Verifying bash script inputs - bash

I've just started using Linux as part of my computer science degree.
I'm writing some very simple Bash scripts and I've become a tad bit stuck.
I would like the script I'm attempting to write to be able to differentiate between "non valid inputs ie letters" from "valid inputs ie numbers from a specific range"
Currently the script "works" although I'm having troubles with another echo that I would like only to "echo" when the below line is "not true", is there a simple way to write this? I'm not specifically looking for efficient code, just code that I can learn from and understand at my amateur level.
So, long story short, is it possible to obtain information from the command line below, so that I can have a simple "not true" variable that I can use in another "else" or "elif" command?
For reference line 1 is to detect alphabetical inputs, and line 2 being the line of code I would like to write as "not true" for use in another part of my script.
let xx=$a+1-1 2>/dev/null; ret=$?
if [ $a -ge 7 ] && [ $a -le 70 ] && [ $xx -eq $xx ] && [ $ret -eq 0 ]
I'm not sure I'm explaining it very well, so any help would be appreciated. :)

Welcome to Stack Overflow. :)
Start by reading the docs. I don't mean that in any way to be mean - it's just the best way to go about this.
c.f. this manual
Then read through the BashFAQs
Also, this site is really your friend. Start by familiarizing yourself with how to ask a question well.
For your question, if I read it right:
typeset -i xx # accepts only digits now.
If the input is foo, the value defaults to 0, so now just check the range.
if (( xx >= 7 && xx <= 70 )); then : value is ok
else echo "Value must be a number from 7 to 70"
exit 1
fi
Good luck. :)

One problem with the "variable with integer attribute" is that it still doesn't protect you from invalid input:
$ declare -i aNumber
$ aNumber=1234X
bash: 1234X: value too great for base (error token is "1234X")
See 6.5 Shell Arithmetic for how bash interprets values to be numbers (scroll down to the paragraph starting with "Integer constants follow the C language definition")
In my experience, the best way to check for valid numeric input is with string-oriented pattern matching.
if [[ $1 =~ ^[+-]?[0-9]+$ ]]; then
echo "input $1 is an integer"
fi
In addition to extended regular expressions, bash's advanced pattern matching can also be used within [[...]]
if [[ $1 == ?([+-])+([0-9]) ]]; then
echo "input $1 is an integer"
fi
((...)) is preferred over let. See the let builtin
command for details.
Also the shellcheck wiki entry.

Related

How to make Bash interpret variable as string for comparison? [duplicate]

This question already has answers here:
How to use double or single brackets, parentheses, curly braces
(9 answers)
Closed 3 years ago.
I have a bash script that stores the output of a file comparison. The variable becomes something like: thing="/path/to/file - differ: byte 2, line 3".
In later lines I want to check that thing is not empty. However, when I try comparing them, it interprets thing as a command and not simply as a string.
My code is somewhat as follows:
#!/bin/bash
thing="/path/to/file - differ: byte 2, line 3"
if ["$thing" != ""]; then
echo
echo "Something went wrong"
else
echo "Everything worked"
fi
Rather than saying thing is not an empty string, I get an error message that says something like
bash: [/path/to/thing - differs: byte2, line 3: No such file or directory.
How can I ensure that a comparison is happening between strings and that thing is not being interpreted as a command?
Try this instead:
#!/bin/bash
thing="/path/to/file - differ: byte 2, line 3"
if [[ -n "$thing" ]]; then
echo "Something went wrong"
else
echo "Everything worked"
fi
Notice the double [[]] pairs. That will make use of Bash's internal test.
Use help test on Bash's command prompt to get more information.
EDIT: As explained by Jonathan in the comments, using any of [ … ] or [[ … ]] or test uses a built-in internal command in Bash. Section CONDITIONAL EXPRESSIONS of man bash (or Conditional Expressions, Conditional Constructs and Bourne Shell Built-ins) explains that. However, while [ … ] and test are logically the same (the only difference is that [ expects its last argument to be ] but test has no analogous expectation), the tests implemented by [[ … ]] are subject to different parsing rules from [ … ] and test, and [[ … ]] implements some extra test capabilities missing from the other.

