take replaced variable from replaceAll - java

I have a big string and I wanna take links from that string. I can print link.
Pattern pattern = Pattern.compile(".*(?<=overlay-link\" href=\").*?(?=\">).*");
with that code. Example output:
<a title="TITLE" class="overlay-link" href="LINK HERE"></a>
when I try string.replaceAll, regex deleting link and printing another variables.
EX: <a title="TITLE" class="overlay-link" href=""></a>
I am new on regex. Can you help me?
Here is full code :
String content;
Pattern pattern = Pattern.compile(".*(?<=overlay-link\" href=\").*?(?=\">).*");
try {
Scanner scanner = new Scanner(new File("sourceCode.txt"));
while (scanner.hasNext()) {
content = scanner.nextLine();
if (pattern.matcher(content).matches()) {
System.out.println(content.replaceAll("(?<=overlay-link\" href=\").*?(?=\">)", ""));
}
}
} catch (IOException ex) {
Logger.getLogger(SourceCodeExample.class.getName()).log(Level.SEVERE, null, ex);
}


If I understand your question correctly you are looking to pull out just the link specified in the href tag.
To do this you should use a capture group in your regex itself instead of trying to replaceAll.
The replaceAll method is accurately finding the link and replacing it with an empty string and returning the full resulting string as per the docs which is not the desired result.
The regex you should use is as such: .*(?<=overlay-link\" href=\")(.*?)(?=\">).* Notice the capture group () around the link.
This will allow you to find the matches and access the capture group 1. I found a good example of how to do this in this other question. (important snippet pasted below)
String line = "This order was placed for QT3000! OK?"; //<a> tag string
Pattern pattern = Pattern.compile("(.*?)(\\d+)(.*)"); //insert regex provided above
Matcher matcher = pattern.matcher(line);
while (matcher.find()) {
System.out.println("group 1: " + matcher.group(1)); //This will be your link
System.out.println("group 2: " + matcher.group(2));
System.out.println("group 3: " + matcher.group(3));
}
Comments added by me
Note: index 0 represents the whole Pattern

Related

Regex to remove line break within double quote in CSV

Hi I have a csv file with an error in it.so i want it to correct with regular expression, some of the fields contain line break, Example as below
"AHLR150","CDS","-1","MDCPBusinessRelationshipID",,,"Investigating","1600 Amphitheatre Pkwy
California",,"Mountain View",,"United States",,"California",,,"94043-1351","9958"
the above two lines should be in one line
"AHLR150","CDS","-1","MDCPBusinessRelationshipID",,,"Investigating","1600 Amphitheatre PkwyCalifornia",,"Mountain View",,"United States",,"California",,,"94043-1351","9958"
I tried to use the below regex but it didnt help me
%s/\\([^\"]\\)\\n/\\1/

Try this:
public static void main(String[] args) {
String input = "\"AHLR150\",\"CDS\",\"-1\",\"MDCPBusinessRelationshipID\","
+ ",,\"Investigating\",\"1600 Amphitheatre Pkwy\n"
+ "California\",,\"Mountain View\",,\"United\n"
+ "States\",,\"California\",,,\"94043-1351\",\"9958\"\n";
Matcher matcher = Pattern.compile("\"([^\"]*[\n\r].*?)\"").matcher(input);
Pattern patternRemoveLineBreak = Pattern.compile("[\n\r]");
String result = input;
while(matcher.find()) {
String quoteWithLineBreak = matcher.group(1);
String quoteNoLineBreaks = patternRemoveLineBreak.matcher(quoteWithLineBreak).replaceAll(" ");
result = result.replaceFirst(quoteWithLineBreak, quoteNoLineBreaks);
}
//Output
System.out.println(result);
}
Output:
"AHLR150","CDS","-1","MDCPBusinessRelationshipID",,,"Investigating","1600 Amphitheatre Pkwy California",,"Mountain View",,"United States",,"California",,,"94043-1351","9958"

Create a RegEx surrounding the text you want to keep by parentheses and that will create a group of matched characters. Then replace the string using the group index to compose as you wish.
String test = "\"AHLR150\",\"CDS\",\"-1\",\"MDCPBusinessRelationshipID\","
+ ",,\"Investigating\",\"1600 Amphitheatre Pkwy\n"
+ "California\",,\"Mountain View\",,\"United\n"
+ "States\",,\"California\",,,\"94043-1351\",\"9958\"\n";
System.out.println(test.replaceAll("(\"[^\"]*)\n([^\"]*\")", "$1$2"));
So when we replace the matching string ("United\nStates") by $1$2 we are removing the line break because it not belongs to any group:
$1 => the first group (\"[^\"]*) that will match "United
$2 => the second group ([^\"]*\")" that will match States"

Based on this you can try with:
/\r?\n|\r/
I checked it here and seems to be fine

How can I get non-matching groups using a Matcher in Java?

