I'm working on a simple bot for discord and the first pattern reading works fine and I get the results I'm looking for, but the second one doesn't seem to work and I can't figure out why.
Any help would be appreciated
public void onMessageReceived(MessageReceivedEvent event) {
if (event.getMessage().getContent().startsWith("!")) {
String output, newUrl;
String word, strippedWord;
String url = "http://jisho.org/api/v1/search/words?keyword=";
Pattern reading;
Matcher matcher;
word = event.getMessage().getContent();
strippedWord = word.replace("!", "");
newUrl = url + strippedWord;
//Output contains the raw text from jisho
output = getUrlContents(newUrl);
//Searching through the raw text to pull out the first "reading: "
reading = Pattern.compile("\"reading\":\"(.*?)\"");
matcher = reading.matcher(output);
//Searching through the raw text to pull out the first "english_definitions: "
Pattern def = Pattern.compile("\"english_definitions\":[\"(.*?)]");
Matcher matcher2 = def.matcher(output);
event.getTextChannel().sendMessage(matcher2.toString());
if (matcher.find() && matcher2.find()) {
event.getTextChannel().sendMessage("Reading: "+matcher.group(1)).queue();
event.getTextChannel().sendMessage("Definition: "+matcher2.group(1)).queue();
}
else {
event.getTextChannel().sendMessage("Word not found").queue();
}
}
}
You had to escape the [ character to \\[ (once for the Java String and once for the Regex). You also did forget the closing \".
the correct pattern looks like this:
Pattern def = Pattern.compile("\"english_definitions\":\\[\"(.*?)\"]");
At the output, you might want to readd \" and start/end.
event.getTextChannel().sendMessage("Definition: \""+matcher2.group(1) + "\"").queue();
Related
I'm trying to extract CANseIqFMnf from the URL https://www.instagram.com/p/CANseIqFMnf/ using regex in Android studio. Please help me to get a regex expression eligible for Android Studio.
Here is the code for my method:
String url = "https://www.instagram.com/p/CANseIqFMnf/";
String REGEX = "/p\//";
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(url);
boolean match = matcher.matches();
if (match){
Log.e("success", "start = " + matcher.start() + " end = " + matcher.end() );
}else{
Log.e("failed", "failed");
}
But it gives me failed in return!
Method 1
You just need to use replaceAll method in String, no need to compile a pattern and complicate things:
String input = "https://www.instagram.com/p/CANseIqFMnf/";
String output = input.replaceAll("https://www.instagram.com/p/", "").replaceAll("/", "");
Log.v(TAG, output);
Note that the first replaceAll is to remove the url and the second replaceAll is to remove any slashes /
Method 2
Pattern pattern = Pattern.compile("https://www.instagram.com/p/(.*?)/");
Matcher matcher = pattern.matcher("https://www.instagram.com/p/CANseIqFMnf/");
while(matcher.find()) {
System.out.println(matcher.group(1));
}
Note that if matcher.find() returns true then if you used modifiers like this in your REGEX (.*?) then the part found there will be in group(1), and group(0) will hold the entire regex match which is in your case the entire url.
Alternate option w/o regex can be implemented in a simpler manner as below using java.nio.file.Paths APIs
public class Url {
public static void main(String[] args) {
String url = "https://www.instagram.com/p/CANseIqFMnf/";
String name = java.nio.file.Paths.get(url).getFileName().toString();
System.out.println(name);
}
}
I want to remove the content between <script></script>tags. I'm manually checking for the pattern and iterating using while loop. But, I'm getting StringOutOfBoundException at this line:
String script = source.substring(startIndex,endIndex-startIndex);
Below is the complete method:
public static String getHtmlWithoutScript(String source) {
String START_PATTERN = "<script>";
String END_PATTERN = " </script>";
while (source.contains(START_PATTERN)) {
int startIndex=source.lastIndexOf(START_PATTERN);
int endIndex=source.indexOf(END_PATTERN,startIndex);
String script=source.substring(startIndex,endIndex);
source.replace(script,"");
}
return source;
}
Am I doing anything wrong here? And I'm getting endIndex=-1. Can anyone help me to identify, why my code is breaking.
String text = "<script>This is dummy text to remove </script> dont remove this";
StringBuilder sb = new StringBuilder(text);
String startTag = "<script>";
String endTag = "</script>";
//removing the text between script
sb.replace(text.indexOf(startTag) + startTag.length(), text.indexOf(endTag), "");
System.out.println(sb.toString());
If you want to remove the script tags too add the following line :
sb.toString().replace(startTag, "").replace(endTag, "")
UPDATE :
If you dont want to use StringBuilder you can do this:
String text = "<script>This is dummy text to remove </script> dont remove this";
String startTag = "<script>";
String endTag = "</script>";
//removing the text between script
String textToRemove = text.substring(text.indexOf(startTag) + startTag.length(), text.indexOf(endTag));
text = text.replace(textToRemove, "");
System.out.println(text);
You can use a regex to remove the script tag content:
public String removeScriptContent(String html) {
if(html != null) {
String re = "<script>(.*)</script>";
Pattern pattern = Pattern.compile(re);
Matcher matcher = pattern.matcher(html);
if (matcher.find()) {
return html.replace(matcher.group(1), "");
}
}
return null;
}
You have to add this two imports:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
I know I'm probably late to the party. But I would like to give you a regex (really tested solution).
