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Perl to change value based on set of rules

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# 1  
Old 08-04-2017
Perl to change value based on set of rules
In the perl there is a default rule that sets f[55] to VUS, and then a seris of rules that will change f[55] based on the result that is
obtained from the rule. The code below is a rule that is supposed to be applicable to lines 2-4 because this rule just looks at the digit in f[8]. So in line 2 f[8] is 27
and that value is greater than 10, so f[55] would be Likely Benign. Since the symbol before the digit could be either a > or + or - in the regex I use
\D to look for any non-digit before the number.
The else portion of the rule is supposed to be applicable to lines 1 and 5 as it uses the regex to parse out the digit after the - ot + or *in the string
that begins with NM_ in [ICODE] in f[8]. I am currently only getting the second line's f[55] value to be correct and I am not sure what I am doing incorrect. I have tried
changing the regex but not to the correct one (maybe there is something else I am missing). Thank you Smilie.

file
Code:
R_Index Chr Start End Ref Alt Func.refGene Gene.refGene GeneDetail.refGene Inheritence ExonicFunc.refGene AAChange.refGene avsnp147 PopFreqMax 1000G_ALL 1000G_AFR 1000G_AMR 1000G_EAS 1000G_EUR 1000G_SAS ExAC_ALL ExAC_AFR ExAC_AMR ExAC_EAS ExAC_FIN ExAC_NFE ExAC_OTH ExAC_SAS ESP6500siv2_ALL ESP6500siv2_AA ESP6500siv2_EA CG46 SIFT_score SIFT_pred Polyphen2_HDIV_score Polyphen2_HDIV_pred Polyphen2_HVAR_score Polyphen2_HVAR_pred LRT_score LRT_pred MutationTaster_score MutationTaster_pred MutationAssessor_score MutationAssessor_pred dpsi_max_tissue dpsi_zscore CLINSIG CLNDBN CLNACC CLNDSDB CLNDSDBID Quality Reads Zygosity Score Classification HGMD Sanger
28 chr2 149246946 149246946 T C splicing MBD5 NM_018328:exon12:c.3055-9T>C . . . rs370173652 0.0043 0.0008 0.003 . . . . 0.0003 0.003 0.0004 . . 0.0001 . . 0.0015 0.0043 . . . . . . . . . . . . . . -2.3896 -2.011 other|Benign|Uncertain significance "not_specified|Mental_retardation,_autosomal_dominant_1|Intellectual_Disability,_Dominant" RCV000188062.2|RCV000230037.1|RCV000392347.1 MedGen|Gene:MedGen:OMIM:Orphanet|MedGen CN169374|100820633:C1969562:156200:ORPHA228402|CN239282 GOOD 174 het 9 . . .
211 chr15 68522107 68522107 C G upstream CLN6 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . GOOD 5 het . . . .
212 chr15 68522115 68522115 A G upstream CLN6 35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . GOOD 6 het . . . .
43 chr2 166930214 166930214 T A splicing SCN1A >50 . . . rs566839 1 0.99 0.95 0.99 1 1 1 . . . . . . . . . . . 1 . . . . . . . . . . . . 1.4402 1.752 . . . . . GOOD 108 hom 31 . . .
60 chr3 11078886 11078886 C A UTR3 SLC6A1 NM_003042:c.*234C>A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . GOOD 8 het . Likely Benign n .