I'm trying to write a java regex to catch some groups of words from a String using a Matcher.
Say i got this string: "Hello, we are #happy# to see you today".
I would like to get 2 group of matches, one having
Hello, we are
to see you today
and the other
happy
So far, I was only able to match the word between the #s using this Pattern:
Pattern p = Pattern.compile("#(.+?)#");
I've read about negative lookahead and lookaround, played a bit with it but without success.
I assume I should do some sort of negation of the regex so far, but I couldn't come up with anything.
Any help would be really appreciated, thank you.

From comment:
I may incur in a string where I got more than one instances of words wrapped by #, such as "#Hello# kind #stranger#"
From comment:
I need to apply some different style format to both the text inside and outside.
Since you need to apply different stylings, the code need to process each block of text separately, and needs to know if the text is inside or outside a #..# section.
Note, in the following code, it will silently skip the last #, if there is an odd number of them.
String input = ...
for (Matcher m = Pattern.compile("([^#]+)|#([^#]+)#").matcher(input); m.find(); ) {
if (m.start(1) != -1) {
String outsideText = m.group(1);
System.out.println("Outside: \"" + outsideText + "\"");
} else {
String insideText = m.group(2);
System.out.println("Inside: \"" + insideText + "\"");
}
}
Output for input = "Hello, we are #happy# to see you today"
Outside: "Hello, we are "
Inside: "happy"
Outside: " to see you today"
Output for input = "#Hello# kind #stranger#"
Inside: "Hello"
Outside: " kind "
Inside: "stranger"
Output for input = "This #text# has unpaired # characters"
Outside: "This "
Inside: "text"
Outside: " has unpaired "
Outside: " characters"

The best I could do is splitting in 3 groups, then merging the group 1 and 4 :
(^.*)(\#(.+?)\#)(.*)
Test it here
EDIT: Taking remarks from the comments :
(^[^\#]*)(?:\#(.+?)\#)([^\#]*)
Thanks to #Lino we don't capture the useless group with # anymore, and we capture anything except #, instead of any non whitespace character in the 1st and 2nd groups.
Test it here

Is this solution fine?
Pattern pattern =
Pattern.compile("([^#]+)|#([^#]*)#");
Matcher matcher =
pattern.matcher("Hello, we are #happy# to see you today");
List<String> notBetween = new ArrayList<>(); // not surrounded by #
List<String> between = new ArrayList<>(); // surrounded by #
while (matcher.find()) {
if (Objects.nonNull(matcher.group(1))) notBetween.add(matcher.group(1));
if (Objects.nonNull(matcher.group(2))) between.add(matcher.group(2));
}
System.out.println("Printing group 1");
for (String string :
notBetween) {
System.out.println(string);
}
System.out.println("Printing group 2");
for (String string :
between) {
System.out.println(string);
}

Using Regular Expression in Java to extract information from a String

I have one input String like this:
"I am Duc/N Ta/N Van/N"
String "/N" present it is the Name of one person.
The expected output is:
Name: Duc Ta Van
How can I do it by using regular expression?

You can use Pattern and Matcher like this :
String input = "I am Duc/N Ta/N Van/N";
Pattern pattern = Pattern.compile("([^\\s]+)/N");
Matcher matcher = pattern.matcher(input);
String result = "";
while (matcher.find()) {
result+= matcher.group(1) + " ";
}
System.out.println("Name: " + result.trim());
Output
Name: Duc Ta Van
Another Solution using Java 9+
From Java9+ you can use Matcher::results like this :
String input = "I am Duc/N Ta/N Van/N";
String regex = "([^\\s]+)/N";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
String result = matcher.results().map(s -> s.group(1)).collect(Collectors.joining(" "));
System.out.println("Name: " + result); // Name: Duc Ta Van

Here is the regex to use to capture every "name" preceded by a /N
(\w+)\/N
Validate with Regex101
Now, you just need to loop on every match in that String and concatenate the to get the result :
String pattern = "(\\w+)\\/N";
String test = "I am Duc/N Ta/N Van/N";
Matcher m = Pattern.compile(pattern).matcher(test);
StringBuilder sbNames = new StringBuilder();
while(m.find()){
sbNames.append(m.group(1)).append(" ");
}
System.out.println(sbNames.toString());
Duc Ta Van
It is giving you the hardest part. I let you adapt this to match your need.
Note :
In java, it is not required to escape a forward slash, but to use the same regex in the entire answer, I will keep "(\\w+)\\/N", but "(\\w+)/N" will work as well.

I've used "[/N]+" as the regular expression.
Regex101
[] = Matches characters inside the set
\/ = Matches the character / literally (case sensitive)
+ = Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)

Regex expression to get the file name

I want to extract only filename from the complete file name + time stamp . below is the input.
String filePath = "fileName1_20150108.csv";
expected output should be: "fileName1"
String filePath2 = "fileName1_filedesc1_20150108_002_20150109013841.csv"
And expected output should be: "fileName1_filedesc1"
I wrote a below code in java to get the file name but it is working for first part (filePath) but not for filepath2.
Pattern pattern = Pattern.compile(".*.(?=_)");
String filePath = "fileName1_20150108.csv";
String filePath2 = "fileName1_filedesc1_20150108_002_20150109013841.csv";
Matcher matcher = pattern.matcher(filePath);
while (matcher.find()) {
System.out.print("Start index: " + matcher.start());
System.out.print(" End index: " + matcher.end() + " ");
System.out.println(matcher.group());
}
Can somebody please help me to correct the regex so i can parse both filepath using same regex?
Thanks

Anchor the start, and make the .* non-greedy:
^.*?(_\D.*?)?(?=[_.])
Update: change the second group (for fileDesc) to optional, and enforce that it starts with a non-digit character. This will work as long as your fileDesc strings never start with numbers.

