I'm trying to come up with a search/replace expression that will convert from Oracle style inserts with timestamp fields to insert statements for another database.
Basically, I want to convert strings like:
to_timestamp('13-SEP-09 12.00.00.000000000 PM','DD-MON-RR HH.MI.SS.FF AM')
to just:
'13-SEP-09 12.00.00.000000000 PM'
I've tried several expressions in the IDEA's search/replace box, but I still can't quite get it. This one:
to_timestamp(.[^,]*,.[^)]*)
replaced with $1 ends up matching the string I want except the close parenthesis, but then only deletes the first part. I end up with:
('13-SEP-09 12.00.00.000000000 PM','DD-MON-RR HH.MI.SS.FF AM')
I really don't understand what's happening here.
Looks like you want:
to_timestamp\(('[^']*')[^)]*\)
Breaking it down:
to_timestamp is obvious
\( matches the opening paren
( starts capturing
'[^']' matches the first quoted string
) stops capturing
[^)]*\) matches the remaining text
If that's the whole string you're matching, and not just part of a larger text, you can use .* instead of [^)]*\) for the last part; you don't really care what comes after the closing '.
Related
I'm doing a search and replace in Notepad++ and am looking for a regex that will literally give me the first ( in a given string, so I can replace it.
I am not interested in any preceding or succeeding characters, literally just the first (.
An example string is:
"starLan(11), -- Deprecated via RFC3635 ethernetCsmacd (6) should be used instead
I'd like to find the first ( (near starLan(11) in this case) so I can replace that character with something else.
It should not match any other ( in the same line, so in this case it should not match the second ( near (6).
All of the examples I've come across seem to be returning everything up to and including the given character, which is not what I'm after in this case.
I would match the following pattern :
^([^(]*)\((.*)$
And replace it with this :
\1X\2
Where X is the text you want to replace your ( with.
It uses back-references to refer to the parts before and after the first (.
Edit : as mentioned by OP, matching ^([^(]*)\( and replacing with \1X is enough.
you can use this
^(.*?)\(
the text captured inside () will be available in back reference $1. so you can replace it like:
$1someText
where someText is the text you want to put in place of removed '('
if you want the text after removed '(' to remain intact as well, you can use:
^(.*?)\((.*)
and replacement as:
$1someText$2
In Autoit script Iam unable to do Regular expression for the below string Here the numbers will get changed always.
Actual String = _WinWaitActivate("RX_IST2_AM [PID:942564 NPID:10991 SID:498702881] sbivvrwm060.dev.ib.tor.Test.com:30000","")
Here the PID, NPID & SID : will be changing and rest of the things are always constant.
What i have tried below is
_WinWaitActivate("RX_IST2_AM [PID:'([0-9]{1,6})' NPID:'([0-9]{1,5})' SID:'([0-9]{1,9})' sbivvrwm060.dev.ib.tor.Test.com:30000","")
Can someone please help me
As stated in the documentation, you should write the prefix REGEXPTITLE: and surround everything with square brackets, but "escape" all including ones as the dots (.) and spaces () with a backslash (\) and instead of [0-9] you might use \d like "[REGEXPTITLE:RX_IST2_AM\ \[PID:(\d{1,6})\ NPID:(\d{1,5})\ SID:(\d{1,9})\] sbivvrwm060\.dev\.ib\.tor\.Test\.com:30000]" as your parameter for the Win...(...)-Functions.
You can even omit the round brackets ((...)) but keep their content if you don't want to capture the content to process it further like with StringRegExp(...) or StringRegExpReplace(...) - using the _WinWaitActivete(...)-Function it won't make sense anyways as it is only matching and not replacing or returning anything from your regular expression.
According to regex101 both work, with the round brackets and without - you should always use a tool like this site to confirm that your expression is actually working for your input string.
Not familiar with autoit, but remember that regex has to completely match your string to capture results. For example, (goat)s will NOT capture the word goat if your string is goat or goater.
You have forgotten to add a ] in your regex, so your pattern doesn't match the string and capture groups will not be extracted. Also I'm not completely sold on the usage of '. Based on this page, you can do something like StringRegExp(yourstring, 'RX_IST2_AM [PID:([0-9]{1,6}) NPID:([0-9]{1,5}) SID:([0-9]{1,9})]', $STR_REGEXPARRAYGLOBALMATCH) and $1, $2 and $3 would be your results respectively. But maybe your approach works too.
