how to split the data using string using java - java

String str = sum(target_details.target_1_n)
as target details
above String how to split the word using target
str.split("target")
but it can divided into sum(target,_details.target_1_n) as,details
but i need on individal word split like sum(target_details.target_1_n) as ,details

Im not sure actually what you want to do but if you wish to separate the string by a token like "." or "_" you can do this:
String alfa = "something.stuff";
String[] result = alfa.split(".");
And it will return an array like this:
result[0] = "something"
result[1] = "stuff"

Related

Remove parts of String? [duplicate]

I want to remove a part of string from one character, that is:
Source string:
manchester united (with nice players)
Target string:
manchester united
There are multiple ways to do it. If you have the string which you want to replace you can use the replace or replaceAll methods of the String class. If you are looking to replace a substring you can get the substring using the substring API.
For example
String str = "manchester united (with nice players)";
System.out.println(str.replace("(with nice players)", ""));
int index = str.indexOf("(");
System.out.println(str.substring(0, index));
To replace content within "()" you can use:
int startIndex = str.indexOf("(");
int endIndex = str.indexOf(")");
String replacement = "I AM JUST A REPLACEMENT";
String toBeReplaced = str.substring(startIndex + 1, endIndex);
System.out.println(str.replace(toBeReplaced, replacement));
String Replace
String s = "manchester united (with nice players)";
s = s.replace(" (with nice players)", "");
Edit:
By Index
s = s.substring(0, s.indexOf("(") - 1);
Use String.Replace():
http://www.daniweb.com/software-development/java/threads/73139
Example:
String original = "manchester united (with nice players)";
String newString = original.replace(" (with nice players)","");
originalString.replaceFirst("[(].*?[)]", "");
https://ideone.com/jsZhSC
replaceFirst() can be replaced by replaceAll()
Using StringBuilder, you can replace the following way.
StringBuilder str = new StringBuilder("manchester united (with nice players)");
int startIdx = str.indexOf("(");
int endIdx = str.indexOf(")");
str.replace(++startIdx, endIdx, "");
You should use the substring() method of String object.
Here is an example code:
Assumption: I am assuming here that you want to retrieve the string till the first parenthesis
String strTest = "manchester united(with nice players)";
/*Get the substring from the original string, with starting index 0, and ending index as position of th first parenthesis - 1 */
String strSub = strTest.subString(0,strTest.getIndex("(")-1);
I would at first split the original string into an array of String with a token " (" and the String at position 0 of the output array is what you would like to have.
String[] output = originalString.split(" (");
String result = output[0];
Using StringUtils from commons lang
A null source string will return null. An empty ("") source string will return the empty string. A null remove string will return the source string. An empty ("") remove string will return the source string.
String str = StringUtils.remove("Test remove", "remove");
System.out.println(str);
//result will be "Test"
If you just need to remove everything after the "(", try this. Does nothing if no parentheses.
StringUtils.substringBefore(str, "(");
If there may be content after the end parentheses, try this.
String toRemove = StringUtils.substringBetween(str, "(", ")");
String result = StringUtils.remove(str, "(" + toRemove + ")");
To remove end spaces, use str.trim()
Apache StringUtils functions are null-, empty-, and no match- safe
Kotlin Solution
If you are removing a specific string from the end, use removeSuffix (Documentation)
var text = "one(two"
text = text.removeSuffix("(two") // "one"
If the suffix does not exist in the string, it just returns the original
var text = "one(three"
text = text.removeSuffix("(two") // "one(three"
If you want to remove after a character, use
// Each results in "one"
text = text.replaceAfter("(", "").dropLast(1) // You should check char is present before `dropLast`
// or
text = text.removeRange(text.indexOf("("), text.length)
// or
text = text.replaceRange(text.indexOf("("), text.length, "")
You can also check out removePrefix, removeRange, removeSurrounding, and replaceAfterLast which are similar
The Full List is here: (Documentation)
// Java program to remove a substring from a string
public class RemoveSubString {
public static void main(String[] args) {
String master = "1,2,3,4,5";
String to_remove="3,";
String new_string = master.replace(to_remove, "");
// the above line replaces the t_remove string with blank string in master
System.out.println(master);
System.out.println(new_string);
}
}
You could use replace to fix your string. The following will return everything before a "(" and also strip all leading and trailing whitespace. If the string starts with a "(" it will just leave it as is.
str = "manchester united (with nice players)"
matched = str.match(/.*(?=\()/)
str.replace(matched[0].strip) if matched

Java String TRIM function not working

I am receiving a string from server trailing one or two lines of spaces like below given string.
String str = "abc*******
********";
Consider * as spaces after my string
i have tried a few methods like
str = str.trim();
str = str.replace(String.valueOf((char) 160), " ").trim();
str = str.replaceAll("\u00A0", "");
but none is working.
Why i am not able to remove the space?
You should try like this:
str = str.replaceAll("\n", "").trim();
You can observe there is a new line in that string . first replace new line "\n" with space("") and than trim
You should do:
str = str.replaceAll("\n", "");
In my case use to work the function trim()
Try this:
str = str.replaceAll("[.]*[\\s\t]+$", "");
I have tried your 3 methods, and them all work. I think your question describing not correctly or complete, in fact, a String in java would not like
String str = "abc*******
********";
They must like
String str = "abc*******"
+ "********";
So I think you should describe your question better to get help.

