I am trying to create a regular expression to determine if a string contains a number for an SQL statement. If the value is numeric, then I want to add 1 to it. If the number is not numeric, I want to return a 1. More or less. Here is the SQL:
SELECT
field,
CASE
WHEN regexp_like(field, '^ *\d*\.?\d* *$') THEN dec(field) + 1
ELSE 1
END nextnumber
FROM mytable
This actually works, and returns something like this:
INVALID 1
00000 1
00001E 1
00379 380
00013 14
99904 99905
But to push the envelope of understanding, what if I wanted to cover negative numbers, or those with a positive sign. The sign would have to immediately precede or follow the number, but not both, and I would not want to allow white space between the sign and the number.
I came up with a conditional expression with a capture group to capture the sign on the front of the number to determine if a sign was allowed on the end, but it seems a little awkward to handle given I don't really need a yes-pattern.
Here is the modified regex: ^ ([+-]?)*\d*\.?\d*(?(1) *|[+-]? *)$
This works at regex101.com, but in order for it to work I need to have something before the pipe, so I have to duplicate the next pattern in both the yes-pattern and the no-pattern.
All that background for this question: How can I avoid that duplication?
EDIT: DB2 for i uses International Components for Unicode to provide regular expression processing. It turns out that this library does not support conditionals like PRCE, so I changed the tags on this question. The answer given by Wiktor Stribiżew provides a working alternative to the conditional by using a negative lookahead.
You do not have to duplicate the end pattern, just move it outside the conditional:
^ *([+-])?\d*\.?\d*(?(1)|[+-]?) *$
See the regex demo. So, the yes-part is empty, and the no-part has an optional pattern.
You may also solve it with a mere negative lookahead:
^ *([+-](?!.*[-+]))?\d*\.?\d*[+-]? *$
See another regex demo. Here, ([+-](?!.*[-+]))? matches (optionally) a + or - that are not followed with any 0+ char followed with another + or -.
How to use matlab regexprep , for multiple expression and replacements?
file='http:xxx/sys/tags/Rel/total';
I want to replace 'sys' with sys1 and 'total' with 'total1'. For a single expression a replacement it works like this:
strrep(file,'sys', 'sys1')
and want to have like
strrep(file,'sys','sys1','total','total1') .
I know this doesn't work for strrep
Why not just issue the command twice?
file = 'http:xxx/sys/tags/Rel/total';
file = strrep(file,'sys','sys1')
strrep(file,'total','total1')
To solve it you need substitute functionality with regex, try to find in matlab's regexes something similar to this in php:
$string = 'http:xxx/sys/tags/Rel/total';
preg_replace('/http:(.*?)\//', 'http:${1}1/', $string);
${1} means 1st match group, that is what in parenthesis, (.*?).
http:(.*?)\/ - match pattern
http:${1}1/ - replace pattern with second 1 as you wish to add (first 1 is a group number)
http:xxx/sys/tags/Rel/total - input string
The secret is that whatever is matched by (.*?) (whether xxx or yyyy or 1234) will be inserted instead of ${1} in replace pattern, and then replace instead of old stuff into the input string. Welcome to see more examples on substitute functionality in php.
As documented in the help page for regexprep, you can specify pairs of patterns and replacements like this:
file='http:xxx/sys/tags/Rel/total';
regexprep(file, {'sys' 'total'}, {'sys1' 'total1'})
ans =
http:xxx/sys1/tags/Rel/total1
It is even possible to use tokens, should you be able to define a match pattern for everything you want to replace:
regexprep(file, '/([st][yo][^/$]*)', '/$11')
ans =
http:xxx/sys1/tags/Rel/total1
However, care must be taken with the first approach under certain circumstances, because MATLAB replaces the pairs one after another. That is to say if, say, the first pattern matches a string and replaces it with something that is subsequently matched by a later pattern, then that will also be replaced by the later replacement, even though it might not have matched the later pattern in the original string.
Example:
regexprep('This\is{not}LaTeX.', {'\\' '([{}])'}, {'\\textbackslash{}' '\\$1'})
ans =
This\textbackslash\{\}is\{not\}LaTeX.
=> This\{}is{not}LaTeX.
and
regexprep('This\is{not}LaTeX.', {'([{}])' '\\'}, {'\\$1' '\\textbackslash{}'})
ans =
This\textbackslash{}is\textbackslash{}{not\textbackslash{}}LaTeX.
=> This\is\not\LaTeX.
Both results are unintended, and there seems to be no way around this with consecutive replacements instead of simultaneous ones.
I need a regex to match any number between 0 to 100 including decimal numbers example:
my expression should match 1,2,2.3 ,40,40.12 ,100,100.00 like this ..thanks in advance?
