I am attempting to search through an array of strings (new_string) and check if it includes any 'operators'
where am I going wrong?
def example
operators = ["+", "-"]
string = "+ hi"
new_string = string.split(" ")
if new_string.include? Regexp.union(operators)
print "true"
else
print "false"
end
end
You can use any? instead, which takes a pattern:
pattern = Regexp.union(['+', '-']) #=> /\+|\-/
['foo', '+', 'bar'].any?(pattern) #=> true
But since you already have a string, you can skip the splitting and use match?:
'foo + bar'.match?(pattern) #=> true
You wish to determine if a string (string) contains at least one character in a given array of characters (operators). The fact that those characters are '+' and '-' is not relevant; the same methods would be used for any array of characters. There are many ways to do that. #Stefan gives one. Here are a few more. None of them mutate (modify) string.
string = "There is a + in this string"
operators = ["+", "-"]
The following is used in some calculations.
op_str = operators.join
#=> "+-"
#1
r = /[#{ op_str }]/
#=> /[+-]/
string.match?(r)
#=> true
[+-] is a character class. It asserts that the string matches any character in the class.
#2
string.delete(op_str).size < string.size
#=> true
See String#delete.
#3
string.tr(op_str, '').size < string.size
#=> true
See String#tr.
#4
string.count(op_str) > 0
#=> true
See String#count.
#5
(string.chars & operators).any?
#=> true
See Array#&.
In SML, is it possible for you to have multiple patterns in one case statement?
For example, I have 4 arithmetic operators express in string, "+", "-", "*", "/" and I want to print "PLUS MINUS" of it is "+" or "-" and "MULT DIV" if it is "*" or "/".
TL;DR: Is there somewhere I can simplify the following to use less cases?
case str of
"+" => print("PLUS MINUS")
| "-" => print("PLUS MINUS")
| "*" => print("MULT DIV")
| "/" => print("MULT DIV")
Given that you've tagged your question with the smlnj tag, then yes, SML/NJ supports this kind of patterns. They call it or-patterns and it looks like this:
case str of
("+" | "-") => print "PLUS MINUS"
| ("*" | "/") => print "MULT DIV"
Notice the parentheses.
The master branch of MLton supports it too, as part of their Successor ML effort, but you'll have to compile MLton yourself.
val str = "+"
val _ =
case str of
"+" | "-" => print "PLUS MINUS"
| "*" | "/" => print "MULT DIV"
Note that MLton does not require parantheses. Now compile it using this command (unlike SML/NJ, you have to enable this feature explicitly in MLton):
mlton -default-ann 'allowOrPats true' or-patterns.sml
In Standard ML, no. In other dialects of ML, such as OCaml, yes. You may in some cases consider splitting pattern matching up into separate cases/functions, or skip pattern matching in favor of a shorter catch-all expression, e.g.
if str = "+" orelse str = "-" then "PLUS MINUS" else
if str = "*" orelse str = "/" then "MULT DIV" else ...
Expanding upon Ionuț's example, you can even use datatypes with other types in them, but their types (and identifier assignments) must match:
datatype mytype = COST as int | QUANTITY as int | PERSON as string | PET as string;
case item of
(COST n|QUANTITY n) => print Int.toString n
|(PERSON name|PET name) => print name
If the types or names don't match, it will get rejected:
case item of
(COST n|PERSON n) => (* fails because COST is int and PERSON is string *)
(COST n|QUANTITY q) => (* fails because the identifiers are different *)
And these patterns work in function definitions as well:
fun myfun (COST n|QUANTITY n) = print Int.toString n
|myfun (PERSON name|PET name) = print name
;
I want to write a simple regex, in vim, that will find all strings lexicographically smaller than another string.
Specifically, I want to use this to compare dates formatted as 2014-02-17. These dates are lexicographically sortable, which is why I use them.
My specific use case: I'm trying to run through a script and find all the dates that are earlier than today's today.
I'm also OK with comparing these as numbers, or any other solution.
