I have a problem in getting the correct Regular expression.I have below xml as string
<user_input>
<UserInput Question="test Q?" Answer=<value>0</value><sam#testmail.com>"
</user_input>
Now I need to remove the xml character from Answer attribute only.
So I need the below:-
<user_input>
<UserInput Question="test Q?" Answer=value0value sam#testmail.com"
</user_input>
I have tried the below regex but did not worked out:-
str1.replaceAll("Answer=.*?<([^<]*)>", "$1");
its removing all the text before..
Can anyone help please?
You need to put ? within the first group to make it none greedy, also you dont need Answer=.*?:
str1.replaceAll("<([^<]*?)>", "$1")
DEMO
httpRequest.send("msg="+data+"&TC="+TC); try like this
Although variable width look-behinds are not supported in Java, you can work around it with .{0,1000} that should suffice.
Please check out this approach using 2 regexes, or 1 regex and 1 replace. Choose the one that suits best (I removed the \n line break from the first input string to show the flaw with using simple replace):
String input = "<user_input><UserInput Question=\"test Q?\" Answer=<value>0</value><sam#testmail.com>\"\n</user_input>";
String st = input.replace("><", " ").replaceAll("(?<=Answer=.{0,1000})[<>/]+(?=[^\"]*\")", "");
String st1 = input.replaceAll("(?<=Answer=.{0,1000})><(?=[^\"]*\")", " ").replaceAll("(?<=Answer=.{0,1000})[<>/]+(?=[^\"]*\")", "");
System.out.println(st + "\n" + st1);
Output of a sample program:
<user_input UserInput Question="test Q?" Answer=value0value sam#testmail.com"
</user_input>
<user_input><UserInput Question="test Q?" Answer=value0value sam#testmail.com"
</user_input>
First off, in your sample above, there is a trailing " after the email and > which I do not know if it was placed by error.
However, I will keep it there as according to your expected result, you need it to still be present.
This is my hack.
(Answer=)(<)(value)(>)(.+?([^<]*))(</)(value)(><)(.+?([^>]*))(>) to replace it with
$1$3$5$8 $10
The explanation...
(Answer=)(<)(value)(>) matches from Answer to the start of the value 0
(.+?([^<]*) matches the result from 0 or more right to the beginning < which starts the closing value tag
(</) here, I still select this since it was dropped in the previous expression
(><) I will later replace this with a space
(.+?([^>]*) This matches from the start of the email and excludes the > after the .com
(>) this one selects the last > which I will later drop when replacing.
The trailing " is not selected as I will rather not touch it as requested.
Related
Input -
String ipXmlString = "<root>"
+ "<accntNoGrp><accntNo>1234567</accntNo></accntNoGrp>"
+ "<accntNoGrp><accntNo>6663823</accntNo></accntNoGrp>"
+ "</root>";
Tried follwing things using to mask values within using
String op = ipXmlString .replaceAll("<accntNo>(.+?)</accntNo>", "######");
But above code masks all the values
<root><accntNoGrp>######</accntNoGrp><accntNoGrp>######</accntNoGrp></root>
Expected Output:
<root><accntNoGrp><accntNo>#####67</accntNo></accntNoGrp><accntNoGrp><accntNo>#####23</accntNo></accntNoGrp></root>
How to achieve this using java regex ?Could someone help
Your replacement is wrong, you need to include the <accntNo> tag in the actual replacement. Also, it appears that you want to show the last two characters/numbers of the account number. In this case, we can capture this information during the match and use it in the replacement.
Code:
String op = ipXmlString.replaceAll("<accntNo>(?:.+?)(.{2})</accntNo>", "<accntNo>######$1</accntNo>");
Explanation:
<accntNo> match an opening tag
(?:.+?) match, but do not capture, anything up until the first
(.{2}) two characters before closing tag (and capture this)
</accntNo> match a closing tag
Note here that by using ?: inside a parenthesis in the pattern, we tell the regex engine to not capture it. There is no point in capturing anything before the last two characters of the account number because we don't want to us it.