perl
Code:
     if ($FuncrefGene !~ /exonic/i && $GeneDetailrefGene=~/(\D\d+)/) {   # capture the digits after any non-digit into $1
                if ($1 > 11) {   # 
                      $1 //= 0;  # Give it a value of zero if no numeric value was found.
                        $classification = 'Likely Benign';  # Reclassify intronic variants (with distance only) based on distance to exon > 10 to Likely Benign
         }
     }
   else {
                 if ($FuncrefGene !~ /exonic/i) {
                    my ($transcript) = ($GeneDetailrefGene) =~ /(?:\.\d+[+*-])(\d+)/;   # Get a numeric value if exists using (.) and (+/-) and capture digits into $transcript.
                             $transcript //= 0;  # Give it a value of zero if no numeric value was found.
                                $classification = 'Likely Benign' if $transcript > 11; # Reclassify intronic variants (following c. nomenclature) to Likely Benign if distance greater than 10
                 }
           }

desired output in f[55]
Code:
VUS
Likely Benign
Likely Benign
Likely Benign
Likely Benign
Likely Benign

current output in f[55]
Code:
Likely Benign
Likely Benign
VUS
VUS
VUS
VUS

# 2  
Old 08-08-2017
Quote:
Originally Posted by cmccabe
...
...
The code below is a rule that is supposed to be applicable to lines 2-4 because this rule just looks at the digit in f[8].
...
...
That's wrong!
The regex looks for "a non-digit followed by one or more digits" ("\D\d+").
Line numbers 2 and 3 do not have that, so it will not match.
Line number 4 has that, so it will match.


Quote:
Originally Posted by cmccabe
...
...
So in line 2 f[8] is 27
and that value is greater than 10, so f[55] would be Likely Benign.
...
...
That's wrong again.
Line # 2 will never match "\D\d+" so 27 will never be extracted and hence never compared to anything.

Quote:
Originally Posted by cmccabe
...
...
Since the symbol before the digit could be either a > or + or - in the regex I use
\D to look for any non-digit before the number.
...
...
Wrong again.
There are cases where a symbol does not exist in the first place.
For example, lines 2 and 3 do not have the symbol at all.
You did not do anything for those cases hence those lines fail to match your regex.
Line 5 does have the symbol, so it matches your regex.


Quote:
Originally Posted by cmccabe
...
...
The else portion of the rule is supposed to be applicable to lines 1 and 5 as it uses the regex to parse out the digit after the - ot + or *in the string
that begins with NM_ in [ICODE] in f[8].
...
...
Yes, but before the control goes to the "else" portion, it will go to the "if" portion.
And the "if" portion will match your lines 1 and 5 because both of them have "a non-digit followed by one or more digit" ("\D\d+") in their f[8] values.

So the "else" portion will not even get a chance to execute for lines 1 and 5.

Here's some diagnostic output for your data file:
Code:
$ 
$ cat -n process_files.pl
     1    #!perl
     2    
     3    $file = $ARGV[0];
     4    open(FH, '<', $file) or die "Can't open $file: $!";
     5    while (<FH>) {
     6        next if $. == 1;
     7        chomp;
     8        @x = split /\s+/;
     9        $FuncrefGene = $x[6];
    10        $GeneDetailrefGene = $x[8];
    11        printf("Row no. = [%d]\n", $.);
    12        printf("FuncrefGene = [%s]\n", $FuncrefGene);
    13        printf("GeneDetailrefGene = [%s]\n", $GeneDetailrefGene);
    14        if ($FuncrefGene !~ /exonic/i && $GeneDetailrefGene=~/(\D\d+)/) {
    15            printf("In IF branch:\n");
    16            printf("==> \$1 = [%s]\n", $1);
    17            if ($1 > 11) {
    18                printf("==> Inside \$1 > 11\n");
    19                $1 //= 0;
    20                $classification = 'Likely Benign';
    21            }
    22            printf("==> Last line in IF branch: classification = [%s]\n", $classification);
    23        } else {
    24            printf("In ELSE branch:\n");
    25            if ($FuncrefGene !~ /exonic/i) {
    26                my ($transcript) = ($GeneDetailrefGene) =~ /(?:\.\d+[+*-])(\d+)/;
    27                printf("==> transcript = [%s]\n", $transcript);
    28                $transcript //= 0;
    29                $classification = 'Likely Benign' if $transcript > 11;
    30            }
    31            printf("==> Last line in ELSE branch: classification = [%s]\n", $classification);
    32        }
    33        printf("%s\n\n", "="x60);
    34    }
    35    close(FH) or die "Can't close $file: $!";
    36    
$ 
$ perl process_files.pl data.txt
Row no. = [2]
FuncrefGene = [splicing]
GeneDetailrefGene = [NM_018328:exon12:c.3055-9T>C]
In IF branch:
==> $1 = [_018328]
==> Last line in IF branch: classification = []
============================================================