I don't think there is anyway to do this easily in regex. For matching any date earlier than the current date you can use run the function below (Some of the stuff was stolen from benjifisher)
function! Convert_to_char_class(cur)
if a:cur =~ '[2-9]'
return '[0-' . (a:cur-1) . ']'
endif
return '0'
endfunction
function! Match_number_before(num)
let branches = []
let init = ''
for i in range(len(a:num))
if a:num[i] =~ '[1-9]'
call add(branches, init . Convert_to_char_class(a:num[i]) . repeat('\d', len(a:num) - i - 1))
endif
let init .= a:num[i]
endfor
return '\%(' . join(branches, '\|') .'\)'
endfunction
function! Match_date_before(date)
if a:date !~ '\v\d{4}-\d{2}-\d{2}'
echo "invalid date"
return
endif
let branches =[]
let parts = split(a:date, '-')
call add(branches, Match_number_before(parts[0]) . '-\d\{2}-\d\{2}')
call add(branches, parts[0] . '-' . Match_number_before(parts[1]) . '-\d\{2}')
call add(branches, parts[0] . '-' . parts[1] . '-' .Match_number_before(parts[2]))
return '\%(' . join(branches, '\|') .'\)'
endfunction
To use you the following to search for all matches before 2014-02-24.
/<C-r>=Match_date_before('2014-02-24')
You might be able to wrap it in a function to set the search register if you wanted to.
The generated regex for dates before 2014-02-24 is the following.
\%(\%([0-1]\d\d\d\|200\d\|201[0-3]\)-\d\{2}-\d\{2}\|2014-\%(0[0-1]\)-\d\{2}\|2014-02-\%([0-1]\d\|2[0-3]\)\)
It does not do any validation of dates. It assumes if you are in that format you are a date.
Equivalent set of functions for matching after the passed in date.
function! Convert_to_char_class_after(cur)
if a:cur =~ '[0-7]'
return '[' . (a:cur+1) . '-9]'
endif
return '9'
endfunction
function! Match_number_after(num)
let branches = []
let init = ''
for i in range(len(a:num))
if a:num[i] =~ '[0-8]'
call add(branches, init . Convert_to_char_class_after(a:num[i]) . repeat('\d', len(a:num) - i - 1))
endif
let init .= a:num[i]
endfor
return '\%(' . join(branches, '\|') .'\)'
endfunction
function! Match_date_after(date)
if a:date !~ '\v\d{4}-\d{2}-\d{2}'
echo "invalid date"
return
endif
let branches =[]
let parts = split(a:date, '-')
call add(branches, Match_number_after(parts[0]) . '-\d\{2}-\d\{2}')
call add(branches, parts[0] . '-' . Match_number_after(parts[1]) . '-\d\{2}')
call add(branches, parts[0] . '-' . parts[1] . '-' .Match_number_after(parts[2]))
return '\%(' . join(branches, '\|') .'\)'
endfunction
The regex generated was
\%(\%([3-9]\d\d\d\|2[1-9]\d\d\|20[2-9]\d\|201[5-9]\)-\d\{2}-\d\{2}\|2014-\%([1-9]\d\|0[3-9]\)-\d\{2}\|2014-02-\%([3-9]\d\|2[5-9]\)\)
You do not say how you want to use this; are you sure that you really want a regular expression? Perhaps you could get away with
if DateCmp(date, '2014-02-24') < 0
" ...
endif
In that case, try this function.
