Input string contains multiple key[with some value], which we need to replace with key[with some value],val[value which is same as key].
Input string:
...key[102]...key[108]... key[211]...
Output string:
... key[102],val[102]...key[108],val[108]...key[211],val[211]...
Basically I need to replace all the key with values inside square braces with key[value],val[same value].
E.g. key[102] → key[102],val[102], and key[108] → key[108],val[108].
You need to use capturing groups.( http://www.regular-expressions.info/brackets.html )
key\[(.*?)\]
Debuggex Demo
Example java code (i couldn't test it):
var str = "...key[102]...key[108]... key[211]...";
System.out.println( (str.replaceAll("key\\[(.*?)\\]", "key[$1],val[$1]") );
Related
I use string.format(str, regex) of LUA to fetch some key word.
local RICH_TAGS = {
"texture",
"img",
}
--\[((img)|(texture))=
local START_OF_PATTER = "\\[("
for index = 1, #RICH_TAGS - 1 do
START_OF_PATTER = START_OF_PATTER .. "(" .. RICH_TAGS[index]..")|"
end
START_OF_PATTER = START_OF_PATTER .. "("..RICH_TAGS[#RICH_TAGS].."))"
function RichTextDecoder.decodeRich(str)
local result = {}
print(str, START_OF_PATTER)
dump({string.find(str, START_OF_PATTER)})
end
output
hello[img=123] \[((texture)|(img))
dump from: [string "utils/RichTextDecoder.lua"]:21: in function 'decodeRich'
"<var>" = {
}
The output means:
str = hello[img=123]
START_OF_PATTER = \[((texture)|(img))
This regex works well with some online regex tools. But it find nothing in LUA.
Is there any wrong using in my code?
You cannot use regular expressions in Lua. Use Lua's string patterns to match strings.
See How to write this regular expression in Lua?
Try dump({str:find("\\%[%("))})
Also note that this loop:
for index = 1, #RICH_TAGS - 1 do
START_OF_PATTER = START_OF_PATTER .. "(" .. RICH_TAGS[index]..")|"
end
will leave out the last element of RICH_TAGS, I assume that was not your intention.
Edit:
But what I want is to fetch several specific word. For example, the
pattern can fetch "[img=" "[texture=" "[font=" any one of them. With
the regex string I wrote in my question, regex can do the work. But
with Lua, the way to do the job is write code like string.find(str,
"[img=") and string.find(str, "[texture=") and string.find(str,
"[font="). I wonder there should be a way to do the job with a single
pattern string. I tryed pattern string like "%[%a*=", but obviously it
will fetch a lot more string I need.
You cannot match several specific words with a single pattern unless they are in that string in a specific order. The only thing you could do is to put all the characters that make up those words into a class, but then you risk to find any word you can build from those letters.
Usually you would match each word with a separate pattern or you match any word and check if the match is one of your words using a look up table for example.
So basically you do what a regex library would do in a few lines of Lua.
I need to parse an input string that has the format of
AB~11111, AB~22222, AB~33333, AB~44444
into separate strings:
AB~11111, AB~22222, AB~33333, and AB~44444
Here is my attempted Regex:
range = "([^~,\n]+~[^,]+,)?";
non_delimiter = "[^,\n;]+";
range_regex = new RegExp(this.range + this.non_delimiter, 'g');
But somehow this regex would only parse the input string into
AB~11111, AB~22222 and AB~33333, AB~44444
instead of parsing the input string into individual strings.
Maybe this is missing the boat, but from your input what about something like:
AB~\d+
This should match each of the strings from the above: https://regex101.com/r/vVFDIG/1. And if there's variation (i.e., it can be other letters) then maybe something like:
[A-Z]{2}~\d+
Or whatever it would need to be but using the negative character class seems like quite a roundabout way of doing it. If that's the case, you could just do:
[^ ,]+
You should use a regex split here on ,\s*:
var input = "AB~11111, AB~22222, AB~33333, AB~44444";
var parts = input.split(/,\s*/);
console.log(parts);
If you need to check that the input also consists of CSV list of AB~11111 terms, then you may use test to assert that:
var input = "AB~11111, AB~22222, AB~33333, AB~44444";
console.log(/^[A-Z]{2}~\d{5}(?:,\s*[A-Z]{2}~\d{5})*$/.test(input));
I have to extract some substrings, this is like an XML markup in a plain text doc, like
lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf
Can i extract this pattern in a single command?
In a case like this, I tried to use a matcher, the group command to extract this single match.
I don't want to do something like
String pattern = /<AAA>(.*)<\/AAA>/;
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher("lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf");
if (m.find( )) {
System.out.println("Found value: " + m.group(0) );
}
There must be a more elegant way.
