinfodesc $f) # Find the string I want

This is not a correct variable assignment. Do not include spaces before or after
the  '='  character. Instead do something like this:

    stringZ=$(some command)

>   echo $stringZ # Let's see the string
> pos1 = `expr index "$string" \>` + 1 # Position of first character after ">"

Putting aside the incorrect assignment,
here the  '+ 1' is outside of the command substitution (the stuff between
backticks). I guess you intended to have an additional command substitution
wrapping the  "expr index"  one. Keep in mind that nesting command substitutions
using backticks is not pretty, e.g.

  $ name=John; x=`echo hello \`echo "$name"\``; echo "$x"
  hello John

So I'd use the  $(...)  command substitution form instead, it's much easier to
nest.

  $ name=John; x=$(echo hello $(echo "$name")); echo "$x"
  hello John

> len = (pos2 - pos1) # How long will the substring be?

This is not how you do integer substraction with Bash. You have a few options,
e.g.

   # Start with these values
   $ pos2=10 pos1=5

   # Option 1
   $ echo "$((pos2 - pos1))"
   5

   # Option 2
   $ ((diff = pos2 - pos1))
   $ echo "$diff"
   5


In addition, please consider enabling the xtrace option (set -x) at the
beginning of your script, which will cause Bash to output a trace of commands as
it executes them. This comes in handy when troubleshooting, as you can review if
bash is executing commands in the way you expect.


[1] <https://mywiki.wooledge.org/ParsingLs>

[2] A parameter expansion is how a construct like  $foo  is called. When you
omit wrapping the expansion in double quotes, the shell will do additional
processing on the expanded value, like expanding file name patterns.