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How can I extract substring from the given url using regex in Android Studio

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);
}
}

JAVA regex to find string

i have a string like this:
font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman;
How can I get the value of the color and the value of background-color?
color:#ffffff;
background-color:#ff0000;
i have tried the following code but the result is not my expected.
Pattern pattern = Pattern.compile("^.*(color:|background-color:).*;$");
The result will display:
font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;

If you want to have multiple matches in a string, don't assert ^ and $ because if those matches, then the whole string matches, which means that you can't match it again.
Also, use a lazy quantifier like *?. This will stop matching as soon as it finds some string that matches the pattern after it.
This is the regex you should use:
(color:|background-color:)(.*?);
Group 1 is either color: or background-color:, group 2 is the color code.
Demo

To do this you should use the (?!abc) expression in regex. This finds a match but doesn't select it. After that you can simply select the hexcode, like this:
String s = "font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman";
Pattern pattern = Pattern.compile("(?!color:)#.{6}");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
System.out.println(matcher.group());
}

Pattern pattern = Pattern.compile("color\\s*:\\s*([^;]+)\\s*;\\s*background-color\\s*:\\s*([^;]+)\\s*;");
Matcher matcher = pattern.matcher("font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;");
if (matcher.find()) {
System.out.println("color:" + matcher.group(1));
System.out.println("background-color:" + matcher.group(2));
}

No need to describe the whole input, only the relevant part(s) that you're looking to extract.
The regex color:(#[\\w\\d]+); does the trick for me:
String input = "font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman;";
String regex = "color:(#[\\w\\d]+);";
Matcher m = Pattern.compile(regex).matcher(input);
while (m.find()) {
System.out.println(m.group(1));
}
Notice that m.group(1) returns the matching group which is inside the parenthesis in the regex. So the regex actually matches the whole color:#ffffff; and color:#ff0000; parts, but the print only handles the number itself.

Use a CSS parser like ph-css
String input = "font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;";
final CSSDeclarationList cssPropertyList =
CSSReaderDeclarationList.readFromString(input, ECSSVersion.CSS30);
System.out.println(cssPropertyList.get(1).getProperty() + " , "
+ cssPropertyList.get(1).getExpressionAsCSSString());
System.out.println(cssPropertyList.get(2).getProperty() + " , "
+ cssPropertyList.get(2).getExpressionAsCSSString());
Prints:
color , #ffffff
background-color , #ff0000
Find more about ph-css on github

How do I use regex in Java to pull this from html?

I'm trying to pull data from the ESPN box scores, and one of the html files has:
<td style="text-align:left" nowrap>Channing Frye, PF</td>
and I'm only interested in grabbing the name (Channing Frye) and the position (PF)
Right now, I've been using Pattern.quote(start) + "(.*?)" + Pattern.quote(end) to grab text in between start and end, but I'm not sure how I'm supposed to grab text that starts with pattern .../http://espn.go.com/nba/player/_/id/ and then can contain (any integer)/anyfirst-anylast"> then grab the name I need (Channing Frye), then </a>, and then grab the position I need (PF) and ends with pattern </td>
Thanks!

Here is the pattern:
http://espn.go.com/nba/player/_/id/(\d+)/([\w-]+)">(.*?)</a>,\s*(\w+)</td>
You can use this tool - http://www.regexplanet.com/advanced/java/index.html for verifying regular expressions.

You could use this pattern:
\\/nba\\/player\\/_\\/.*\\\">(.*)<.+>,\\s(.*)<
This will match any link in the html that contains `/nba/player/
String re = "\\/nba\\/player\\/_\\/.*\\">(.*)<.+>,\\s(.*)<";
String str = "<td style=\"text-align:left\" nowrap>Channing Frye, PF</td>";
Pattern p = Pattern.compile(re, Pattern.MULTILINE | Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(str);
example: http://regex101.com/r/hA3uV0

Use this regex:
[A-Z\sa-z0-9]+(?=</a>)|\w+(?=</td>)

Here is one regex:
. is used for any item, .+ is used for any 1+ items
.* means o or more items
\s is used for space
String str = "<td style=\"text-align:left\" nowrap>Channing Frye, PF</td>";
Pattern pattern = Pattern.compile("<td.+>.*<a.+>(.+)</a>[\\s,]+(.+)</td>");
Matcher matcher = pattern.matcher(str);
while(matcher.find()){
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}

You can use :
String lString = "<td style=\"text-align:left\" nowrap>Channing Frye, PF</td>";
Pattern lPattern = Pattern.compile("<td.+><a.+id/\\d+/.+\\-.+>(.+)</a>, (.+)</td>");
Matcher lMatcher = lPattern.matcher(lString);
while(lMatcher.find()) {
System.out.println(lMatcher.group(1));
System.out.println(lMatcher.group(2));
}
This will give you :
Channing Frye
PF

Splitting a string java

I have a string in format:
<+923451234567>: Hi here is the text.
Now I want to get the mobile number(without any non-alphanumeric characters) ie 923451234567 in the start of the string in-between < > symbols, and also the text ie Hi here is the text.
Now I can place a hardcoded logic, which I am currently doing.
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
String[] splitted = cpaMessage.getText().split(">: ", 2);
String mobileNumber=MyUtils.removeNonDigitCharacters(splitted[0]);
String text=splitted[1];
How can I neatly get the required strings from the string with regular expression? So that I don't have to change the code whenever the format of the string changes.

