Hi I am new to perl and trying to write a regex to find a match for specific number range and strings in a line inside the file, i need to find the lines("Document has 15 rows and 2 columns").
I know I am missing something, but the code I have tried so far is :
if(/^[a-zA-Z\d]+(has\s[1-9][0-9]$)\srows.*columns/)
{
print "$_\n";
}
It would be really helpful if anyone let me know what is wrong here!
The other answers here are good, but to explain what was wrong with the regex you used:
if(/^[a-zA-Z\d]+(has\s[1-9][0-9]$)\srows.*columns/)
First problem: the expression does not specify any whitespace between the beginning of the string and the word has, so there is no way for this pattern to match the space in Document has...
Second problem: the $ character in a regular expression means "match if the line ends here." It's almost always a mistake to use the $ anchor in the middle of a regex; the only way this would match would be in a multiline string like
Documenthas 15
rows and 7 columns
Making those two changes to your expression makes it work:
if(/^[a-zA-Z\d]+\s(has\s[1-9][0-9])\srows.*columns/)
{
print "$_\n";
}
Easy regex to use:
/Document has [0-9]+ row(s?) and [0-9]+ column(s?)/
If the s is only used when there is more than one row/column
I'm assuming you want to capture the numbers.
if ( /^Document has (\d+) rows and (\d+) columns/ ) {
my $rows = $1;
my $cols = $2;
my $line = "Document has 15 rows and 2 columns"
if ($line =~ /^Document has (\d+) rows? and (\d+) columns?/)
{
print "rows = $1\n";
print "cols = $2\n";
}
If you just want the number of rows, use this:
if (/(\d+)\s+rows/) {
print "$1\n";
}
If you want rows and columns (and they are always in that order), use:
if (/(\d+)\s+rows\s+and\s+(\d+)\s+columns/) {
print "$1 rows and $2 columns\n";
}
If you think it is necessary, you can be more restrictive if you need to: restricting the number of digits, forcing non-leading zeros, etc.
Also, I assume you are either using "-n" on the command line or have a loop around this.
Related
I am using Perl to do some prototyping.
I need an expression to replace e by [ee] if the string is exactly 2 chars and finishes by "e".
le -> l [ee]
me -> m [ee]
elle -> elle : no change
I cannot test the length of the string, I need one expression to do the whole job.
I tried:
`s/(?=^.{0,2}\z).*e\z%/[ee]/g` but this is replacing the whole string
`s/^[c|d|j|l|m|n|s|t]e$/[ee]/g` same result (I listed the possible letters that could precede my "e")
`^(?<=[c|d|j|l|m|n|s|t])e$/[ee]/g` but I have no match, not sure I can use ^ on a positive look behind
EDIT
Guys you're amazing, hours of search on the web and here I get answers minutes after I posted.
I tried all your solutions and they are working perfectly directly in my script, i.e. this one:
my $test2="le";
$test2=~ s/^(\S)e$/\1\[ee\]/g;
print "test2:".$test2."\n";
-> test2:l[ee]
But I am loading these regex from a text file (using Perl for proto, the idea is to reuse it with any language implementing regex):
In the text file I store for example (I used % to split the line between match and replace):
^(\S)e$% \1\[ee\]
and then I parse and apply all regex like that:
my $test="le";
while (my $row = <$fh>) {
chomp $row;
if( $row =~ /%/){
my #reg = split /%/, $row;
#if no replacement, put empty string
if($#reg == 0){
push(#reg,"");
}
print "reg found, reg:".$reg[0].", replace:".$reg[1]."\n";
push #regs, [ #reg ];
}
}
print "orgine:".$test."\n";
for my $i (0 .. $#regs){
my $p=$regs[$i][0];
my $r=$regs[$i][1];
$test=~ s/$p/$r/g;
}
print "final:".$test."\n";
This technique is working well with my other regex, but not yet when I have a $1 or \1 in the replace... here is what I am obtaining:
final:\1\ee\
PS: you answered to initial question, should I open another post ?
Something like s/(?i)^([a-z])e$/$1[ee]/
Why aren't you using a capture group to do the replacement?
`s/^([c|d|j|l|m|n|s|t])e$/\1 [ee]/g`
If those are the characters you need and if it is indeed one word to a line with no whitespace before it or after it, then this will work.
