I am getting comma sepeated string in below format:
String codeList1 = "abc,pqr,100101,P101001,R108972";
or
String codeList2 = "mno, 100101,108972";
Expected Result : Check if code is numeric after removing first alphabet. If yes, remove prefix and return. If no, still return the code.
codeList1 = "abc,pqr,100101,101001,108972";
or
codeList2 = "mno, 100101,108972";
As you can see, I can get codes (P101001 or 101001) and (R108972 ,108972) format. There is will be only one prefix only.
If I am getting(P101001), I want to remove 'P' prefix and return number 101001.
If I am getting 101001, do nothing.
Below is the working code. But is there any easier or more efficient way of achieving this. Please help
for (String code : codeList.split(",")) {
if(StringUtils.isNumeric(code)) {
codes.add(code);
} else if(StringUtils.isNumeric(code.substring(1))) {
codes.add(Integer.toString(Integer.parseInt(code.substring(1))));
} else {
codes.add(code);
}
}
If you want to remove prefixes from the numbers you can easilly use :
String[] codes = {"abc,pqr,100101,P101001,R108972", "mno, 100101,108972"};
for (String code : codes){
System.out.println(
code.replaceAll("\\b[A-Z](\\d+)\\b", "$1")
);
}
Outputs
abc,pqr,100101,101001,108972
mno, 100101,108972
If you are using Java 8+, and want to extract only the numbers, you can just use :
String codeList1 = "abc,pqr,100101,P101001,R108972";
List<Integer> results = Arrays.stream(codeList1.split("\\D")) //split with non degits
.filter(c -> !c.isEmpty()) //get only non empty results
.map(Integer::valueOf) //convert string to Integer
.collect(Collectors.toList()); //collect to results to list
Outputs
100101
101001
108972
You can use regex to do it
String str = "abc,pqr,100101,P101001,R108972";
String regex = ",?[a-zA-Z]{0,}(\\d+)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while(matcher.find()){
System.out.println(matcher.group(1));
}
Output
100101
101001
108972
Updated：
For your comment(I want to add add the codes. If single alphabet prefix found , remove it and add remaining ),you can use below code:
String str = "abc,pqr,100101,P101001,R108972";
String regex = "(?=,?)[a-zA-Z]{0,}(?=\\d+)|\\s";// \\s is used to remove space
String[] strs = str.replaceAll(regex,"").split(",");
Output:
abc
pqr
100101
101001
108972
How about this:
String codeList1 = "abc,pqr,100101,P101001,R108972";
String[] codes = codeList1.split(",");
for (String code : codes) {
if (code.matches("[A-Z]?\\d{6}")) {
String codeF = code.replaceAll("[A-Z]+", "");
System.out.println(codeF);
}
}
100101
101001
108972
Demo
Related
I have a string (which is an URL) in this pattern https://xxx.kflslfsk.com/kjjfkskfjksf/v1/files/media/93939393hhs8.jpeg
now I want to clip it to this
media/93939393hhs8.jpeg
I want to remove all the characters before the second last slash /.
i'm a newbie in java but in swift (iOS) this is how we do this:
if let url = NSURL(string:"https://xxx.kflslfsk.com/kjjfkskfjksf/v1/files/media/93939393hhs8.jpeg"), pathComponents = url.pathComponents {
let trimmedString = pathComponents.suffix(2).joinWithSeparator("/")
print(trimmedString) // "output = media/93939393hhs8.jpeg"
}
Basically, I'm removing everything from this Url expect of last 2 item and then.
I'm joining those 2 items using /.
String ret = url.substring(url.indexof("media"),url.indexof("jpg"))
Are you familiar with Regex? Try to use this Regex (explained in the link) that captures the last 2 items separated with /:
.*?\/([^\/]+?\/[^\/]+?$)
Here is the example in Java (don't forget the escaping with \\:
Pattern p = Pattern.compile("^.*?\\/([^\\/]+?\\/[^\\/]+?$)");
Matcher m = p.matcher(string);
if (m.find()) {
System.out.println(m.group(1));
}
Alternatively there is the split(..) function, however I recommend you the way above. (Finally concatenate separated strings correctly with StringBuilder).