" Compare formatted date strings:
" #param String date1, date2
" dates in YYYY-MM-DD format, e.g. '2014-02-24'
" #return Integer
" negative, zero, or positive according to date1 < date2, date1 == date2, or
" date1 > date2
function! DateCmp(date1, date2)
let [year1, month1, day1] = split(a:date1, '-')
let [year2, month2, day2] = split(a:date2, '-')
if year1 != year2
return year1 - year2
elseif month1 != month2
return month1 - month2
else
return day1 - day2
endif
endfun
If you really want a regular expression, then try this:
" Construct a pattern that matches a formatted date string if and only if the
" date is less than the input date. Usage:
" :echo '2014-02-24' =~ DateLessRE('2014-03-12')
function! DateLessRE(date)
let init = ''
let branches = []
for c in split(a:date, '\zs')
if c =~ '[1-9]'
call add(branches, init . '[0-' . (c-1) . ']')
endif
let init .= c
endfor
return '\d\d\d\d-\d\d-\d\d\&\%(' . join(branches, '\|') . '\)'
endfun
Does that count as a "simple" regex? One way to use it would be to type :g/ and then CRTL-R and = and then DateLessRE('2014-02-24') and Enter, followed by the rest of your command. In other words,
:g/<C-R>=DateLessRE('2014-02-24')<CR>/s/foo/bar
EDIT: I added a concat (:help /\&) that matches a complete "formatted date string". Now, there is no need to anchor the pattern.
Use nested subpatterns. It starts simple, with the century:
[01]\d\d\d-\d\d-\d\d|20
As for each digit to follow, use one of the following patterns; you may want to replace .* by an appropriate sequence of \d and -.
for 0: (0
for 1: (0.*|1
for 2: ([01].*|2
for 3: ([0-2].*|3
for 4: ([0-3].*|4
for 5: ([0-4].*|5
for 6: ([0-5].*|6
for 7: ([0-6].*|7
for 8: ([0-7].*|8
for 9: ([0-8].*|9
For the last digit, you only need the digit range, e.g.:
[0-6]
Finally, all parentheses should be closed:
)))))
In the example of 2014-02-17, this becomes:
[01]\d\d\d-\d\d-\d\d|20
(0\d-\d\d-\d\d|1
([0-3]-\d\d-\d\d|4
-
(0
([01]-\d\d|2
-
(0\d|1
[0-6]
)))))
Now in one line:
[01]\d\d\d-\d\d-\d\d|20(0\d-\d\d-\d\d|1([0-3]-\d\d-\d\d|4-(0([01]-\d\d|2-(0\d|1[0-6])))))
For VIM, let's not forget to escape (, ) and |:
[01]\d\d\d-\d\d-\d\d\|20\(0\d-\d\d-\d\d\|1\([0-3]-\d\d-\d\d\|4-\(0\([01]-\d\d\|2-\(0\d\|1[0-6]\)\)\)\)\)
Would be best to try and generate this (much like in FDinoff's answer), rather than write it yourself...
Update:
Here is a sample AWK script to generate the correct regex for any date yyyy-mm-dd.
#!/usr/bin/awk -f
BEGIN { # possible overrides for non-VIM users
switch (digit) {
case "ascii" : digit = "[0-9]"; break;
case "posix" : digit = "[:digit:]"; break;
default : digit = "\\d";
}
switch (metachar) {
case "unescaped" : escape = ""; break;
default : escape = "\\";
}
}
/^[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]$/ {
print BuildRegex($0);
}
function BuildRegex(s) {
if (s ~ /^[1-9][^1-9]*$/) {
regex = LessThanOnFirstDigit(s);
}
else {
regex = substr(s, 1, 1) BuildRegex(substr(s, 2)); # recursive call
if (s ~ /^[1-9]/) {
regex = escape "(" LessThanOnFirstDigit(s) escape "|" regex escape ")";
}
}
return regex;
}
function LessThanOnFirstDigit(s) {
first = substr(s, 1, 1) - 1;
rest = substr(s, 2);
gsub(/[0-9]/, digit, rest);
return (first ? "[0-" first "]" : "0") rest;
}
Call it like this:
echo 2014-02-17 | awk -f genregex.awk
Of course, you can write such a simple generator in any language you like.
Would be nice to do it in Vimscript, but I have no experience with that, so I will leave that as a home assignment.
If you wanted to search for all dates that were less than 2014-11-23, inclusive, you would use the following regex.
2014-(?:[1-9]|1[0-1])-(?:[1-9]|1[0-9]|2[0-3])
for a better explanation of the regex visit regex101.com and paste the regex in. You can also test it by using that site.