You can get the characters before the first underscode, the first underscore, and then the characters until the next underscore:
^[^_]*_[^_]*

This should work: "^(.*?)_([0-9_]*)\\.([^.]*)$"
It will return you 3 groups:
the base name (assuming not a single part will be all numbers)
the timestamp info
the extension.
You can test here: http://fiddle.re/v0hne6 (RegexPlanet)

Regex composion

I want to parse a line from a CSV(comma separated) file, something like this:
Bosh,Mark,mark#gmail.com,"3, Institute","83, 1, 2",1,21
I have to parse the file, and instead of the commas between the apostrophes I wanna have ';', like this:
Bosh,Mark,mark#gmail.com,"3; Institute","83; 1; 2",1,21
I use the following Java code but it doesn't parse it well:
Pattern regex = Pattern.compile("(\"[^\\]]*\")");
Matcher matcher = regex.matcher(line);
if (matcher.find()) {
String replacedMatch = matcher.group();
String gr1 = matcher.group(1);
gr1.trim();
replacedMatch = replacedMatch.replace(",", ";");
line = line.replace(matcher.group(), replacedMatch);
}
the output is:
Bosh,Mark,mark#gmail.com,"3; Institute";"83; 1; 2",1,21
anyone have any idea how to fix this?

This is my solution to replace , inside quote to ;. It assumes that if " were to appear in a quoted string, then it is escaped by another ". This property ensures that counting from start to the current character, if the number of quotes " is odd, then that character is inside a quoted string.
// Test string, with the tricky case """", which resolves to
// a length 1 string of single quote "
String line = "Bosh,\"\"\"\",mark#gmail.com,\"3, Institute\",\"83, 1, 2\",1,21";
Pattern pattern = Pattern.compile("\"[^\"]*\"");
Matcher matcher = pattern.matcher(line);
int start = 0;
StringBuilder output = new StringBuilder();
while (matcher.find()) {
// System.out.println(m.group() + "\n " + m.start() + " " + m.end());
output
.append(line.substring(start, matcher.start())) // Append unrelated contents
.append(matcher.group().replaceAll(",", ";")); // Append replaced string
start = matcher.end();
}
output.append(line.substring(start)); // Append the rest of unrelated contents
// System.out.println(output);
Although I cannot find any case that will fail the method of replace the matched group like you did in line = line.replace(matcher.group(), replacedMatch);, I feel safer to rebuild the string from scratch.

Here's a way:
import java.util.regex.*;
class Main {
public static void main(String[] args) {
String in = "Bosh,Mark,mark#gmail.com,\"3, \"\" Institute\",\"83, 1, 2\",1,21";
String regex = "[^,\"\r\n]+|\"(\"\"|[^\"])*\"";
Matcher matcher = Pattern.compile(regex).matcher(in);
StringBuilder out = new StringBuilder();
while(matcher.find()) {
out.append(matcher.group().replace(',', ';')).append(',');
}
out.deleteCharAt(out.length() - 1);
System.out.println(in + "\n" + out);
}
}
which will print:
Bosh,Mark,mark#gmail.com,"3, "" Institute","83, 1, 2",1,21
Bosh,Mark,mark#gmail.com,"3; "" Institute","83; 1; 2",1,21
Tested on Ideone: http://ideone.com/fCgh7

Here is the what you need
String line = "Bosh,Mark,mark#gmail.com,\"3, Institute\",\"83, 1, 2\",1,21";
Pattern regex = Pattern.compile("(\"[^\"]*\")");
Matcher matcher = regex.matcher(line);
while(matcher.find()){
String replacedMatch = matcher.group();
String gr1 = matcher.group(1);
gr1.trim();
replacedMatch = replacedMatch.replace(",", ";");
line = line.replace(matcher.group(), replacedMatch);
}
line will have value you needed.

Have you tried to make the RegExp lazy?
Another idea: inside the [] you should use a " too. If you do that, you should have the expected output with global flag set.

Your regex is faulty. Why would you want to make sure there are no ] within the "..." expression? You'd rather make the regex reluctant (default is eager, which means it catches as much as it can).
"(\"[^\\]]*\")"
should be
"(\"[^\"]*\")"
But nhadtdh is right, you should use a proper CSV library to parse it and replace , to ; in the values the parser returns.
I'm sure you'll find a parser when googling "Java CSV parser".

Shouldn't your regex be ("[^"]*") instead? In other words, your first line should be:
Pattern regex = Pattern.compile("(\"[^\"]*\")");
Of course, this is assuming you can't have quotes in the quoted values of your input line.

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