String part[] = string.split("/");
int l = part.length;
StringBuilder sb = new StringBuilder();
String result = sb.append(part[l-2]).append("/").append(part[l-1]).toString();
Both giving the same result: media/93939393hhs8.jpeg
string result=url.substring(url.substring(0,url.lastIndexOf('/')).lastIndexOf('/'));
or
Use Split and add last 2 items
string[] arr=url.split("/");
string result= arr[arr.length-2]+"/"+arr[arr.length-1]
public static String parseUrl(String str) {
return (str.lastIndexOf("/") > 0) ? str.substring(1+(str.substring(0,str.lastIndexOf("/")).lastIndexOf("/"))) : str;
}
I have to split a string using comma(,) as a separator and ignore any comma that is inside quotes(")
fieldSeparator : ,
fieldGrouper : "
The string to split is : "1","2",3,"4,5"
I am able to achieve it as follows :
String record = "\"1\",\"2\",3,\"4,5\"";
String[] tokens = record.split(",(?=([^\"]*\"[^\"]*\")*[^\"]*$)");
Output :
"1"
"2"
3
"4,5"
Now the challenge is that the fieldGrouper(") should not be a part of the split tokens. I am unable to figure out the regex for this.
The expected output of the split is :
1
2
3
4,5
Update:
String[] tokens = record.split( "(,*\",*\"*)" );
Result:
Initial Solution:
( doesn't work # .split method )
This RexEx pattern will isolate the sections you want:
(?:\\")(.*?)(?:\\")
It uses non-capturing groups to isolate the pairs of escaped quotes,
and a capturing group to isolate everything in between.
Check it out here:
Live Demo
My suggestion:
"([^"]+)"|(?<=,|^)([^,]*)
See the regex demo. It will match "..." like strings and capture into Group 1 only what is in-between the quotes, and then will match and capture into Group 2 sequences of characters other than , at the start of a string or after a comma.
Here is a Java sample code:
String s = "value1,\"1\",\"2\",3,\"4,5\",value2";
Pattern pattern = Pattern.compile("\"([^\"]+)\"|(?<=,|^)([^,]*)");
Matcher matcher = pattern.matcher(s);
List<String> res = new ArrayList<String>();
while (matcher.find()){ // Run the matcher
if (matcher.group(1) != null) { // If Group 1 matched
res.add(matcher.group(1)); // Add it to the resulting array
} else {
res.add(matcher.group(2)); // Add Group 2 as it got matched
}
}
System.out.println(res); // => [value1, 1, 2, 3, 4,5, value2]
I would try with this kind of workaround:
String record = "\"1\",\"2\",3,\"4,5\"";
record = record.replaceAll("\"?(?<!\"\\w{1,9999}),\"?|\""," ");
String[] tokens = record.trim().split(" ");
for(String str : tokens){
System.out.println(str);
}
Output:
1
2
3
4,5
My proposition:
record = record.replaceAll("\",", "|");
record = record.replaceAll(",\\\"", "|");
record = record.replaceAll("\"", "");
String[] tokens = record.split("\\|");
for (String token : tokens) {
System.out.println(token);
}
I am trying to get a regex to match, then get the value with it. For example, I want to check for 1234 as an id and if present, get the status (which is 0 in this case). Basically its id:status. Here is what I am trying:
String topicStatus = "1234:0,567:1,89:2";
String someId = "1234";
String regex = "\\b"+someId+":[0-2]\\b";
if (topicStatus.matches(regex)) {
//How to get status?
}
Not only do I not know how to get the status without splitting and looping through, I don't know why it doesn't match the regex.
Any help would be appreciated. Thanks.
Use the Pattern class
String topicStatus = "1234:0,567:1,89:2";
String someId = "1234";
String regex = "\\b"+someId+":[0-2]\\b";
Pattern MY_PATTERN = Pattern.compile(regex);
Matcher m = MY_PATTERN.matcher(topicStatus);
while (m.find()) {
String s = m.group(1);
System.out.println(s);
}
The key here is to surround the position you want [0-2] in parenthesis which means it will be saved as the first group. You then access it through group(1)
I made some assumptions that your pairs we're always comma separate and then delimited by a colon. Using that I just used split.
String[] idsToCheck = topicStatus.split(",");
for(String idPair : idsToCheck)
{
String[] idPairArray = idPair.split(":");
if(idPairArray[0].equals(someId))
{
System.out.println("id : " + idPairArray[0]);
System.out.println("status: " + idPairArray[1]);
}
}
I can have this string as below :
String s = "chapterId=c_1§ionId=s_24666&isHL=1&cssFileName=haynes";
or
String s = "chapterId=c_1§ionId=s_24666";
I need to get the number ("24666" in the examples).
String res = s.substring(s.lastIndexOf("s_")+ 2) this returns me the number + chars till the end of the string(the second example is ok). But I need to stop after the number ends. How can I do that.? Thanks
You can use regExp
String s = "chapterId=c_1§ionId=s_24666";
//OR
//String s = "chapterId=c_1§ionId=s_24666&isHL=1&cssFileName=haynes";
s=s.replaceAll(".*?s_(\\d+).*","$1");
System.out.println(s);
OUTPUT:
24666
Where,
.*?s_ means anything before s_ (s_ inclusive)
(\\d+) means one or more digits () used for group
$1 means group 1 which is digits after s_
Note:Assumed that your every string follows specific format which includes s_ and number after s_.
