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regular expression to replaceall substrings embedded in open curling brackets and followed by equal sign and digits

In the follwing String
String toBeFormatted= "[[LngLatAlt{longitude=-7.125924901999952, latitude=33.831783175000055, altitude=NaN},
LngLatAlt{longitude=-5.401396163999948, latitude=35.92213140900003, altitude=NaN}]]"
1- I need to replace all "LngLatAlt{longitude=" with open bracket "["
2- also need to replace all the intermediate ", latitude=33.831783175000055, altitude=NaN}" with ",33.831783175000055]"
That way my string result :
"[[[-7.125924901999952,33.831783175000055],[-5.401396163999948,35.92213140900003]]]"
try it the following reg exp :
String regexTarget = "(\\[\\[LngLatAlt\\{longitude=)";
toBeFormatted.replaceAll(regexTarget, "\\[\\[\\[");
String regexTarget0 = "(, altitude=NaN\\}, LngLatAlt\\{longitude=)";
toBeFormatted.replaceAll(regexTarget0, "],\\[");
String regexTarget1 = "(, latitude=)";
toBeFormatted.replaceAll(regexTarget1, " ,");
String regexTarget2 = "(, altitude=NaN\\})";
toBeFormatted.replaceAll(regexTarget2, "]");
but it seems not working.
Thank you for your help.

try something like:
String result = toBeFormatted.replaceAll("LngLatAlt\\{longitude=([^,]+), latitude=([^,]+), ([^}]+)\\}", "[$1, $2]");
System.out.println(result);

Two separate patterns and matchers (java)

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();

complex regular expression 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)

Need help to form a regex in java

I want to find a regx and occurrences of it in the page source using language Java. The value I am trying to search is as given in the program below.
There might be one or more spaces between tags. I am not able to form a regx for this value. Can some one please help me to find the regx for this value?
My program which checks regx is as given below-
String regx=""<img height=""1"" width=""1"" style=""border-style:none;"" alt="""" src=""//api.adsymptotic.com/api/s/trackconversion?_pid=12170&_psign=3841da8d95cc1dbcf27a696f27ccab0b&_aid=1376&_lbl=RT_LampsPlus_Retargeting_Pixel""/>";
WebDrive driver = new FirefoxDriver();
driver.navigate().to("abc.xom");
int count=0, found=0;
source = driver.getPageSource();
source = source.replaceAll("\\s+", " ").trim();
pattern = Pattern.compile(regx);
matcher = pattern.matcher(source);
while(matcher.find())
{
count++;
found=1;
}
if(found==0)
{
System.out.println("Maximiser not found");
pixelData[rowNumber][2] = String.valueOf(count) ;
pixelData[rowNumber][3] = "Fail";
}
else
{
System.out.println("Maximiser is found" + count);
pixelData[rowNumber][2] = String.valueOf(count) ;
pixelData[rowNumber][3] = "Pass";
}
count=0; found=0;

Hard to tell without the original text and expected result, but your Pattern clearly won't compile as is.
You should single-escape double quotes (\") and double-escape special characters (i.e. \\?) for your code and your Pattern to compile.
Something in the lines of:
String regx="<img height=\"1\" width=\"1\" style=\"border-style:none;\" " +
"alt=\"\" src=\"//api.adsymptotic.com/api/s/trackconversion" +
"\\?_pid=12170&_psign=3841da8d95cc1dbcf27a696f27ccab0b" +
"&_aid=1376&_lbl=RT_LampsPlus_Retargeting_Pixel\"/>";
Also consider scraping markup with appropriate framework (i.e. JSoup for HTML) instead of regex.

I want to check if a word or a set of words exists in a String

My requirement is to check if a group of words or a single word is present in a larger string. I tried using String.contains() method but this fails in case the larger string has new line character. Currently I am using a regex mentioned below. But this works for only one word. The searched text is a user entered value and can contain more than one word. This is an android application.
String regex = ".*.{0}" + searchText + ".{0}.*";
Pattern pattern = Pattern.compile(regex);
pattern.matcher(largerString).find();
Sample String
String largerString ="John writes about this, and John writes about that," +
" and John writes about everything. ";
String searchText = "about this";

Why not just replace line breaks with spaces, and on top of that, convert it all to lower case?
String s = "hello";
String originalString = "Does this contain \n Hello?";
String formattedString = originalString.toLowerCase().replace("\n", " ");
System.out.println(formattedString.contains(s));
Edit: Thinking about it, I don't really understand how line breaks make a difference...
Edit 2: I was right. Line breaks don't matter.
String s = "hello";
String originalString = "Does this contain \nHello?";
String formattedString = originalString.toLowerCase();
System.out.println(formattedString.contains(s));

here is code not using regex.
String largerString = "John writes about this, and John writes about that," +" and John writes about everything. ";
String searchText = "about this";
Pattern pattern = Pattern.compile(searchText);
Matcher m = pattern.matcher(largerString);
if(m.find()){
System.out.println(m.group().toString());
}
Result:
about this
I hope it will help you.