why in an 'if' statement 'then' has to be in the next line in bash?

if is followed by then in bash but I don't understand why then cannot be used in the same line like if [...] then it has to be used in the next line. Does that remove some ambiguity from the code? or bash is designed like that? what is the underlying reason for it?
I tried to write if and then in the same line but it gave the error below:
./test: line 6: syntax error near unexpected token \`fi'
./test: line 6: \`fi'
the code is:
#!/bin/bash
if [ $1 -gt 0 ] then
echo "$1 is positive"
fi
It has to be preceded by a separator of some description, not necessarily on the next line(a). In other words, to achieve what you want, you can simply use:
if [[ $1 -gt 0 ]] ; then
echo "$1 is positive"
fi
As an aside, for one-liners like that, I tend to prefer:
[[ $1 -gt 0 ]] && echo "$1 is positive"
But that's simply because I prefer to see as much code on screen as possible. It's really just a style thing which you can freely ignore.
(a) The reason for this can be found in the Bash manpage (my emphasis):
RESERVED WORDS: Reserved words are words that have a special meaning to the shell. The following words are recognized as reserved when unquoted and either the first word of a simple command (see SHELL GRAMMAR below) or the third word of a case or for command:
! case coproc do done elif else esac fi for function if in select then until while { } time [[ ]]
Note that, though that section states it's the "first word of a simple command", the manpage seems to contradict itself in the referenced SHELL GRAMMAR section:
A simple command is a sequence of optional variable assignments followed by blank-separated words and redirections, and terminated by a control operator. The first word specifies the command to be executed, and is passed as argument zero.
So, whether you consider it part of the next command or a separator of some sort is arguable. What is not arguable is that it needs a separator of some sort (newline or semicolon, for example) before the then keyword.
The manpage doesn't go into why it was designed that way but it's probably to make the parsing of commands a little simpler.
Here's another way to explain the need for a line break or semicolon before then: the thing that goes between if and then is a command (or sequence of commands); if the then just came directly after the command without a delimiter, it'd be ambiguous whether it should be treated as a shell keyword or just an argument to the command.
For instance, this is a perfectly valid command:
echo This prints a phrase ending with then
...which prints "This prints a phrase ending with then". Now, consider this one:
if echo This prints a phrase ending with then
should that print "This prints a phrase ending with then" and look for a then keyword later on, or should it just print "This prints a phrase ending with" and treat the then as a keyword?
In order to settle this ambiguity, shell syntax says it should treat "then" as an argument to echo, and in order to get it treated as a keyword you need a command delimiter (line break or semicolon) to mark the end of the command.
Now, you might think that your if condition [ $1 -gt 0 ], already has a perfectly good delimiter, namely the ]. But in shell syntax, that's really just an argument to the [ command (yes, that's a command). Try this command:
[ 1 -gt 0 ] then
...and you'll probably get an error like "-bash: [: missing ']'", because the [ command checked its last argument to make sure it was "]", found that it was "then" instead, and panicked.
Perhaps it helps to understand why this is so by way of a few examples. The argument to if is a sequence of commands; so you can say e.g.
if read -r -p "What is your name?" name
[ "$name" -eq "tripleee" ]
then
echo "I kneel before thee"
fi
or even a complex compound like
while read -r -p "Favorite number?" number
case $number in
42) true; break;;
*) false;;
esac
do
echo "Review your preferences, then try again"
done
This extremely powerful but potentially confusing feature of the shell is probably one of its most misunderstood constructs. The ability to pass a sequence of commands to the flow control statements can make for very elegant scripts, but is often missed entirely (see e.g. Why is testing "$?" to see if a command succeeded or not, an anti-pattern?)
If it helps, you can use semi-colons
if [ $1 -gt 0 ]; then
echo "$1 is positive"
fi
# or even
if [ $1 -gt 0 ]; then echo "$1 is positive"; fi
As for why, it helps me to think of if, then, else, and fi as bash commands, and just like all other commands, they need to be at the start of a line (or after a semi-colon).