You can split the string by the character & to get the parameters, and split each parameter with the = to get the parameter name and parameter value. And now look for the parameter name "sectionId", and cut the first 2 characters of its value to get the number, and you can use Integer.parseInt() if you need it as an int.
Note that this solution is flexible enough to process all parameters, not just the one you're currently interested in:
String s = "chapterId=c_1§ionId=s_24666&isHL=1&cssFileName=haynes";
String[] params = s.split("&");
for (String param : params) {
String[] nameValue = param.split("=");
if ("sectionId".equals(nameValue[0])) {
int number = Integer.parseInt(nameValue[1].substring(2));
System.out.println(number); // Prints 24666
// If you don't care about other parameters, this will skip the rest:
break;
}
}
Note:
You might want to put Integer.parseInt() into a try-catch block in case an invalid number would be passed from the client:
try {
int number = Integer.parseInt(nameValue[1].substring(2));
} catch (Exception e) {
// Invalid parameter value, not the expected format!
}
Try this:
I use a check in the substring() method - if there is no "&isHL" in the string (meaning its type 2 you showed us), it will just read until the string ends. otherwise, it will cut the string before the "&isHL". Hope this helps.
Code:
String s = "chapterId=c_1§ionId=s_**24666**";
int endIndex = s.indexOf("&isHL");
String answer = s.substring(s.lastIndexOf("s_") + 2, endIndex == -1 ? s.length() : endIndex);
Try following:
String s = "chapterId=c_1§ionId=s_24666&isHL=1&cssFileName=haynes";
String tok[]=s.split("&");
for(String test:tok){
if(test.contains("s_")){
String next[]=test.split("s_");
System.out.println(next[1]);
}
}
Output :
24666
Alternatively you can simply remove all other words if they are not required as below
String s="chapterId=c_1§ionId=s_24666&isHL=1&cssFileName=haynes";
s=s.replaceAll(".*s_(\\d+).*","$1");
System.out.println(s);
Output :
24666
The dig over here is splitting your string using a Regular Expression to further divide the string into parts and get what is required. For more on Regular Expressions visit this link.
You could sue this regex : (?<=sectionId=s_)(\\d+) This uses positive look-behind.
demo here
Following code will work even if there is multiple occurrence of integer in given string
String inputString = "chapterId=c_a§ionId=s_24666&isHL=1&cssFileName=haynes_45";
String[] inputParams = inputString.split("&");
for (String param : inputParams)
{
String[] nameValue = param.split("=");
try {
int number = Integer.parseInt(getStringInt(nameValue[1]));
System.out.println(number);
}
catch(IllegalStateException illegalStateException){
}
}
private String getStringInt(String inputString)
{
Pattern onlyInt = Pattern.compile("\\d+");
Matcher matcher = onlyInt.matcher(inputString);
matcher.find();
String inputInt = matcher.group();
return inputInt;
}
OUTPUT
2466
1
45
Use split method as
String []result1 = s.split("&");
String result2 = tempResult[1];
String []result3 = result2.split("s_");
Now to get your desire number you just need to do
String finalResult = result3[1];
INPUT :
String s = "chapterId=c_1§ionId=s_24666&isHL=1&cssFileName=haynes";
OUPUT :
24666
I want to split this String to give my desired output
sinXcos(b+c)
Gives output as
sinX
cos(b+c)
I know how to split a string like
200XY
using
token = 200XY;
String[] mix_token = token.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
But how can I use something like this on a string like
sinXcos(b+c)
or a String like
sinXcos(b+c)tan(z)
This will work..
public static void main(String[] args) {
String text = "sinXcos(b+c)tan(z)";
String patternString1 = "(sin|cos|tan)(?![a-z])\\(?\\w(\\+\\w)?\\)?";
Pattern pattern = Pattern.compile(patternString1);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
System.out.println(matcher.group());
}
}
O/P:
sinX
cos(b+c)
tan(z)
2. Input :"sinabc(X+y)cos(b+c)tan(z)";
O/P :
cos(b+c)
tan(z)
Explaination :
S
tring patternString1 = "(sin|cos|tan)(?![a-z])\\(?\\w(\\+\\w)?\\)?";
1. (sin|cos|tan) -->start with (sin or cos or tan)
2. (?:![a-z]) --> negative lookahead. check if the next character is not in between [a to z].
3. \\(?\\w(\\+\\w)?\\)?--> an optional brace followed by an alphabet followed by a "+" and another alphabet.