The basics of the regex are to search all dates that:
start with 2014-
either contain a single character from 1 - 9
or a 1 and a single character from 0 - 1, i.e. numbers from 1 - 11
finished by - and numbers from 1 - 23 done in the same style as the second term
I am having a code as
$wrk = OC192-1-1-1;
#temp = split (/-/, $wrk);
if ($temp1[3] =~ /101 || 102 /)
{
print "yes";
} else {
print "no";
}
Output :
yes
Need to know why this is printing yes. I know for regular expression | is supported for OR operator. But need to know why || is giving as "yes" as output
It is because || will make regex match succeed by matching with nothing all the time.
So it is essentially matching $temp1[3] (which doesn't exist) with anyone of the following
"101 "
""
" 102 "
I added double quotes just for explanation.
/101 || 102 / regex tries to match '101 ', or '' (empty string), or ' 102 '.
Since empty string can always be matched, it always returns true in your condition.
In addition to the regex-relevant answer from #anubhava, note that: OC192-1-1-1 is same as 0-1-1-1, which is just "-3", therefore #temp evaluates to ( "", "3" )
And of course there's no such thing as $temp1
I'm looking for a regular expression to remove a single parameter from a query string, and I want to do it in a single regular expression if possible.
Say I want to remove the foo parameter. Right now I use this:
/&?foo\=[^&]+/
That works as long as foo is not the first parameter in the query string. If it is, then my new query string starts with an ampersand. (For example, "foo=123&bar=456" gives a result of "&bar=456".) Right now, I'm just checking after the regex if the query string starts with ampersand, and chopping it off if it does.
Example edge cases:
Input | Expected Output
-------------------------+--------------------
foo=123 | (empty string)
foo=123&bar=456 | bar=456
bar=456&foo=123 | bar=456
abc=789&foo=123&bar=456 | abc=789&bar=456
Edit
OK as pointed out in comments there are there are way more edge cases than I originally considered. I got the following regex to work with all of them:
/&foo(\=[^&]*)?(?=&|$)|^foo(\=[^&]*)?(&|$)/
This is modified from Mark Byers's answer, which is why I'm accepting that one, but Roger Pate's input helped a lot too.
Here is the full suite of test cases I'm using, and a Javascript snippet which tests them:
$(function() {
var regex = /&foo(\=[^&]*)?(?=&|$)|^foo(\=[^&]*)?(&|$)/;
var escapeHtml = function (str) {
var map = {
'&': '&',
'<': '<',
'>': '>',
'"': '"',
"'": '''
};
return str.replace(/[&<>"']/g, function(m) { return map[m]; });
};
//test cases
var tests = [
'foo' , 'foo&bar=456' , 'bar=456&foo' , 'abc=789&foo&bar=456'
,'foo=' , 'foo=&bar=456' , 'bar=456&foo=' , 'abc=789&foo=&bar=456'
,'foo=123' , 'foo=123&bar=456' , 'bar=456&foo=123' , 'abc=789&foo=123&bar=456'
,'xfoo' , 'xfoo&bar=456' , 'bar=456&xfoo' , 'abc=789&xfoo&bar=456'
,'xfoo=' , 'xfoo=&bar=456' , 'bar=456&xfoo=' , 'abc=789&xfoo=&bar=456'
,'xfoo=123', 'xfoo=123&bar=456', 'bar=456&xfoo=123', 'abc=789&xfoo=123&bar=456'
,'foox' , 'foox&bar=456' , 'bar=456&foox' , 'abc=789&foox&bar=456'
,'foox=' , 'foox=&bar=456' , 'bar=456&foox=' , 'abc=789&foox=&bar=456'
,'foox=123', 'foox=123&bar=456', 'bar=456&foox=123', 'abc=789&foox=123&bar=456'
];
//expected results