Here's another option depending on what you are looking for. It will match a two character string consisting of one a-z character followed by one 'e' on its own line with possible whitespace before or after. It will replace this will the single a-z character followed by ' [ee]'
`s/^\s*([a-z])e\s*$/\1 [ee]/`
^(\S)e$
Try this.Replace by $1 [ee].See demo.
https://regex101.com/r/hR7tH4/28
I'd do something like this
$word =~ s/^(\w{1})(e)$/$1$2e/;
You can use following regex which match 2 character and then you can replace it with $1\[$2$2\]:
^([a-zA-Z])([a-zA-Z])$
Demo :
$my_string =~ s/^([a-zA-Z])([a-zA-Z])$/$1[$2$2]/;
See demo https://regex101.com/r/iD9oN4/1
I have this little script:
my #list = ('R3_05_foo.txt','T3_12_foo_bar.txt','01.txt');
foreach (#list) {
s/(\d{2}).*\.txt$/$1.txt/;
s/^0+//;
print $_ . "\n";
}
The expected output would be
5.txt
12.txt
1.txt
But instead, I get
R3_05.txt
T3_12.txt
1.txt
The last one is fine, but I cannot fathom why the regex gives me the string start for $1 on this case.
Try this pattern
foreach (#list) {
s/^.*?_?(?|0(\d)|(\d{2})).*\.txt$/$1.txt/;
print $_ . "\n";
}
Explanations:
I use here the branch reset feature (i.e. (?|...()...|...()...)) that allows to put several capturing groups in a single reference ( $1 here ). So, you avoid using a second replacement to trim a zero from the left of the capture.
To remove all from the begining before the number, I use :
.*? # all characters zero or more times
# ( ? -> make the * quantifier lazy to match as less as possible)
_? # an optional underscore
Note that you can ensure that you have only 2 digits adding a lookahead to check if there is not a digit that follows:
s/^.*?_?(?|0(\d)|(\d{2}))(?!\d).*\.txt$/$1.txt/;
(?!\d) means not followed by a digit.
The problem here is that your substitution regex does not cover the whole string, so only part of the string is substituted. But you are using a rather complex solution for a simple problem.
It seems that what you want is to read two digits from the string, and then add .txt to the end of it. So why not just do that?
my #list = ('R3_05_foo.txt','T3_12_foo_bar.txt','01.txt');
for (#list) {
if (/(\d{2})/) {
$_ = "$1.txt";
}
}
To overcome the leading zero effect, you can force a conversion to a number by adding zero to it:
$_ = 0+$1 . ".txt";
I would modify your regular expression. Try using this code:
my #list = ('R3_05_foo.txt','T3_12_foo_bar.txt','01.txt');
foreach (#list) {
s/.*(\d{2}).*\.txt$/$1.txt/;
s/^0+//;
print $_ . "\n";
}
The problem is that the first part in your s/// matches, what you think it does, but that the second part isn't replacing what you think it should. s/// will only replace what was previously matched. Thus to replace something like T3_ you will have to match that too.
s/.*(\d{2}).*\.txt$/$1.txt/;
My string is $tables="newdb1.table1:100,db2.table2:90,db1.table1:90". My search string is db1.table1 and my aim is to extract the value after : (i.e 90 in this case).
I am using:
if ($tables =~ /db1.table1:(\d+)/) { print $1; }
but the problem is it is matching newdb1.table1:100 and printing 100.
Can you please give my a regular expression to match a string which either starts with a newline or has comma before it.
Use word boundaries:
if ($tables =~ /\bdb1.table1:(\d+)/) { print $1; }
here __^^
if ($tables =~ /(^|,)db1.table1:(\d+)/) { print $2; }
To answer your exact question, that is to match just after the start of the string or a comma, you want a positive look-behind assertion. You may be tempted to write a pattern of
/(?<=^|,)db1\.table1:(\d+)/
but that may fail with an error of
Variable length lookbehind not implemented in regex m/(?<=^|,)db1\.table1:(\d+)/ ...
So hold the regex engine’s hand a bit by making the alternatives equal in length—tricky to do in the general case but workable here.
/(?<=^d|,)db1\.table1:(\d+)/
While we are locking it down, let’s be sure to bracket the end with a look-ahead assertion.
while ($tables =~ /(?<=^d|,)db1\.table1:(\d+)(?=,|$)/g) {
print "[$1]\n";
}
Output:
[90]
You could also use \b for a regex word boundary, which has the same output.