I'm trying to use regex to find single quotes (so I can turn them all into double quotes) anywhere in a line that starts with mySqlQueryToArray (a function that makes a query to a SQL DB). I'm doing the regex in Sublime Text 3 which I'm pretty sure uses Perl Regex. I would like to have my regex match with every single quote in a line so for example I might have the line:
mySqlQueryToArray($con, "SELECT * FROM Template WHERE Name='$name'");
I want the regex to match in that line both of the quotes around $name but no other characters in that line. I've been trying to use (?<=mySqlQueryToArray.*)' but it tells me that the look behind assertion is invalid. I also tried (?<=mySqlQueryToArray)(?<=.*)' but that's also invalid. Can someone guide me to a regex that will accomplish what I need?
To find any number of single quotes in a line starting with your keyword you can use the \G anchor ("end of last match") by replacing:
(^\h*mySqlQueryToArray|(?!^)\G)([^\n\r']*)'
With \1\2<replacement>: see demo here.
Explanation
( ^\h*mySqlQueryToArray # beginning of line: check the keyword is here
| (?!^)\G ) # if not at the BOL, check we did match sth on this line
( [^\n\r']* ) ' # capture everything until the next single quote
The general idea is to match everything until the next single quote with ([^\n\r']*)' in order to replace it with \2<replacement>, but do so only if this everything is:
right after the beginning keyword (^mySqlQueryToArray), or
after the end of the last match ((?!^)\G): in that case we know we have the keyword and are on a relevant line.
\h* accounts for any started indent, as suggested by Xælias (\h being shortcut for any kind of horizontal whitespace).
https://stackoverflow.com/a/25331428/3933728 is a better answer.
I'm not good enough with RegEx nor ST to do this in one step. But I can do it in two:
1/ Search for all mySqlQueryToArray strings
Open the search panel: ⌘F or Find->Find...
Make sure you have the Regex (.* ) button selected (bottom left) and the wrap selector (all other should be off)
Search for: ^\s*mySqlQueryToArray.*$
^ beginning of line
\s* any indentation
mySqlQueryToArray your call
.* whatever is behind
$ end of line
Click on Find All
This will select every occurrence of what you want to modify.
2/ Enter the replace mode
⌥⌘F or Find->Replace...
This time, make sure that wrap, Regex AND In selection are active .
Them search for '([^']*)' and replace with "\1".
' are your single quotes
(...) si the capturing block, referenced by \1 in the replace field
[^']* is for any character that is not a single quote, repeated
Then hit Replace All
I know this is a little more complex that the other answer, but this one tackles cases where your line would contain several single-quoted string. Like this:
mySqlQueryToArray($con, "SELECT * FROM Template WHERE Name='$name' and Value='1234'");
If this is too much, I guess something like find: (?<=mySqlQueryToArray)(.*?)'([^']*)'(.*?) and replace it with \1"\2"\3 will be enough.
You can use a regex like this:
(mySqlQueryToArray.*?)'(.*?)'(.*)
Working demo
Check the substitution section.
You can use \K, see this regex:
mySqlQueryToArray[^']*\K'(.*?)'
Here is a regex demo.
I have this dataset: (about 10k times)
<Id>HOW2SING</Id>
<PopularityRank>1</PopularityRank>
<Title><![CDATA[Superior Singing Method - Online Singing Course]]></Title>
<Description><![CDATA[High Quality Vocal Improvement Product With High Conversions. Online Singing Lessons Course Converts Like Crazy Using Content Packed Sales Video. You Make 75% On Every Sale Including Front End, Recurring, And 1-click Upsells!]]></Description>
<HasRecurringProducts>true</HasRecurringProducts>
<Gravity>45.9395</Gravity>
<PercentPerSale>74.0</PercentPerSale>
<PercentPerRebill>20.0</PercentPerRebill>
<AverageEarningsPerSale>74.9006</AverageEarningsPerSale>
<InitialEarningsPerSale>70.1943</InitialEarningsPerSale>
<TotalRebillAmt>16.1971</TotalRebillAmt>
<Referred>75.0</Referred>
<Commission>75</Commission>
<ActivateDate>2011-06-23</ActivateDate>
</Site>
I am trying to do the following:
Get the data from within the tags, and use it to create a URL, so in this example it should make
http://www.reviews.how2sing.domain.com
also, all other data has to go, i want to perform a REGEX function that will just give me a list of URLS.