Assuming you have to allow for a leading sign, you are best off writing
if ( /(?<![-+.\d])([-+]?\d+(?:\.\d*)?(?![-+.\d])/ and $1 >= 0 and $1 <= 100 ) { .. }
But if you are forced into using a regex, then you need
if ( /(?<![-+.\d])(([-+]?(?:100|\d\d)(?:\.\d*)?(?![-+.\d])/ ) { .. }
These pattern may well be more complex than necessary because they allow for the number appearing anywhere in the string. If you are simply checking an entire string to see if it matches the criteria then it could be much shorter
This would work:
(100(\.0+))|([0-9]{1,2}(\.[0-9]+)?)
match either "100" (with optional dot plus one or more zeroes) or one or two digits, optionally followed by a dot and at least one digit.
EDITED!!!
This problem was much more difficult than I initially realized. With some amount of effort, I have produced a new regex that is without error. Enjoy.
/(?<!\d)(?<!\.)(100(?:(?!\.)|(?:\.0*+|\.))(?=\D)|[0-9]?[0-9](?:\.|\.[0-9]*+)?(?=[\D]))/
This pattern will capture in $1
So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ do?
What's pack+q,c*,, doing?
This is a variant on “Just another Perl hacker”, a Perl meme. As JAPHs go, this one is relatively tame.
The first thing you need to do is figure out how to parse the perl program. It lacks parentheses around function calls and uses the + and quote-like operators in interesting ways. The original program is this:
print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34
pack is a function, whereas print and map are list operators. Either way, a function or non-nullary operator name immediately followed by a plus sign can't be using + as a binary operator, so both + signs at the beginning are unary operators. This oddity is described in the manual.
If we add parentheses, use the block syntax for map, and add a bit of whitespace, we get:
print(+pack(+q,c*,,
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
The next tricky bit is that q here is the q quote-like operator. It's more commonly written with single quotes:
print(+pack(+'c*',
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
Remember that the unary plus is a no-op (apart from forcing a scalar context), so things should now be looking more familiar. This is a call to the pack function, with a format of c*, meaning “any number of characters, specified by their number in the current character set”. An alternate way to write this is
print(join("", map {chr($.+=$_)} (74, …, -34)))
The map function applies the supplied block to the elements of the argument list in order. For each element, $_ is set to the element value, and the result of the map call is the list of values returned by executing the block on the successive elements. A longer way to write this program would be
#list_accumulator = ();
for $n in (74, …, -34) {
$. += $n;
push #list_accumulator, chr($.)
}
print(join("", #list_accumulator))
The $. variable contains a running total of the numbers. The numbers are chosen so that the running total is the ASCII codes of the characters the author wants to print: 74=J, 74+43=117=u, 74+43-2=115=s, etc. They are negative or positive depending on whether each character is before or after the previous one in ASCII order.
For your next task, explain this JAPH (produced by EyesDrop).
''=~('(?{'.('-)#.)#_*([]#!#/)(#)#-#),#(##+#)'
^'][)#]`}`]()`#.#]#%[`}%[#`#!##%[').',"})')
Don't use any of this in production code.
The basic idea behind this is quite simple. You have an array containing the ASCII values of the characters. To make things a little bit more complicated you don't use absolute values, but relative ones except for the first one. So the idea is to add the specific value to the previous one, for example:
74 -> J
74 + 43 -> u
74 + 42 + (-2 ) -> s
Even though $. is a special variable in Perl it does not mean anything special in this case. It is just used to save the previous value and add the current element:
map($.+=$_, ARRAY)
Basically it means add the current list element ($_) to the variable $.. This will return a new array with the correct ASCII values for the new sentence.
The q function in Perl is used for single quoted, literal strings. E.g. you can use something like
q/Literal $1 String/
q!Another literal String!
q,Third literal string,
This means that pack+q,c*,, is basically pack 'c*', ARRAY. The c* modifier in pack interprets the value as characters. For example, it will use the value and interpret it as a character.
It basically boils down to this:
#!/usr/bin/perl
use strict;
use warnings;
my $prev_value = 0;
my #relative = (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34);
my #absolute = map($prev_value += $_, #relative);
print pack("c*", #absolute);
Is there any reason to use a scalar comma operator anywhere other than in a for loop?
Since the Perl scalar comma is a "port" of the C comma operator, these comments are probably apropos:
Once in a while, you find yourself in
a situation in which C expects a
single expression, but you have two
things you want to say. The most
common (and in fact the only common)
example is in a for loop, specifically
the first and third controlling
expressions. What if (for example) you
want to have a loop in which i counts
up from 0 to 10 at the same time that
j is counting down from 10 to 0?
So, your instinct that it's mainly useful in for loops is a good one, I think.
I occasionally use it in the conditional (sometimes erroneously called "the ternary") operator, if the code is easier to read than breaking it out into a real if/else:
my $blah = condition() ? do_this(), do_that() : do_the_other_thing();
It could also be used in some expression where the last result is important, such as in a grep expression, but in this case it's just the same as if a semicolon was used:
my #results = grep { setup(), condition() } #list;