The $1 quantity in the replacement refers to the first capture group. In this case, it is the last two characters of the account number. Hence, we build the replacement string you want this way.
Demo here:
Rextester
Try this code:
public static void main(String[] args) {
String ipXmlString = "<root>"
+ "<accntNoGrp><accntNo>1234567</accntNo></accntNoGrp>"
+ "<accntNoGrp><accntNo>6663823</accntNo></accntNoGrp>"
+ "</root>";
String replaceAll = ipXmlString.replaceAll("\\d+", "######");
System.out.println(replaceAll);
}
Prints:
<root><accntNoGrp><accntNo>######</accntNo></accntNoGrp><accntNoGrp><accntNo>######</accntNo></accntNoGrp></root>
Does Java have a built-in way to escape arbitrary text so that it can be included in a regular expression? For example, if my users enter "$5", I'd like to match that exactly rather than a "5" after the end of input.
Since Java 1.5, yes:
Pattern.quote("$5");
Difference between Pattern.quote and Matcher.quoteReplacement was not clear to me before I saw following example
s.replaceFirst(Pattern.quote("text to replace"),
Matcher.quoteReplacement("replacement text"));
It may be too late to respond, but you can also use Pattern.LITERAL, which would ignore all special characters while formatting:
Pattern.compile(textToFormat, Pattern.LITERAL);
I think what you're after is \Q$5\E. Also see Pattern.quote(s) introduced in Java5.
See Pattern javadoc for details.
First off, if
you use replaceAll()
you DON'T use Matcher.quoteReplacement()
the text to be substituted in includes a $1
it won't put a 1 at the end. It will look at the search regex for the first matching group and sub THAT in. That's what $1, $2 or $3 means in the replacement text: matching groups from the search pattern.
I frequently plug long strings of text into .properties files, then generate email subjects and bodies from those. Indeed, this appears to be the default way to do i18n in Spring Framework. I put XML tags, as placeholders, into the strings and I use replaceAll() to replace the XML tags with the values at runtime.
I ran into an issue where a user input a dollars-and-cents figure, with a dollar sign. replaceAll() choked on it, with the following showing up in a stracktrace:
java.lang.IndexOutOfBoundsException: No group 3
at java.util.regex.Matcher.start(Matcher.java:374)
at java.util.regex.Matcher.appendReplacement(Matcher.java:748)
at java.util.regex.Matcher.replaceAll(Matcher.java:823)
at java.lang.String.replaceAll(String.java:2201)
In this case, the user had entered "$3" somewhere in their input and replaceAll() went looking in the search regex for the third matching group, didn't find one, and puked.
Given:
// "msg" is a string from a .properties file, containing "<userInput />" among other tags
// "userInput" is a String containing the user's input
replacing
msg = msg.replaceAll("<userInput \\/>", userInput);
with
msg = msg.replaceAll("<userInput \\/>", Matcher.quoteReplacement(userInput));
solved the problem. The user could put in any kind of characters, including dollar signs, without issue. It behaved exactly the way you would expect.
To have protected pattern you may replace all symbols with "\\\\", except digits and letters. And after that you can put in that protected pattern your special symbols to make this pattern working not like stupid quoted text, but really like a patten, but your own. Without user special symbols.
public class Test {
public static void main(String[] args) {
String str = "y z (111)";
String p1 = "x x (111)";
String p2 = ".* .* \\(111\\)";
p1 = escapeRE(p1);
p1 = p1.replace("x", ".*");
System.out.println( p1 + "-->" + str.matches(p1) );
//.*\ .*\ \(111\)-->true
System.out.println( p2 + "-->" + str.matches(p2) );
//.* .* \(111\)-->true
}
public static String escapeRE(String str) {
//Pattern escaper = Pattern.compile("([^a-zA-z0-9])");
//return escaper.matcher(str).replaceAll("\\\\$1");
return str.replaceAll("([^a-zA-Z0-9])", "\\\\$1");
}
}
Pattern.quote("blabla") works nicely.