I prefer to do it using notepad++ but i suck at regex, any help would be welome
To keep the regex relatively simple you can just use:
.*?<id>(.+?)</id>
Replace with:
http://www.reviews.\1.domain.com\n
That will search and replace all instances of Id tag and preceding text. You can then just remove the last manually.
Make sure matches newline is selected.
Regex is straightforward, only slightly tricky part is that it uses +? and *? which are non-greedy. This prevents the whole file from being matched. The () indicate a capture group that is used in the replacement, i.e. \1.
If you want to a regex that will include replacing the last part then use:
.*?(?:(<id>)?(.+?)</id>).+?(?:<id>|\Z)
This is a bit more tricky, it uses:
?:. A non-capturing group.
| OR
\Z end of file
Basically, the first time it will match everything up to the end of the first </id> and replace up to and including the next <id>. After that it will have replaced the starting <id> so everything before </id> goes in the group. On the last match it will match the end of file \Z.
If you only want the Id values, you can do:
'<Id>([^<]*)<\/Id>'
Then you can get the first captured group \1 which is the Id text value and then create a link from it.
Here is a demo:
http://regex101.com/r/jE9qN8
[UPDATE]
To get rid of all other lines, match this regex: '.*<Id>([^<]*)<\/Id>.*' and replace by first captured group \1. Note for the regex match, since there are multiple lines, you will need to have the DOTALL or /s flag activated to also match newlines.
Hope that helps.
How can I create a regular expression that will grab delimited text from a string? For example, given a string like
text ###token1### text text ###token2### text text
I want a regex that will pull out ###token1###. Yes, I do want the delimiter as well. By adding another group, I can get both:
(###(.+?)###)
/###(.+?)###/
if you want the ###'s then you need
/(###.+?###)/
the ? means non greedy, if you didn't have the ?, then it would grab too much.
e.g. '###token1### text text ###token2###' would all get grabbed.
My initial answer had a * instead of a +. * means 0 or more. + means 1 or more. * was wrong because that would allow ###### as a valid thing to find.
For playing around with regular expressions. I highly recommend http://www.weitz.de/regex-coach/ for windows. You can type in the string you want and your regular expression and see what it's actually doing.
Your selected text will be stored in \1 or $1 depending on where you are using your regular expression.
In Perl, you actually want something like this:
$text = 'text ###token1### text text ###token2### text text';
while($text =~ m/###(.+?)###/g) {
print $1, "\n";
}
Which will give you each token in turn within the while loop. The (.*?) ensures that you get the shortest bit between the delimiters, preventing it from thinking the token is 'token1### text text ###token2'.
Or, if you just want to save them, not loop immediately:
#tokens = $text =~ m/###(.+?)###/g;
Assuming you want to match ###token2### as well...
/###.+###/
Use () and \x. A naive example that assumes the text within the tokens is always delimited by #:
text (#+.+#+) text text (#+.+#+) text text
The stuff in the () can then be grabbed by using \1 and \2 (\1 for the first set, \2 for the second in the replacement expression (assuming you're doing a search/replace in an editor). For example, the replacement expression could be:
token1: \1, token2: \2
For the above example, that should produce:
token1: ###token1###, token2: ###token2###
If you're using a regexp library in a program, you'd presumably call a function to get at the contents first and second token, which you've indicated with the ()s around them.
Well when you are using delimiters such as this basically you just grab the first one then anything that does not match the ending delimiter followed by the ending delimiter. A special caution should be that in cases as the example above [^#] would not work as checking to ensure the end delimiter is not there since a singe # would cause the regex to fail (ie. "###foo#bar###). In the case above the regex to parse it would be the following assuming empty tokens are allowed (if not, change * to +):
###([^#]|#[^#]|##[^#])*###