while ($tables =~ /\bdb1\.table1:(\d+)(?=,|$)/g) {
print "[$1]\n";
}
For the most natural solution, follow the rule of thumb proposed by Randal Schwartz, author of Learning Perl. Use capturing when you know what you want to keep and split when you know what you want to throw away. In your case you have a mixture: you want to discard the comma separators, and you want to keep the digits after the colon for a certain table. Write that as
for (split /\s*,\s*/, $tables) { # / to fix Stack Overflow highlighting
if (my($value) = /^db1\.table1:(\d+)$/) {
print "[$value]\n";
}
}
Output:
[90]
I want to use a regular expression to check a string to make sure 4 and 5 are in order. I thought I could do this by doing
'$string =~ m/.45./'
I think I am going wrong somewhere. I am very new to Perl. I would honestly like to put it in an array and search through it and find out that way, but I'm assuming there is a much easier way to do it with regex.
print "input please:\n";
$input = <STDIN>;
chop($input);
if ($input =~ m/45/ and $input =~ m/5./) {
print "works";
}
else {
print "nata";
}
EDIT: Added Info
I just want 4 and 5 in order, but if 5 comes before at all say 322195458900023 is the number then where 545 is a problem 5 always have to come right after 4.
Assuming you want to match any string that contains two digits where the first digit is smaller than the second:
There is an obscure feature called "postponed regular expressions". We can include code inside a regular expression with
(??{CODE})
and the value of that code is interpolated into the regex.
The special verb (*FAIL) makes sure that the match fails (in fact only the current branch). We can combine this into following one-liner:
perl -ne'print /(\d)(\d)(??{$1<$2 ? "" : "(*FAIL)"})/ ? "yes\n" :"no\n"'
It prints yes when the current line contains two digits where the first digit is smaller than the second digit, and no when this is not the case.
The regex explained:
m{
(\d) # match a number, save it in $1
(\d) # match another number, save it in $2
(??{ # start postponed regex
$1 < $2 # if $1 is smaller than $2
? "" # then return the empty string (i.e. succeed)
: "(*FAIL)" # else return the *FAIL verb
}) # close postponed regex
}x; # /x modifier so I could use spaces and comments
However, this is a bit advanced and masochistic; using an array is (1) far easier to understand, and (2) probably better anyway. But it is still possible using only regexes.
Edit
Here is a way to make sure that no 5 is followed by a 4:
/^(?:[^5]+|5(?=[^4]|$))*$/
This reads as: The string is composed from any number (zero or more) characters that are not a five, or a five that is followed by either a character that is not a four or the five is the end of the string.
This regex is also a possibility:
/^(?:[^45]+|45)*$/
it allows any characters in the string that are not 4 or 5, or the sequence 45. I.e., there are no single 4s or 5s allowed.
You just need to match all 5 and search fails, where preceded is not 4:
if( $str =~ /(?<!4)5/ ) {
#Fail
}
I am looking to do a 2-step regular expression look-up in Perl, I have text that looks like this:
here is some text 9337 more text AA 2214 and some 1190 more BB stuff 8790 words
I also have a hash with the following values:
%my_hash = ( 9337 => 'AA', 2214 => 'BB', 8790 => 'CC' );
Here's what I need to do:
Find a number
Look up the text code for the number using my_hash
Check if the text code appears within 50 characters of the identified number, and if true print the result
So the output I'm looking for is:
Found 9337, matches 'AA'
Found 2214, matches 'BB'
Found 1190, no matches
Found 8790, no matches
Here's what I have so far:
while ( $text =~ /(\d+)(.{1,50})/g ) {
$num = $1;
$text_after_num = $2;
$search_for = $my_hash{$num};
if ( $text_after_num =~ /($search_for)/ ) {
print "Found $num, matches $search_for\n";
}
else {
print "Found $num, no matches\n";
}
This sort of works, except that the only correct match is 9337; the code doesn't match 2214. I think the reason is that the regular expression match on 9337 is including 50 characters after the number for the second-step match, and then when the regex engine starts again it is starting from a point after the 2214. Is there an easy way to fix this? I think the \G modifier can help me here, but I don't quite see how.
Any suggestions or help would be great.
You have a problem with greediness. The 1,50 will consume as much as it can. Your regex should be /(\d+)(.+?)(?=($|\d))/
To explain, the question mark will make the multiple match non-greedy (it will stop as soon as the next pattern is matched - the next pattern gets precedence). The ?= is a lookahead operator to say "check if the next element is a digit. If so, match but do not consume." This allows the first digit to get picked up by the beginning of the regex and be put into the next matched pattern.
[EDIT]
I added an optional end value to the lookahead so that it wouldn't die on the last match.
Just use :
/\b\d+\b/g
Why match everything if you don't need to? You should use other functions to determine where the number is :
/(?=9337.{1,50}AA)/
This will fail if AA is further than 50 chars away from the end of 9337. Of course you will have to interpolate your variables to match your hashe's keys and values. This was just an example for your first key/value pair.