I woud like to know if there's a way (Using XPath) to verify if dcterms:created has been created after May 22th 1990 ? The expression should give a match if it's the case. I've tryed several things but I can't get it to work. Ex: //dcterms:created[text() > "1990-05-22"] The problem is that the comparison opertors don't seem to work because the date format isn't a recognized value.
Code that I want to search:
<dcterms:created>1990-05-25</dcterms:created>
Thanks for your help!
You are comparing strings, convert them first to a date type and than compare them:
//dcterms:created[xs:date(text()) > xs:date("1990-05-22")]
Related
I'm new here, so please don't scold me for misspellings etc.
What I need to do is to rename a bunch of files with a date in different formats at the beginning of their names, like:
05.07.2020-abc.pdf
2020.07.05-pqr.pdf
Instead of writing a different expression for each formatting, eg.
^(\d{2})\.(\d{2}).(\d{4})(.+) => $3-$2-$1$4
Example
02.11.2022-abc.pdf => 2022-11-02-abc.pdf
I'd like to do it in one fell swoop using the OR operator "|" but I have no idea how to formulate the groupings etc. Can one have nested groupings in regex?
Any ideas? Thank in advance!
#The fourth bird:
No (.+) needed. You're right, I condensed my actual expression and could have taken it out.
The different date 'formats' I mean are dd.mm.yyyy and yyyy.mm.dd respectively, and I need to convert both to yyyy-mn-dd
So,if the format is dd.mm.yyyy I have to flip the string, so to say, else I just need to replace the dots by hyphens.
The OS is Android, and for this operation I use Solid Explorer multi search & replace using regex.
I hope I made myself clear this time around ;-)
I would like to extract a line of strings but am having difficulties using the correct RegEx. Any help would be appreciated.
String to extract: KSEA 122053Z 21008KT 10SM FEW020 SCT250 17/08 A3044 RMK AO2 SLP313 T01720083 50005
For Some reason StackOverflow wont let me cut and paste the XML data here since it includes "<>" characters. Basically I am trying to extract data between "raw_text" ... "/raw_text" from a xml that will always be formatted like the following: http://www.aviationweather.gov/adds/dataserver_current/httpparam?dataSource=metars&requestType=retrieve&format=xml&hoursBeforeNow=3&mostRecent=true&stationString=PHNL%20KSEA
However, the Station name, in this case "KSEA" will not always be the same. It will change based on user input into a search variable.
Thanks In advance
if I can assume that every strings that you want starts with KSEA, then the answer would be:
.*(KSEA.*?)KSEA.*
using ? would let .* match as less as possible.
I was just wondering if there is a shorter way of writing an XPath query to find all HREF values containing at least one of many search values?
What I currently have is the following:
//a[contains(#href, 'value1') or contains(#href, 'value2')]
But it seems quite ugly, especially if I were to have more values.
First of all, in many cases you have to live with the "ugliness" or long-windedness of expressions if only XPath 1.0 is at your disposal. Elegance is something introduced with version 2.0, I'd daresay.
But there might be ways to improve your expression: Is there a regularity to the href attributes you'd like to find? For instance, if it is sufficient as a rule to say that the said href attribute values must start with "value", then the expression could be
//a[starts-with(#href,'value')]
I know that "value1" and "value2" are most probably not your actual attribute values but there might be something else that uniquely identifies the group of a elements you're after. Post your HTML input if this is something you want us to help you with.
Personally, I do not find your expression ugly. There is just one or operator and the expression is quite short and readable. I take
if I were to have more values.
to mean that currently, there are only two attribute values you are interested in and that your question therefore is a theoretical one.
In case you're using XPath 2 and would like to have exact matches instead of also matches only containing part of a search value, you can shorten with
//a[#href = ('value1', 'value2')]
For contains() this syntax wouldn't work as the second argument of contains() is only allowed to be 0 or 1 value.
In XPath 2 you could also use
//a[some $s in ('value1', 'value2') satisfies contains(#href, $s)]
or
//a[matches(#href, "value1|value2")]
I got the following regular expression:
=([^"]*)
Basically I want to extract a value between = and " (example: "City=Paris", output "Paris"). But for some reason this expression wont work in PowerCenter.
You got any idea how to implement it?
Thank you
Thomas
You can try this
REG_EXTRACT('"City=Paris"','(.*\w+=)(\w+)(".*)',2)
This basically breaks up the string in three groups of characters and returns the second group.
I have these 2 things I am working with:
CSV.foreach('datafile.csv','r') {|row| D_Location << row[0]}
puts Date.new(2003,05,02).cwday
In the first line I would like to change the datafile.csv to something like a string so I can change one string and it changes for all of these codes. I have many, each controlling 1 csv column.
In the second one I would like to replace the actual date written, and replace it with a string. This is so that can be automatic, because the string will be generated based on other criteria.
I trust the mods will ban me if I'm being too much of a noob hehe. Then I'll toughen up and find these answers myself eventually. But so far I've solved a lot, but not this. Thanks in advance!
Make a function which takes in a string representing a weekday, and returns a number. Call this function later in your code:
Date.new(2003, 05, yourfun('Tuesday')).cwday
For the first part of your question, you're already working with a string. I think what you mean is that you want it to be in a variable:
csv_file = 'datafile.csv'
CSV.foreach(csv_file,'r') {|row| D_Location << row[0]}
For the second part of your question, Date.parse() works with strings, but they need to be in a format that it can recognize. If your date strings use commas, you can replace them with hyphens:
date_str = "2003,05,02"
Date.parse(date_str.gsub(",", "-")).cwday # => 5
It's not clear where your date strings will be coming from or what format they'll be in, but the general concepts you need to understand are that you can use variables, and that you can transform strings.