Row no. = [3]
FuncrefGene = [upstream]
GeneDetailrefGene = [27]
In ELSE branch:
==> transcript = []
==> Last line in ELSE branch: classification = []
============================================================

Row no. = [4]
FuncrefGene = [upstream]
GeneDetailrefGene = [35]
In ELSE branch:
==> transcript = []
==> Last line in ELSE branch: classification = []
============================================================

Row no. = [5]
FuncrefGene = [splicing]
GeneDetailrefGene = [>50]
In IF branch:
==> $1 = [>50]
==> Last line in IF branch: classification = []
============================================================

Row no. = [6]
FuncrefGene = [UTR3]
GeneDetailrefGene = [NM_003042:c.*234C>A]
In IF branch:
==> $1 = [_003042]
==> Last line in IF branch: classification = []
============================================================

$ 
$

This User Gave Thanks to durden_tyler For This Post:
cmccabe
# 3  
Old 08-29-2017
Thank you for the diagnostics, they help, is it more or less I am trying to capture to many conditions with the regex? What would you recommend? Thank you Smilie.
# 4  
Old 08-30-2017
Quote:
Originally Posted by cmccabe
...
...
is it more or less I am trying to capture to many conditions with the regex?
What would you recommend?
...
...
Yes, from your other Perl related posts, I do get the impression that you are trying to use the regexes for too many things. That should be avoided.
However, for this particular piece of code, I think, you may want to deepen your understanding of regexes.

You have two types of data in F[8] column.

Type 1:
Code:
27
35
>50

and

Type 2:
Code:
NM_018328:exon12:c.3055-9T>C
NM_003042:c.*234C>A

So use regular expressions that work specifically with each type of data.
Your regex "\D\d+" is meant for Type 1, but it will actually match Type 2 as well.
Why?
Because "\D" means "non-digit character" and so it matches the "_" after "NM".
And then that is followed by "\d+" - "one or more digits". That's why the regex doesn't work the way you want.

Here's a demonstration:
Code:
$ perl -le '$x = "NM_018328:exon12:c.3055-9T>C"; if ($x =~ /(\D)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = _
\d+ or $2 = 018328

And for line # 5:
Code:
$ perl -le '$x = "NM_003042:c.*234C>A"; if ($x =~ /(\D)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = _
\d+ or $2 = 003042

As you can see, the regex meant for Type 2 data is working on Type 1 data as well.

So, determine what exactly is there in Type 1 and Type 2 data that differentiates them? Here are a few observations:

(1) Type 1 has "\d+" - "one or more digits"
(2) Type 1 may or may not have a non-digit at the front. This non-digit could be ">", "+" or "-". But nothing else.
(3) If there is a non-digit at the front, there is only one such non-digit. There cannot be more than one. So you need: "zero or one non-digit". For that, you could use "\D{0,1}" or "\D?".

Let's test this on the one-liner above.
First, notice that "\D\d+" will not work on both ">50" and "50".

Code:
$
$ perl -le '$x = ">50"; if ($x =~ /(\D)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = >
\d+ or $2 = 50
$
$ perl -le '$x = "50"; if ($x =~ /(\D)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$

That's because there is nothing before "50" in the second case, but the regex "\D\d+" demands exactly one non-digit at the beginning.
Since there was no non-digit, the match failed.