Edit :
Thank you time_yates, i was looking for something like that.
Could you explain a little why you use [0][1] on the result of
def extract = (input =~ '<AAA>(.+?)</AAA>')[0][1]
Answer by tim_yates :
=~ returns a Matcher, and so [0] gets the first match, which is 2 groups, the first is the String that had the match in it (your whole string) the second [1] is the group you defined in your expression
Thank you so much for your help, and thanks to all the readers.
Power of a community !!!
Can't you just do:
def input = 'lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf'
def extract = (input =~ '<AAA>(.+?)</AAA>')[0][1]
assert extract == 'myString'
This is the shortest (not the best) way I can think of without external libs:
String str = "lsdkfjsdklfj sdklfsdklfjsd <AAA>myString</AAA>sdfsdfsdfsdf";
System.out.println(str.substring(str.indexOf(">") + 1, str.lastIndexOf("<")));
Or using StringUtils (which is million times better than my previous sugestion with substring):
StringUtils.substringBetween(str, "<AAA>", "</AAA>");
Still I'd go with matcher() like you proposed among all these.
I need to convert this (date) String "12112014" to "12.11.2014"
What i would like to to is:
Split first 2 Strings "12", add ".",
then split the string from 3-4 to get "11", add "."
at the end split the last 4 strings (or 5-8) to get "2012"
I already found out how to get the first 2 characters ( "^\d{2}" ), but I failed to get characters based on a position.
Whatever be the programming language, You should try to extract the digits from string and then join them with a ".".
In perl, it can be done as :
$_ = '12112014';
s/(\d{2})(\d{2})(\d{4})/$1.$2.$3/;
print "$_";
Without you specifying the language you're after, I've picked javascript:
var s = '12012011';
var s2 = s.replace(/(\d{2})(\d{2})(\d{4})/,'$1.$2.$3'));
console.log(s2); // prints "12.01.2011"
The gist of it is that you use () to specify groups inside your regular expression and then can use the groups in your replace expression.
Same in Java:
String s = "12012011";
String s2 = s.replaceAll("(\\d{2})(\\d{2})(\\d{4})", "$1.$2.$3");
System.out.println(s2);
I dont think that you could do that only with split.
You could expand your expression to:
"(^(\d{2})(\d{2})(\d{4}))"
Then access the groups with the Regex language of your choice and build the string you want.
Note that - besides all regex learning - alternatively you could always parse the original string into strongly typed Date or DateTime variables and output the value using the appropriate locales.
I have the following sample text and I want to replace '[core].' with something else but I only want to replace it when it is not between text markers ' (SQL):
PRINT 'The result of [core].[dbo].[FunctionX]' + [core].[dbo].[FunctionX] + '.'
EXECUTE [core].[dbo].[FunctionX]
The Result shoud be:
PRINT 'The result of [core].[dbo].[FunctionX]' + [extended].[dbo].[FunctionX] + '.'
EXECUTE [extended].[dbo].[FunctionX]
I hope someone can understand this. Can this be solved by a regular expression?
With RegLove
Kevin
Not in a single step, and not in an ordinary text editor. If your SQL is syntactically valid, you can do something like this:
First, you remove every string from the SQL and replace with placeholders. Then you do your replace of [core] with something else. Then you restore the text in the placeholders from step one:
Find all occurrences of '(?:''|[^'])+' with 'n', where n is an index number (the number of the match). Store the matches in an array with the same number as n. This will remove all SQL strings from the input and exchange them for harmless replacements without invalidating the SQL itself.
Do your replace of [core]. No regex required, normal search-and-replace is enough here.
Iterate the array, replacing the placeholder '1' with the first array item, '2' with the second, up to n. Now you have restored the original strings.
The regex, explained:
' # a single quote
(?: # begin non-capturing group
''|[^'] # either two single quotes, or anything but a single quote
)+ # end group, repeat at least once
' # a single quote
JavaScript this would look something like this:
var sql = 'your long SQL code';
var str = [];
// step 1 - remove everything that looks like an SQL string
var newSql = sql.replace(/'(?:''|[^'])+'/g, function(m) {
str.push(m);
return "'"+(str.length-1)+"'";
});
// step 2 - actual replacement (JavaScript replace is regex-only)
newSql = newSql.replace(/\[core\]/g, "[new-core]");
// step 3 - restore all original strings
for (var i=0; i<str.length; i++){
newSql = newSql.replace("'"+i+"'", str[i]);
}
// done.
Here is a solution (javascript):
str.replace(/('[^']*'.*)*\[core\]/g, "$1[extended]");
See it in action