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)
Related
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();
I am writing a small programming language for a game I am making, this language will be for allowing users to define their own spells for the wizard entity outside the internal game code. I have the language written down, but I'm not entirely sure how to change a string like
setSpellName("Fireball")
setSplashDamage(32,5)
into an array which would have the method name and the arguments after it, like
{"setSpellName","Fireball"}
{"setSplashDamage","32","5"}
How could I do this using java's String.split or string regex's?
Thanks in advance.
Since you're only interested in the function name and parameters I'd suggest scanning up to the first instance of ( and then to the last ) for the params, as so.
String input = "setSpellName(\"Fireball\")";
String functionName = input.substring(0, input.indexOf('('));
String[] params = input.substring(input.indexOf(')'), input.length - 1).split(",");
To capture the String
setSpellName("Fireball")
Do something like this:
String[] line = argument.split("(");
Gets you "setSpellName" at line[0] and "Fireball") at line[1]
Get rid of the last parentheses like this
line[1].replaceAll(")", " ").trim();
Build your JSON with the two "cleaned" Strings.
There's probably a better way with Regex, but this is the quick and dirty way.
With String.indexOf() and String.substring(), you can parse out the function and parameters. Once you parse them out, apply the quotes are around each of them. Then combine them all back together delimited by commas and wrapped in curly braces.
public static void main(String[] args) throws Exception {
List<String> commands = new ArrayList() {{
add("setSpellName(\"Fireball\")");
add("setSplashDamage(32,5)");
}};
for (String command : commands) {
int openParen = command.indexOf("(");
String function = String.format("\"%s\"", command.substring(0, openParen));
String[] parameters = command.substring(openParen + 1, command.indexOf(")")).split(",");
for (int i = 0; i < parameters.length; i++) {
// Surround parameter with double quotes
if (!parameters[i].startsWith("\"")) {
parameters[i] = String.format("\"%s\"", parameters[i]);
}
}
String combine = String.format("{%s,%s}", function, String.join(",", parameters));
System.out.println(combine);
}
}
Results:
{"setSpellName","Fireball"}
{"setSplashDamage","32","5"}
This is a solution using regex, use this Regex "([\\w]+)\\(\"?([\\w]+)\"?\\)":
String input = "setSpellName(\"Fireball\")";
String pattern = "([\\w]+)\\(\"?([\\w]+)\"?\\)";
Pattern r = Pattern.compile(pattern);
String[] matches;
Matcher m = r.matcher(input);
if (m.find()) {
System.out.println("Found value: " + m.group(1));
System.out.println("Found value: " + m.group(2));
String[] params = m.group(2).split(",");
if (params.length > 1) {
matches = new String[params.length + 1];
matches[0] = m.group(1);
System.out.println(params.length);
for (int i = 0; i < params.length; i++) {
matches[i + 1] = params[i];
}
System.out.println(String.join(" :: ", matches));
} else {
matches = new String[2];
matches[0] = m.group(1);
matches[1] = m.group(2);
System.out.println(String.join(", ", matches));
}
}
([\\w]+) is the first group to get the function name.
\\(\"?([\\w]+)\"?\\) is the second group to get the parameters.
This is a Working DEMO.
guys, I wanna extract the content in a string, the content is before "&" and after the "=", like this example:
asdfaf=afl10109&adsfjkl
I want to extract "afl10109" out of the string, can anyone teach me how to do this, I am very new to regex expression...
Use replaceAll() to replace the whole input with just what you want:
String target = str.replaceAll(".*=(.*)&.*", "$1");
The target is captured in a group (group number 1), which is then referenced in the replacement string.
try
public static void main(String args[]) {
String input="asdfaf=afl10109&adsfjkl";
Pattern pattern = Pattern.compile("=[^&]*&");
Matcher m = pattern.matcher(input);
while (m.find()) {
String str = m.group();
System.out.println( str.substring(1,str.length()-1));
}
}
This is not regex but you can also use split()
String str = "asdfaf=afl10109&adsfjkl";
System.out.println(str.split("=")[1].split("&")[0]);
Output:
afl10109
Using good old String#substring()
String str = "foo=bar&baz";
int begin = str.indexOf('=');
if (begin != -1) {
int end = str.indexOf('&', begin);
if (end != -1) {
System.out.println(str.substring(begin+1, end)); // bar
}
}
I have been trying to match the following string -
String temp = "[[Wikipedia:Manual of Style#Links|]]" ;
with the regex
boolean a = temp.matches("\\[\\[Wikipedia:[a-zA-Z_0-9]*#[a-zA-Z_0-9]*\\|\\]\\]");
"\\[\\[Wikipedia:(.*?)#(.*?)\\|\\]\\]"
"\\[\\[Wikipedia:(.*)*#(.+)*\\|\\]\\]"
"\\[\\[(.*?)#(.*?)\\|\\]\\]"
But none of them are giving any positive matches.
Straight away I can see a problem: you are using a character class without a space to match input with spaces.
Try this:
boolean a = temp.matches("\\[\\[Wikipedia:[\\w ]*#[\\w ]+\\|\\]\\]");
Note that [a-zA-Z_0-9] can be replaced by [\w] (but would include letters/numbers from all languages, which should be fine)
public static void main(String[] args) {
String temp = "[[Wikipedia:Manual of Style#Links|]]";
Pattern pattern = Pattern.compile("\\[\\[Wikipedia:([\\w ]+)#([\\w ]+)\\|\\]\\]");
Matcher matcher = pattern.matcher(temp);
if(matcher.find()) {
System.out.println("Manual of Style: " + matcher.group(1));
System.out.println("links : " + matcher.group(2));
}
}
or
temp.matches("\\[\\[Wikipedia:([\\w ]+)#([\\w ]+)\\|\\]\\]");
Just add a space to your custom character class:
String temp = "[[Wikipedia:Manual of Style#Links|]]" ;
temp.matches("\\[\\[Wikipedia:[a-zA-Z_0-9 ]*#[a-zA-Z_0-9]*\\|\\]\\]"); //true
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.