Now notice how "\D?\d+" works for both cases:

Code:
$
$ perl -le '$x = ">50"; if ($x =~ /(\D?)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = >
\d+ or $2 = 50
$
$
$ perl -le '$x = "50"; if ($x =~ /(\D?)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 =
\d+ or $2 = 50
$
$

Now, we make the regex more robust. We know that the "non-digit" character at the beginning is one of ">", "+" or "-".
So we use the bracket notation: "[>+-]"
This will match exactly one of the characters inside the brackets.
And since there can be 0 or 1 of such characters, we use "?" after the brackets: "[>+-]?"
In other words, we simply replaced "\D" by "[>+-]"
"\D" matches any non-digit character; it could match "#" or "A" or ">" etc.
"[>+-]" matches only one of the characters inside the brackets.

Testing again:

Code:
$
$ perl -le '$x = ">50"; if ($x =~ /([>+-]?)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = >
\d+ or $2 = 50
$
$ perl -le '$x = "50"; if ($x =~ /([>+-]?)(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 =
\d+ or $2 = 50
$
$

Finally, we only want the sequence of digits at the end.
So we can remove the parentheses around the non-digits at the beginning.
We can also put the "beginning of string anchor", which is "^" to specify that the non-digits are at the beginning of the string.
The updated regex is "^[>+-]?(\d+)"

Testing again:

Code:
 $
$ perl -le '$x = ">50"; if ($x =~ /^[>+-]?(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = 50
\d+ or $2 =
$
$ perl -le '$x = "50"; if ($x =~ /^[>+-]?(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = 50
\d+ or $2 =
$
$

So that takes care of Type 1 data.

Now for Type 2 data.
Your regex "/(?:\.\d+[+*-])(\d+)/" looks for the following:

(1) A single dot character "." followed by
(2) One or more digits "\d+" followed by
(3) Exactly one of the characters "+", "*", "-" followed by
(4) One or more digits "\d+"

It matches (1), (2), (3) together but does not "group" them into $1 (due to "?:" at the beginning).
It matches (4) and groups the sequence of digits into $1.

Now, if you look at your Line # 5:

Code:
NM_003042:c.*234C>A

the data has:
(1) Single dot character "."
(2) But no sequence of digits after the dot!! There is a "*" after the dot "."

Hence your regex fails.
Here's the demonstration:

Code:
$
$ # Matches Line # 1
$ perl -le '$x = "NM_018328:exon12:c.3055-9T>C"; if ($x =~ /(\.\d+[+*-])(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = .3055-
\d+ or $2 = 9
$
$ # But does not match Line # 5
$ perl -le '$x = "NM_003042:c.*234C>A"; if ($x =~ /(\.\d+[+*-])(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$

So what are the special characteristics of Type 2 data that distinguish it from Type 1 data? And how do we create the regex to match Type 2 data?

Firstly, if all Type 2 data start with "NM_", you could use that in your regex. So we have "NM_"

Now, it has a dot ">" at some point further on. So we get the regex "NM_.*\."
Here ".*" passes through "maximum number of characters till it reaches the right-most dot (.) character". It's a greedy search.

The dot character may or may not have a sequence of digits after it. (Line 1 has, Line 5 does not have.) "\d*" matches "zero or more digits" - "more" means "1 or more", so "zero or 1 or more than 1 digits".
So, we get: "NM_.*\.\d*"

After that, we definitely have one of the following characters "+", "*", "-".
So we use "[+*-]" for that. The regex now becomes "NM_.*\.\d*[+*-]"

Finally, that is followed by a sequence of digits that we want to capture.
Sequence of digits is "\d+". So the final regex is: "NM_.*\.\d*[+*-](\d+)"

Let's test this on Line 1 and Line 5 data:

Code:
$
$ # Line 1
$ perl -le '$x = "NM_018328:exon12:c.3055-9T>C"; if ($x =~ /NM_.*\.\d*[+*-](\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = 9
\d+ or $2 =
$
$ # Line 5
$ perl -le '$x = "NM_003042:c.*234C>A"; if ($x =~ /NM_.*\.\d*[+*-](\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
It matches!
\D  or $1 = 234
\d+ or $2 =
$
$