how to match a specific file extension in shellscript

I looked some other posts and learnt to match file extension in the following way but why my code is not working? Thanks.
1 #!/bin/sh
2
3 for i in `ls`
4 do
5 if [[ "$i" == *.txt ]]
6 then
7 echo "$i is .txt file"
8 else
9 echo "$i is NOT .txt file"
10 fi
11 done
eidt:
I realized #!/bin/sh and #!/bin/bash are different, if you are looking at this post later, remember to check which one you are using.
The [[ ]] expression is only available in some shells, like bash and zsh. Some more basic shells, like dash, do no support it. I'm guessing you're running this on a recent version of Ubuntu or Debian, where /bin/sh is actually dash, and hence doesn't recognize [[. And actually, you shouldn't use [[ ]] with a #!/bin/sh shebang anyway, since it's unsafe to depend on a feature that the shebang doesn't request.
So, what to do about it? You'll have the [ ] type of test expression available, but it doesn't do pattern matching (like *.txt). There are a number of alternate ways to do it:
The case statement is available in even basic shells, and has the same pattern matching capability as [[ = ]]. This is the most common way to do this type of thing, especially when you have a list of different patterns to check against.
More indirectly, you can use ${var%pattern} to try remove .txt from the end of the end of the value (see "Remove Smallest Suffix Pattern" here), and then check to see if that changed the value:
if [ "$i" != "${i%.txt}" ]
More explanation: suppose $i is "file.txt"; then this expands to [ "file.txt" != "file" ], so they're not equal, and the test (for !=) succeeds. On the other hand, if $i is "file.pdf", then it expands to [ "file.pdf" != "file.pdf" ], which fails because the strings are the same.
Other notes: when using [ ], use a single equal sign for string comparison, and be sure to properly double-quote all variable references to avoid confusion. Also, if you use anything that has special meaning to the shell (like < or >), you need to quote or escape them.
You could use the expr command's : operator to do regular expression matching. (Regular expressions are a different type of pattern from the basic wildcard or "glob" expression.) You could do this, but don't.
#!/bin/sh
for i in `ls`
do
if [[ "$i" = *".txt" ]] ; then
echo "$i is .txt file"
else
echo "$i is NOT .txt file"
fi
done
You don't have to loop in ls output, and sh implementation might vary among OS distributions.
Consider:
#! /bin/sh
for i in *
do
if [[ "$i" == *.txt ]]
then
echo "$i is txt file"
else
echo "$i is NOT txt file"
fi
done

Shell Scripting - less than operation not working

echo "Enter age"
read $ag
if [[ $ag -lt 30 ]]
then
echo "Age is less than 30"
else
echo "not"
fi
For this program, whatever be the value of 'ag', the output is shown as "Age is less than 30".
What could be reason? and a solution?
Your mistake is:
read $ag
You should write:
read ag
With your current syntax, read stores the user input in a variable whose name is the content of variable ag.
An echo $ag after the read would have told you. (Checking your input and your assumptions is elementary debugging, really.)
read $ag
This resolves $ag (which is undefined), giving:
read
You're reading, but you're not reading into the variable as you expected. $ag is still empty, which resolves to "less than 30".
Fix:
read ag
That's the proper syntax for read.

Trying to check if information written in a text file is equal to a variable within shell script

I'm still a newbie to Linux scripting. My goal is to determine if the information written in a .txt file is equal to a variable within the script. This is the code that I have written so far:
#!/bin/sh
a=-5
if [$(cat < top.txt) -eq $a ]; then
echo "YES!"
else
echo "NNOOO!"
fi
Please let me know what I should be doing. To reiterate, I want top.txt to equal -5. I currently have -5 written in the text but can't seem to get it to work.
Use More Quotes™:
[ "$(cat top.txt)" = "$a" ]
However, beware that this doesn't handle the case where top.txt ends with a newline - $() will chop that right off. So you might be better served with this:
[ "$(cat top.txt; printf x)" = "$a"x ]
Another issue is that you can't check files containing the NUL character this way; variables can not contain NUL characters.
At this point I'd be tempted to ask what you are trying to achieve.
You can use:
[ "$(<top.txt)" -eq "$a" ] && echo "YES!" || echo "NOOOS!"

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