i am trying to concatenate string in perl.
eg:
my $file = $table_name.".sql";
print $file;
I get output like:
Employee
.sql
(considering $table_name =Employee )
please suggest how to make sure the output comes in single lie without blanks.
Your $table_name variable contains a word 'Employee' as well as a new line character.
You can use chomp to remove the newline.
Add this before you concatenate your variables:
chomp $table_name;
The chomp() function will remove newline character from the end of a string.
check http://perlmeme.org/howtos/perlfunc/chomp_function.html
Related
I have a text file. I am trying to search for a comma followed by a date, like ",08/18/2014".
It looks like my code finds it, but it replaces it with a comma and a new line and removes everything after it.
if ($firstline =~ s|\,(\d{2}\/.*)|\,\n|g){
print "$firstline";
How do I add a new line between the comma and the date and not remove my text and date?
Thanks.
You can reference your capture groups in your replace expression. I also expanded the regular expression to match the date exactly:
if ($firstline =~ s|,(\d{2}-\d{2}-\d{4})|,\n$1|g) {
print "$firstline\n";
}
Using look ahead/behind:
s|(?<=,)(?=\d{2}\/)|\n|g
or
s|,\K(?=\d{2}\/)|\n|g # Probably faster, but requires 5.10+
Similar to Hunter's answer, without the extended regex, like so:
(Oh, and I don't think you need the backslash before the comma):
$firstline = "hello world ,10/20/1987, sounds great to me"; # testing...
if ($firstline =~ s|,(\d{2}/)|,\n$1|g){
print "$firstline";
}
output:
hello world ,
10/20/1987, sounds great to me
I am new to Perl Programming.
I have a CSV File with N fields in which Nth field is having Trailing Spaces for all the records. I want to remove all this Trailing Spaces.Please help me in this.
I have used this substitution in a loop. But it has given me empty file
s/\s+$//
Example File
123,ABCD,"AC,BD",21/12/2013
134,CDEF,"CD,BD,ED",23/11/2013
987,TGYH,"HY,-.FDDS",20/11/2013
Output
123,ABCD,"AC,BD",21/12/2013
134,CDEF,"CD,BD,ED",23/11/2013
987,TGYH,"HY,-.FDDS",20/11/2013
Please let me know if you need more details.
Thanks In Advance.
Your regex seems good. You could say:
perl -ple 's/\s+$//' filename
To save the changes in-place to the file, say:
perl -i -ple 's/\s+$//' filename
Steps to remove the leading and trailing spaces in the csv file:
Open the input file
OPEN FILE, "<$input_file";
Looping each and every row in the file to do trim of each row in the csv file.In the while loop use regex to trim the leading and trailing of the spaces for each field
while(my $row = <FILE>)
{
$row =~ 's/\s//g';
}
This will give you the results you are expecting
I am trying to split a string with "." but getting nothing in the array. File name is "Head-First-Java-2nd-edition.pdf" After splitting I want to extract extension, but don't know why it is giving blank array.
my #fileInfo = split(/./, $filename);
&logMsg("Array is: #fileInfo");
The split is giving an empty list because you are splitting on a wildcard .. Period is a meta character, and if you want to split on a literal period, you need to escape it
my #fileInfo = split(/\./, $filename);
Also, the syntax for calling a subroutine is NAME(LIST). Using the & prefix has a certain hidden feature, in that it circumvents prototypes. Read more in perldoc perlsub.
. in a regular expression means any character except \n. To split on a literal ., you need to escape it:
split /\./, $filename;
I have a directory on a Linux host with several property files which I want to edit by replacing hardcoded values with placeholder tags. My goal was to have a perl script which reads a delimited file that contains entries for each of the property files listing the hardcoded value, the placeholder value and the name of the file to edit.
For example, in file.prop I have these values set
<connection targetHostUrl="99.99.99.99"
targetHostPort="9999"
And I want to replace the values with tags as shown below
<connection targetHostUrl="TARGETHOST"
targetHostPort="PORT"
There will be several entries similar to this so I have to match on the unique combination of IP and PORT so I need a multiline match.