String stringReceivedInSms="<+923451234567>: Hi here is the text.";
Pattern pattern = Pattern.compile("<\\+?([0-9]+)>: (.*)");
Matcher matcher = pattern.matcher(stringReceivedInSms);
if(matcher.matches()) {
String phoneNumber = matcher.group(1);
String messageText = matcher.group(2);
}

Use a regex that matches the pattern - <\\+?(\\d+)>: (.*)
Use the Pattern and Matcher java classes to match the input string.
Pattern p = Pattern.compile("<\\+?(\\d+)>: (.*)");
Matcher m = p.matcher("<+923451234567>: Hi here is the text.");
if(m.matches())
{
System.out.println(m.group(1));
System.out.println(m.group(2));
}

You need to use regex, the following pattern will work:
^<\\+?(\\d++)>:\\s*+(.++)$
Here is how you would use it -
public static void main(String[] args) throws IOException {
final String s = "<+923451234567>: Hi here is the text.";
final Pattern pattern = Pattern.compile(""
+ "#start of line anchor\n"
+ "^\n"
+ "#literal <\n"
+ "<\n"
+ "#an optional +\n"
+ "\\+?\n"
+ "#match and grab at least one digit\n"
+ "(\\d++)\n"
+ "#literal >:\n"
+ ">:\n"
+ "#any amount of whitespace\n"
+ "\\s*+\n"
+ "#match and grap the rest of the string\n"
+ "(.++)\n"
+ "#end anchor\n"
+ "$", Pattern.COMMENTS);
final Matcher matcher = pattern.matcher(s);
if (matcher.matches()) {
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
}
I have added the Pattern.COMMENTS flag so the code will work with the comments embedded for future reference.
Output:
923451234567
Hi here is the text.

You can get your phone number by just doing :
stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">"))
So try this snippet:
public static void main(String[] args){
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
System.out.println(stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">")));
}
You don't need to split your String.

How can I extract all substring by matching a regular expression?

I want extract values of all src attribute in this string, how can i do that:
<p>Test
<img alt="70" width="70" height="50" src="/adminpanel/userfiles/image/1.jpg" />
Test
<img alt="70" width="70" height="50" src="/adminpanel/userfiles/image/2.jpg" />
</p>

Here you go:
String data = "<p>Test \n" +
"<img alt=\"70\" width=\"70\" height=\"50\" src=\"/adminpanel/userfiles/image/1.jpg\" />\n" +
"Test \n" +
"<img alt=\"70\" width=\"70\" height=\"50\" src=\"/adminpanel/userfiles/image/2.jpg\" />\n" +
"</p>";
Pattern p0 = Pattern.compile("src=\"([^\"]+)\"");
Matcher m = p0.matcher(data);
while (m.find())
{
System.out.printf("found: %s%n", m.group(1));
}
Most regex flavors have a shorthand for grabbing all matches, like Ruby's scan method or .NET's Matches(), but in Java you always have to spell it out.

Idea - split around the '"' char, look at each part if it contains the attribute name src and - if yes - store the next value, which is a src attribute.
String[] parts = thisString.split("\""); // splits at " char
List<String> srcAttributes = new ArrayList<String>();
boolean nextIsSrcAttrib = false;
for (String part:parts) {
if (part.trim().endsWith("src=") {
nextIsSrcAttrib = true; {
else if (nextIsSrcAttrib) {
srcAttributes.add(part);
nextIsSrcAttrib = false;
}
}
Better idea - feed it into a usual html parser and extract the values of all src attributes from all img elements. But the above should work as an easy solution, especially in non-production code.

sorry for not coding it (short of time)
how about:
1. (assuming that the file size is reasonable)read the entire file to a String.
2. Split the String arround "src=\"" (assume that the resulting array is called strArr)
3. loop over resulting array of Strings and store strArr[i].substring(0,strArr[i].indexOf("\" />")) to some collection of image sources.
Aviad

since you've requested a regex implementation ...
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
private static String input = "....your html.....";
public static void main(String[] args) {
Pattern pattern = Pattern.compile("src=\".*\"");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println(matcher.group());
}
}
}
You may have to tweak the regex if your src attributes are not double quoted