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# 1  
Old 08-19-2012
sed: -e expression #1, char 0: no previous regular expression
Hello All,

I'm trying to extract the lines between two consecutive elements of an array from a file.
My array looks like:
Code:
problem_arr=(PRS111 PRS213 PRS234)

j=0
while [ $j -le ${problem_arr[@]} ]
do
    k=`expr $j + 1`
    sed  -n "/${problem_arr[$j]}/,/${problem_arr[$k]}/p" problemid.txt
    ---some operation goes here----
    j=`expr $j + 1`
done

However, i get
Code:
sed: -e expression #1, char 0: no previous regular expression

error message.

Just to help you narrow down the problem i want to convey u that following line works:
Code:
sed  -n "/${problem_arr[0]}/,/${problem_arr[1]}/p" problemid.txt

Could you please help me how to use array subscript variable in such cases? Thank you so much in advance!

Regards,
Indu


Moderator's Comments:
Mod Comment Please use code tags next time for your code and data.
Last edited by zaxxon; 08-19-2012 at 03:07 PM.. Reason: code tags
# 2  
Old 08-19-2012
Looks like you've got two problems in your script - obviously both with the linewhile [ $j -le ${problem_arr[@]} ]
First: ${problem_arr[@]} will yield ALL array members and the shell should complain when trying to compare. I guess you want it to read ${#problem_arr[@]} which yields the count of members in array.
Second: You correctly start the loop at j=0, but you run it one too far up - ${problem_arr[${#problem_arr[@]}]} will be empty and thus sed complains. make the comparison [ $j -lt ... ] and it will fly.
# 3  
Old 08-19-2012
sed: -e expression #1, char 0: no previous regular expression
Thanks RudiC for your response.
I verified the points you have highlighted, and there is no problem with array size & initialization.Anyways, i can assign array size to another variable and use it in while condition.
I'd want you just focus on sed part of it, as it throws error there.

Hope you got me right!

Regards,
Indu
# 4  
Old 08-19-2012
Quote:
Originally Posted by InduInduIndu
I'd want you just focus on sed part of it, as it throws error there.
That sed error means that the first use of a regular expression is empty. That's happenning because your variable expansion yields nothing.

If you modified your script according to RudiC's suggestions, post it so we can see the current version.

Also, you may find it instructive to insert set -x before the while-loop.

Regards,
Alister
# 5  
Old 08-19-2012
You haven't said what shell you're using, but at least with bash and ksh, setting an array with:
Code:
array=(a b c)

sets ${array[1]} to a, ${array[2]} to b, and ${array[3]} to c. It doesn't set ${array[0]} so your first call to sed
Code:
sed  -n "/${problem_arr[0]}/,/${problem_arr[1]}/p" problemid.txt

expands to:
Code:
sed  -n "//,/PRS111/p" problemid.txt

and since your first RE is //, it is trying to search for the previously used RE (but there is no previously used RE in this invocation of sed.
# 6  
Old 08-20-2012
Quote:
Originally Posted by Don Cragun
You haven't said what shell you're using, but at least with bash and ksh, setting an array with:
Code:
array=(a b c)

sets ${array[1]} to a, ${array[2]} to b, and ${array[3]} to c. It doesn't set ${array[0]} so your first call to sed
Code:
sed  -n "/${problem_arr[0]}/,/${problem_arr[1]}/p" problemid.txt

expands to:
Code:
sed  -n "//,/PRS111/p" problemid.txt

and since your first RE is //, it is trying to search for the previously used RE (but there is no previously used RE in this invocation of sed.
Sorry, disagree. From the bash man page:
Quote:
Arrays are assigned to using compound assignments of the form name=(value1 ... valuen), where each value is of the form [sub- script]=string. Indexed array assignments do not require the bracket and subscript. When assigning to indexed arrays, if the optional brackets and subscript are supplied, that index is assigned to; other- wise the index of the element assigned is the last index assigned to by the statement plus one. Indexing starts at zero.
Running InduInduIndu's code snippet - using ${#...} to correctly get the member count - fails in the fourth loop,
Code:
++ k=4
++ sed -n //,//p problemid.txt
sed: -e expression #1, char 0: no previous regular expression

indicating that he should limit his loop count to 0 - 2. The first loop works fine:
Code:
++ sed -n /PRS111/,/PRS213/p problemid.txt

# 7  
Old 08-20-2012
Quote:
Originally Posted by RudiC
Sorry, disagree. From the bash man page:
Running InduInduIndu's code snippet - using ${#...} to correctly get the member count - fails in the fourth loop,
Code:
++ k=4
++ sed -n //,//p problemid.txt
sed: -e expression #1, char 0: no previous regular expression

indicating that he should limit his loop count to 0 - 2. The first loop works fine:
Code:
++ sed -n /PRS111/,/PRS213/p problemid.txt

Ouch. Yes, you're right. Both bash and ksh use the same notation for setting an indexed array, but bash uses 0 based indexing and ksh uses 1 based indexing. (I guess you can tell that I do most of my shell scripting with ksh. Smilie)
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