To do this I wrote the following script to take the input of the delimited filename, which is delimited with ||. I go get that file from the config directory and read in the values to get the hardcoded value, tag, and filename to edit. Then I read in that property file, do the substitution and then write it out again.
#!/usr/bin/perl
my $config = $ARGV[0];
chomp $config;
my $filename = '/config/' . $config;
my ($hard,$tagg,$prop);
open(DATAFILE, $filename) or die "Could not open DATAFILE $filename.";
while(<DATAFILE>)
{
chomp $_;
($hard,$tagg,$prop) = split('\|\|', $_);
$*=1;
open(INPUT,"</properties/$prop") or die "Could not open INPUT $prop.";
#input_array=<INPUT>;
close(INPUT);
$input_scalar=join("",#input_array);
$input_scalar =~ s/$hard/$tagg/;
open(OUTPUT,">/properties/$prop") or die "Could not open OUTPUT $prop.";
print(OUTPUT $input_scalar);
close(OUTPUT);
}
close DATAFILE;
Inside the config file I have the following entry
<connection targetHostUrl="99.99.999.99"(.|\n)*?targetHostPort="9999"||<connection targetHostUrl="TARGETHOST1"\n targetHostPort="PORT"||file.prop
My output is as shown below. It puts what I hoped to be a newline as a literal \n
<connection targetHostUrl="TARGETHOST"\n targetHostPort="PORT"
I can't find a way to get the \n taken as a newline. At first I thought, no problem, I'll just do a 2nd substitution like
perl -i -pe 's/\\n/\n/o' $prop
and although this works, for some reason it puts ^M characters at the end of every line except the one I did the replacement on. I don't want to do a 3rd replace to strip them out.
I've searched and found other ways of doing the multiline search/replace but they all interpret the \n literally.
Any suggestions?
My output is as shown below. It puts what I hoped to be a newline as a literal \n
Why would it insert a newline when the string doesn't contain one?
I can't find a way to get the \n taken as a newline.
There isn't any. If you want to substitute a newline, you need to provide a newline.
If you used a proper CSV parser like Text::CSV_XS, you could put a newline in your data file.
Otherwise, you'll have to write some code to handle the escape sequences you want your code to handle.
for some reason it puts ^M characters at the end of every line except the one I did the replacement on.
Quite the opposite. It removes it from the one line you did the replacement on.
That's home some programs represent a Carriage Return. You have a file with CR LF line ends. You could use dos2unix to convert it, or you could leave it as is because XML doesn't care.
how to shift the top element from array based on a regular expression using perl? Also these are datarecords, meaning I have the input record separator ($/) set to
$/='#';
for example, the following text file contains this data record.
#dddddddddd
ccccccccccc
eeeeeeeeeee
fffffffffff
I would like to remove the # sign and keep the text, for example:
dddddddddd
ccccccccccc
eeeeeeeeeee
fffffffffff
If you just want to manipulate a text file, a one-liner seems like the best solution. This will edit the file and keep a backup in "inputfile.txt.bak".
perl -pi.bak -we 's/^#//' inputfile.txt
Or you can do a shell redirection:
perl -wpe 's/^#//' inputfile.txt > outputfile.txt
These will try to alter all the lines in the file. If you just want the first line altered you need something different:
perl -wpe 's/^#// if ($. == 0);' inputfile.txt > outputfile.txt
Don't confuse shift with regex substitution.
shift will remove the first element from the array, not string.
A regex substitution can deal with removal of the leading '#' sigil.
The first element of the array would be $array[0].
If a regex substitution is applied to this first element, the '#' is removed:
my #array = ( '#dddddddddd', 'ccccccccccc', 'eeeeeeeeeee', 'fffffffffff' );
$array[0] =~ s/^#//;
print $array[0]; # 'dddddddddd'
This does not seem to be related to arrays. It appears you are just dealing with strings.
This removes a leading hash mark for the string $line:
$line =~ s/^\#//;