The Pattern.quote() works nicely. It encloses the sentence with the characters "\Q" and "\E", and if it does escape "\Q" and "\E".
However, if you need to do a real regular expression escaping(or custom escaping), you can use this code:
String someText = "Some/s/wText*/,**";
System.out.println(someText.replaceAll("[-\\[\\]{}()*+?.,\\\\\\\\^$|#\\\\s]", "\\\\$0"));
This method returns: Some/\s/wText*/\,**
Code for example and tests:
String someText = "Some\\E/s/wText*/,**";
System.out.println("Pattern.quote: "+ Pattern.quote(someText));
System.out.println("Full escape: "+someText.replaceAll("[-\\[\\]{}()*+?.,\\\\\\\\^$|#\\\\s]", "\\\\$0"));
^(Negation) symbol is used to match something that is not in the character group.
This is the link to Regular Expressions
Here is the image info about negation:
I have certain urls that I am trying to shorten. I want to remove all everything after the / of the url except when that url is equal to plus.google.com
For example:
www.somerubbish.com/about/64848372.meh.php will shorten to www.somerubbish.com
plus.google.com/756934692387498237/about will be left untouched
Any ideas on how I can do this?
My failed attempt is below. I know that the | is saying OR so thats why it is matching the / in the first line as well.
\b!(?:plus.google.com\/.*)\b|\b(?:\/.*)\b
http://regexr.com/3cv6n
Ok I have it.
The answer was to use a negative lookbehind and remove the pipe
(?<!plus.google.com)\b(?:\/.*)\b
https://regex101.com/r/pU3hU4/1
What's wrong with:
if( ! url.contains("plus.google.com")) {
url = StringUtils.substringBefore(url, "/");
}
I am having trouble with a regex in salesforce, apex. As I saw that apex is using the same syntax and logic as apex, I aimed this at java developers also.
I debugged the String and it is correct. street equals 'str 3 B'.
When using http://www.regexr.com/, the regex works('\d \w$').
The code:
Matcher hasString = Pattern.compile('\\d \\w$').matcher(street);
if(hasString.matches())
My problem is, that hasString.matches() resolves to false. Can anyone tell me if I did something somewhere wrong? I tried to use it without the $, with difference casing, etc. and I just can't get it to work.
Thanks in advance!
You need to use find instead of matches for partial input match as matches attempts to match complete input text.
Matcher hasString = Pattern.compile("\\d \\w$").matcher(street);
if(hasString.find()) {
// matched
System.out.println("Start position: " + hasString.start());
}
I am trying to remove full stops from a String. I have tried the below
titleAndBodyContainer = titleAndBodyContainer.replaceAll("\\.", " ");
Unfortunately, this removed all other 'dots' as well. My paragraph includes dates like Jan.13, 2014 , words like U.S and numbers like 2.2. How can I remove only the full stops?
I'd do:
titleAndBodyContainer = titleAndBodyContainer.replaceAll("\\.(?=\\s|$)", " ");
If your strings (sentences) are grammatically correct i.e. a sentence following the fullstop starts with a capital letter I think you can create a custom method to check for such fullstop if after the fullstop there is a string having its first letter capitalized.
You can check if any space or newline coming after . .
You can use the regex : (\.(?=[\s\n\r]|$))
Code :
String input = "Unfortunately, this removed all other 'dots' as well. My paragraph includes dates like Jan.13, 2014 , words like U.S and numbers like 2.2. How can I remove only the full stops?\n"+"Somthing is there.\n"+"Hello someone . What is this something ? ";
String REGEX = "(\\.(?=[\\s\\n\\r]|$))";
String x=input.replaceAll(REGEX, " ");
OUTPUT
Unfortunately, this removed all other 'dots' as well My paragraph includes dates like Jan.13, 2014 , words like U.S and numbers like 2.2 How can I remove only the full stops?
Somthing is there
Hello someone What is this something ?
DEMO