var expected = [
'' , 'bar=456' , 'bar=456' , 'abc=789&bar=456'
,'' , 'bar=456' , 'bar=456' , 'abc=789&bar=456'
,'' , 'bar=456' , 'bar=456' , 'abc=789&bar=456'
,'xfoo' , 'xfoo&bar=456' , 'bar=456&xfoo' , 'abc=789&xfoo&bar=456'
,'xfoo=' , 'xfoo=&bar=456' , 'bar=456&xfoo=' , 'abc=789&xfoo=&bar=456'
,'xfoo=123', 'xfoo=123&bar=456', 'bar=456&xfoo=123', 'abc=789&xfoo=123&bar=456'
,'foox' , 'foox&bar=456' , 'bar=456&foox' , 'abc=789&foox&bar=456'
,'foox=' , 'foox=&bar=456' , 'bar=456&foox=' , 'abc=789&foox=&bar=456'
,'foox=123', 'foox=123&bar=456', 'bar=456&foox=123', 'abc=789&foox=123&bar=456'
];
for(var i = 0; i < tests.length; i++) {
var output = tests[i].replace(regex, '');
var success = (output == expected[i]);
$('#output').append(
'<tr class="' + (success ? 'passed' : 'failed') + '">'
+ '<td>' + (success ? 'PASS' : 'FAIL') + '</td>'
+ '<td>' + escapeHtml(tests[i]) + '</td>'
+ '<td>' + escapeHtml(output) + '</td>'
+ '<td>' + escapeHtml(expected[i]) + '</td>'
+ '</tr>'
);
}
});
#output {
border-collapse: collapse;
}
#output tr.passed { background-color: #af8; }
#output tr.failed { background-color: #fc8; }
#output td, #output th {
border: 1px solid black;
padding: 2px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table id="output">
<tr>
<th>Succ?</th>
<th>Input</th>
<th>Output</th>
<th>Expected</th>
</tr>
</table>
If you want to do this in just one regular expression, you could do this:
/&foo(=[^&]*)?|^foo(=[^&]*)?&?/
This is because you need to match either an ampersand before the foo=..., or one after, or neither, but not both.
To be honest, I think it's better the way you did it: removing the trailing ampersand in a separate step.
/(?<=&|\?)foo(=[^&]*)?(&|$)/
Uses lookbehind and the last group to "anchor" the match, and allows a missing value. Change the \? to ^ if you've already stripped off the question mark from the query string.
Regex is still not a substitute for a real parser of the query string, however.
Update: Test script: (run it at codepad.org)
import re
regex = r"(^|(?<=&))foo(=[^&]*)?(&|$)"
cases = {
"foo=123": "",
"foo=123&bar=456": "bar=456",
"bar=456&foo=123": "bar=456",
"abc=789&foo=123&bar=456": "abc=789&bar=456",
"oopsfoo=123": "oopsfoo=123",
"oopsfoo=123&bar=456": "oopsfoo=123&bar=456",
"bar=456&oopsfoo=123": "bar=456&oopsfoo=123",
"abc=789&oopsfoo=123&bar=456": "abc=789&oopsfoo=123&bar=456",
"foo": "",
"foo&bar=456": "bar=456",
"bar=456&foo": "bar=456",
"abc=789&foo&bar=456": "abc=789&bar=456",
"foo=": "",
"foo=&bar=456": "bar=456",
"bar=456&foo=": "bar=456",
"abc=789&foo=&bar=456": "abc=789&bar=456",
}
failures = 0
for input, expected in cases.items():
got = re.sub(regex, "", input)
if got != expected:
print "failed: input=%r expected=%r got=%r" % (input, expected, got)
failures += 1
if not failures:
print "Success"
It shows where my approach failed, Mark has the right of it—which should show why you shouldn't do this with regex.. :P
The problem is associating the query parameter with exactly one ampersand, and—if you must use regex (if you haven't picked up on it :P, I'd use a separate parser, which might use regex inside it, but still actually understand the format)—one solution would be to make sure there's exactly one ampersand per parameter: replace the leading ? with a &.