What you have to note here is that when it comes to regular expressions, their engines are greedy by default. So a search string such as <script>(.*)</script> will match the entire string starting from <script> up until the end of the line, or end of the file depending on the regexp options used. This is due to the fact that the search engine uses greedy matching by default.
Now in order to perform the match that you want to in an accurate manner... you could use "lazy" searching.
Search with Lazy loading
<script>(.*?)<\/script>
Now with that, you will get accurate results.
You can read more about about Regexp Lazy & Greedy in this answer.
This worked for me:
private static String removeScriptTags(String message) {
String scriptRegex = "<(/)?[ ]*script[^>]*>";
Pattern pattern2 = Pattern.compile(scriptRegex);
if(message != null) {
Matcher matcher2 = pattern2.matcher(message);
StringBuffer str = new StringBuffer(message.length());
while(matcher2.find()) {
matcher2.appendReplacement(str, Matcher.quoteReplacement(" "));
}
matcher2.appendTail(str);
message = str.toString();
}
return message;
}
Credit goes to nealvs: https://nealvs.wordpress.com/2010/06/01/removing-tags-from-a-string-in-java/
I have a rather complex (to me it seems rather complex) problem that I'm using regular expressions in Java for:
I can get any text string that must be of the format:
M:<some text>:D:<either a url or string>:C:<some more text>:Q:<a number>
I started with a regular expression for extracting the text between the M:/:D:/:C:/:Q: as:
String pattern2 = "(M:|:D:|:C:|:Q:.*?)([a-zA-Z_\\.0-9]+)";
And that works fine if the <either a url or string> is just an alphanumeric string. But it all falls apart when the embedded string is a url of the format:
tcp://someurl.something:port
Can anyone help me adjust the above reg exp to extract the text after :D: to be either a url or a alpha-numeric string?
Here's an example:
public static void main(String[] args) {
String name = "M:myString1:D:tcp://someurl.com:8989:C:myString2:Q:1";
boolean matchFound = false;
ArrayList<String> values = new ArrayList<>();
String pattern2 = "(M:|:D:|:C:|:Q:.*?)([a-zA-Z_\\.0-9]+)";
Matcher m3 = Pattern.compile(pattern2).matcher(name);
while (m3.find()) {
matchFound = true;
String m = m3.group(2);
System.out.println("regex found match: " + m);
values.add(m);
}
}
In the above example, my results would be:
myString1
tcp://someurl.com:8989
myString2
1
And note that the Strings can be of variable length, alphanumeric, but allowing some characters (such as the url format with :// and/or . - characters
You mention that the format is constant:
M:<some text>:D:<either a url or string>:C:<some more text>:Q:<a number>
Capture groups can do this for you with the pattern:
"M:(.*):D:(.*):C:(.*):Q:(.*)"
Or you can do a String.split() with a pattern of "M:|:D:|:C:|:Q:". However, the split will return an empty element at the first index. Everything else will follow.
public static void main(String[] args) throws Exception {
System.out.println("Regex: ");
String data = "M:<some text>:D:tcp://someurl.something:port:C:<some more text>:Q:<a number>";
Matcher matcher = Pattern.compile("M:(.*):D:(.*):C:(.*):Q:(.*)").matcher(data);
if (matcher.matches()) {
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println(matcher.group(i));
}
}
System.out.println();
System.out.println("String.split(): ");
String[] pieces = data.split("M:|:D:|:C:|:Q:");
for (String piece : pieces) {
System.out.println(piece);
}
}
Results:
Regex:
<some text>
tcp://someurl.something:port
<some more text>
<a number>
String.split():
<some text>
tcp://someurl.something:port
<some more text>
<a number>
To extract the URL/text part you don't need the regular expression. Use
int startPos = input.indexOf(":D:")+":D:".length();
int endPos = input.indexOf(":C:", startPos);
String urlOrText = input.substring(startPos, endPos);
Assuming you need to do some validation along with the parsing:
break the regex into different parts like this:
String m_regex = "[\\w.]+"; //in jsva a . in [] is just a plain dot
String url_regex = "."; //theres a bunch online, pick your favorite.
String d_regex = "(?:" + url_regex + "|\\p{Alnum}+)"; // url or a sequence of alphanumeric characters
String c_regex = "[\\w.]+"; //but i'm assuming you want this to be a bit more strictive. not sure.