Because of the "NM_" at the beginning of the regex, we are guaranteed that it will not match Type 1 data.
But let's confirm that that is really the case:

Code:
$
$ # Line 2. This is Type 1 data. Regex is for Type 2. Must not match.
$ perl -le '$x = "27"; if ($x =~ /NM_.*\.\d*[+*-](\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$ # Line 3. This is Type 1 data. Regex is for Type 2. Must not match.
$ perl -le '$x = "35"; if ($x =~ /NM_.*\.\d*[+*-](\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$ # Line 4. This is Type 1 data. Regex is for Type 2. Must not match.
$ perl -le '$x = ">50"; if ($x =~ /NM_.*\.\d*[+*-](\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$ # Other Type 1 data. Regex is for Type 2. Must not match.
$ perl -le '$x = "+50"; if ($x =~ /NM_.*\.\d*[+*-](\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$

Let's also confirm that the regex for Type 1 data does not match Type 2 data!

Code:
 $
$
$ perl -le '$x = "NM_018328:exon12:c.3055-9T>C"; if ($x =~ /^[>+-]?(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$
$ perl -le '$x = "NM_003042:c.*234C>A"; if ($x =~ /^[>+-]?(\d+)/){printf("It matches!\n\\D  or \$1 = %s\n\\d+ or \$2 = %s\n",$1,$2)} else {print "Does not match!"}'
Does not match!
$
$

Hope that helps.
If you are unable to incorporate the regexes in your script, do post the problem here.
This User Gave Thanks to durden_tyler For This Post:
cmccabe
# 5  
Old 09-11-2017
I think the below will capture lines 2-6, but not line 1 (looks like 018328) is being captured by the regex. Is the syntax correct or is there a better way? Thank you Smilie.

perl
Code:
    if ($FuncrefGene !~ /exonic/i && $GeneDetailrefGene=~/\D(\d+)/) {   # capture the digits into $1
              if ($1 > 11) {   # 
                      $1 ||= 0;  # Give it a value of zero if no numeric value was found.
                        $classification = 'Likely Benign';  # Reclassify intronic variants (with distance only) based on distance to exon > 10 to Likely Benign
         }
     }
    else {
              if ($FuncrefGene !~ /exonic/i && $GeneDetailrefGene=~/(\D\d+)/) {   # capture the digits after any non-digit into $1
                 if ($1 > 11) {   # 
                      $1 ||= 0;  # Give it a value of zero if no numeric value was found.
                        $classification = 'Likely Benign';  # Reclassify intronic variants (with distance only) based on distance to exon > 10 to Likely Benign
         }
     }
    else {
              if ($FuncrefGene !~ /exonic/i) {
                 my ($transcript) = ($GeneDetailrefGene) =~ /(?:\.\d+[+*-])/;   # Get a numeric value if exists using (.) and (+/-) and capture digits into $transcript.
                           $transcript ||= 0;  # Give it a value of zero if no numeric value was found.
                             $classification = 'Likely Benign' if $transcript > 11; # Reclassify intronic variants (following c. nomenclature) to Likely Benign if distance greater than 10
         }
     }

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Need to change a set of lines between two given Pattrens

Hi All I have a Small Requiement I wanted to replace all the Follwing lines as follows Input:: file1 EVALUATE WS-TEMP-ATTR(15:1) WHEN 'D' MOVE DFHDARK TO WS-ATTR-COLOR WHEN OTHER MOVE DFHDFT ... (9 Replies)
Discussion started by: pbsrinivas
9 Replies

10. Shell Programming and Scripting

how to change "set" values in perl, windows...

i am using perl in win2000advanced server... --------------------------- perl -version: --------------------------- This is perl, v5.6.1 built for MSWin32-x86-multi-thread (with 1 registered patch, see perl -V for more detail) Copyright 1987-2001, Larry Wall Binary build 638 provided by... (1 Reply)
Discussion started by: sekar sundaram
1 Replies
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