This gives /&foo(=[^&]*)?(?=&|$)/, which is very straight forward and the best you're going to get. Remove the leading & in the final result (or change it back into a ?, etc.). Modifying the test case to do this uses the same cases as above, and changes the loop to:
failures = 0
for input, expected in cases.items():
input = "&" + input
got = re.sub(regex, "", input)
if got[:1] == "&":
got = got[1:]
if got != expected:
print "failed: input=%r expected=%r got=%r" % (input, expected, got)
failures += 1
if not failures:
print "Success"
Having a query string that starts with & is harmless--why not leave it that way? In any case, I suggest that you search for the trailing ampersand and use \b to match the beginning of foo w/o taking in a previous character:
/\bfoo\=[^&]+&?/
It's a bit silly but I started trying to solve this with a regexp and wanted to finally get it working :)
$str[] = 'foo=123';
$str[] = 'foo=123&bar=456';
$str[] = 'bar=456&foo=123';
$str[] = 'abc=789&foo=123&bar=456';
foreach ($str as $string) {
echo preg_replace('#(?:^|\b)(&?)foo=[^&]+(&?)#e', "'$1'=='&' && '$2'=='&' ? '&' : ''", $string), "\n";
}
the replace part is messed up because apparently it gets confused if the captured characters are '&'s
Also, it doesn't match afoo and the like.
Thanks. Yes it uses backslashes for escaping, and you're right, I don't need the /'s.
This seems to work, though it doesn't do it in one line as requested in the original question.
public static string RemoveQueryStringParameter(string url, string keyToRemove)
{
//if first parameter, leave ?, take away trailing &
string pattern = #"\?" + keyToRemove + "[^&]*&?";
url = Regex.Replace(url, pattern, "?");
//if subsequent parameter, take away leading &
pattern = "&" + keyToRemove + "[^&]*";
url = Regex.Replace(url, pattern, "");
return url;
}
I based myself on your implementation to get a Java impl that seems to work:
public static String removeParameterFromQueryString(String queryString,String paramToRemove) {
Preconditions.checkArgument(queryString != null,"Empty querystring");
Preconditions.checkArgument(paramToRemove != null,"Empty param");
String oneParam = "^"+paramToRemove+"(=[^&]*)$";
String begin = "^"+paramToRemove+"(=[^&]*)(&?)";
String end = "&"+paramToRemove+"(=[^&]*)$";
String middle = "(?<=[&])"+paramToRemove+"(=[^&]*)&";
String removedMiddleParams = queryString.replaceAll(middle,"");
String removedBeginParams = removedMiddleParams.replaceAll(begin,"");
String removedEndParams = removedBeginParams.replaceAll(end,"");
return removedEndParams.replaceAll(oneParam,"");
}
I had troubles in some cases with your implementation because sometimes it did not delete a &, and did it with multiple steps which seems easier to understand.
I had a problem with your version, particularly when a param was in the query string multiple times (like param1=toto¶m2=xxx¶m1=YYY¶m3=ZZZ¶m1....)
it's never too late right
did the thing using conditional lookbehind to be sure it doesn't mess up &s
/(?(?<=\?)(foo=[^&]+)&*|&(?1))/g
if ? is behind we catch foo=bar and trailing & if it exists
if not ? is behind we catch &foo=bar
(?1) represents 1st cathing group, in this example it's the same as (foo=[^&]+)
actually i needed a oneliner for two similar parameters page and per-page
so i altered this expression a bit
/(?(?<=\?)((per-)?page=[^&]+)&*|&(?1))/g
works like charm
You can use the following regex:
[\?|&](?<name>.*?)=[^&]*&?
If you want to do exact match you can replace (?<name>.*?) with a url parameter.
e.g.:
[\?|&]foo=[^&]*&?
to match any variable like foo=xxxx in any URL.
For anyone interested in replacing GET request parameters:
The following regex works for also for more general GET method queries (starting with ?) where the marked answer fails if the parameter to be removed is the first one (after ?)
This (JS flavour) regex can be used to remove the parameter regardless of position (first, last, or in between) leaving the query in well formated state.
So just use a regex replace with an empty string.
/&s=[^&]*()|\?s=[^&]*$|s=[^&]*&/
Basically it matches one of the three cases mentioned above (hence the 2 pipes)