String q_regex = "\\d+"; //what sort of number exactly? assuming any string of digits here
String regex = "M:(?<M>" + m_regex + "):"
+ "D:(?<D>" + d_regex + "):"
+ "C:(?<D>" + c_regex + "):"
+ "Q:(?<D>" + q_regex + ")";
Pattern p = Pattern.compile(regex);
Might be a good idea to keep the pattern as a static field somewhere and compile it in a static block so that the temporary regex strings don't overcrowd some class with basically useless fields.
Then you can retrieve each part by its name:
Matcher m = p.matcher( input );
if (m.matches()) {
String m_part = m.group( "M" );
...
String q_part = m.group( "Q" );
}
You can go even a step further by making a RegexGroup interface/objects where each implementing object represents a part of the regex which has a name and the actual regex. Though you definitely lose the simplicity makes it harder to understand it with a quick glance. (I wouldn't do this, just pointing out its possible and has its own benefits)
Hi, you can see my code below. I have some strings Country, rank and grank in my code, initially they will be null, but if regex is mached, it should change the value. But even if regex is matched it is not changing the value it is always null. If I remove all if statements and append the string it works fine, but if match is not found it is throwing an exception. Please let me know how can I check this in if logic.
System.err.println(content);
Pattern c = Pattern.compile("NAME=\"(.*)\" RANK");
Pattern r = Pattern.compile("\" RANK=\"(.*)\"");
Pattern gr = Pattern.compile("\" TEXT=\"(.*)\" SOURCE");
Matcher co = c.matcher(content);
Matcher ra = r.matcher(content);
Matcher gra = gr.matcher(content);
co.find();
ra.find();
gra.find();
String country = null;
String Rank = null;
String Grank = null;
if (co.matches()) {
country = co.group(1);
}
if (ra.matches()) {
Rank = ra.group(1);
}
if (gra.matches()) {
Grank = gra.group(1);
}
You have to escape a single \ - use double \\ then it should work.
Tried this?
while (co.find()) {
System.out.print("Start index: " + co.start());
System.out.print(" End index: " + co.end() + " ");
System.out.println(co.group());
}
Personally I can't make your program work with / without the if so it's not a problem of logic but just a problem that it doesn't match the string for me
So I changed it to get something working, maybe you can use it :)
String content = "NAME=\"salut\" RANK=\"pouet\" TEXT=\"text\" SOURCE";
System.out.println(content);
System.out.println(content.replaceAll(("NAME=\"(.*)\"\\sRANK=\"(.*)\"\\sTEXT=\"(.*)\" SOURCE"), "$1---$2---$3"));
Output
NAME="salut" RANK="pouet" TEXT="text" SOURCE
salut---pouet---text
Is there a simple solution to parse a String by using regex in Java?
I have to adapt a HTML page. Therefore I have to parse several strings, e.g.:
href="/browse/PJBUGS-911"
=>
href="PJBUGS-911.html"
The pattern of the strings is only different corresponding to the ID (e.g. 911). My first idea looks like this:
String input = "";
String output = input.replaceAll("href=\"/browse/PJBUGS\\-[0-9]*\"", "href=\"PJBUGS-???.html\"");
I want to replace everything except the ID. How can I do this?
Would be nice if someone can help me :)
You can capture substrings that were matched by your pattern, using parentheses. And then you can use the captured things in the replacement with $n where n is the number of the set of parentheses (counting opening parentheses from left to right). For your example:
String output = input.replaceAll("href=\"/browse/PJBUGS-([0-9]*)\"", "href=\"PJBUGS-$1.html\"");
Or if you want:
String output = input.replaceAll("href=\"/browse/(PJBUGS-[0-9]*)\"", "href=\"$1.html\"");
This does not use regexp. But maybe it still solves your problem.
output = "href=\"" + input.substring(input.lastIndexOf("/")) + ".html\"";
This is how I would do it:
public static void main(String[] args)
{
String text = "href=\"/browse/PJBUGS-911\" blahblah href=\"/browse/PJBUGS-111\" " +
"blahblah href=\"/browse/PJBUGS-34234\"";
Pattern ptrn = Pattern.compile("href=\"/browse/(PJBUGS-[0-9]+?)\"");
Matcher mtchr = ptrn.matcher(text);
while(mtchr.find())
{
String match = mtchr.group(0);
String insMatch = mtchr.group(1);
String repl = match.replaceFirst(match, "href=\"" + insMatch + ".html\"");
System.out.println("orig = <" + match + "> repl = <" + repl + ">");
}
}
This just shows the regex and replacements, not the final formatted text, which you can get by using Matcher.replaceAll:
String allRepl = mtchr.replaceAll("href=\"$1.html\"");
If just interested in replacing all, you don't need the loop -- I used it just for debugging/showing how regex does business.