How to get the string before .(dot) and after /(last) slash in Java

I have a string like this:
"core/pages/viewemployee.jsff"
From this code, I need to get "viewemployee". How do I get this using Java?
Suppose that you have that string saved in a variable named myString.
String myString = "core/pages/viewemployee.jsff";
String newString = myString.substring(myString.lastIndexOf("/")+1, myString.indexOf("."));
But you need to make the same control before doing substring in this one, because if there aren't those characters you will get a "-1" from lastIndexOf(), or indexOf(), and it will break your substring invocation.
I suggest looking for the Javadoc documentation.
You can solve this with regex (given you only need a group of word characters between the last "/" and "."):
String str="core/pages/viewemployee.jsff";
str=str.replaceFirst(".*/(\\w+).*","$1");
System.out.println(str); //prints viewemployee
You can split the string first with "/" so that you can have each folder and the file name got separated. For this example, you will have "core", "pages" and "viewemployee.jsff". I assume you need the file name without the extension, so just apply same split action with "." seperator to the last token. You will have filename without extension.
String myStr = "core/pages/viewemployee.bak.jsff";
String[] tokens = myStr.split("/");
String[] fileNameTokens = tokens[tokens.length - 1].split("\\.");
String fileNameStr = "";
for(int i = 0; i < fileNameTokens.length - 1; i++) {
fileNameStr += fileNameTokens[i] + ".";
}
fileNameStr = fileNameStr.substring(0, fileNameStr.length() - 1);
System.out.print(fileNameStr) //--> "viewemployee.bak"
These are file paths. Consider using File.getName(), especially if you already have the File object:
File file = new File("core/pages/viewemployee.jsff");
String name = file.getName(); // --> "viewemployee.jsff"
And to remove the extension:
String res = name.split("\\.[^\\.]*$")[0]; // --> "viewemployee"
With this we can handle strings like "../viewemployee.2.jsff".
The regex matches the last dot, zero or more non-dots, and the end of the string. Then String.split() treats these as a delimiter, and ignores them. The array will always have one element, unless the original string is ..
The below will get you viewemployee.jsff:
int idx = fileName.replaceAll("\\", "/").lastIndexOf("/");
String fileNameWithExtn = idx >= 0 ? fileName.substring(idx + 1) : fileName;
To remove the file Extension and get only viewemployee, similarly:
idx = fileNameWithExtn.lastIndexOf(".");
String filename = idx >= 0 ? fileNameWithExtn.substring(0,idx) : fileNameWithExtn;

Extract text from string Java

With this string "ADACADABRA". how to extract "CADA" From string "ADACADABRA" in java.
and also how to extract the id between "/" and "?" from the link below.
http://www.youtube-nocookie.com/embed/zaaU9lJ34c5?rel=0
output should be: zaaU9lJ34c5
but should use "/" and "?" in the process.
and also how to extract the id between "/" and "?" from the link below.
http://www.youtube-nocookie.com/embed/zaaU9lJ34c5?rel=0
output should be: zaaU9lJ34c5
Should be :
String url = "http://www.youtube-nocookie.com/embed/zaaU9lJ34c5?rel=0";
String str = url.substring(url.lastIndexOf("/") + 1, url.indexOf("?"));
String s = "ADACADABRA";
String s2 = s.substring(3,7);
Here 3 specifies the beginning index, and 7 specifies the stopping point.
The string returned contains all the characters from the beginning index, up to, but not including, the ending index.
I'm not entirely sure what you mean by extract, so I've provided the code to remove it from the String, I'm not certain if this is what you want.
public static void main (String args[]){
String string = "ADACADABRA";
string = string.replace("CADA", "");
System.out.println(string);
}
This is untested but something like this may help for the youtube part:
String youtubeUrl = "http://www.youtube-nocookie.com/embed/zaaU9lJ34c5?rel=0";
String[] urlParts = youtubeUrl.split("/");
String videoId = urlParts[urlParts.length - 1];
videoId = videoId.substring(0, videoId.indexOf("?"));
Extracting CADA from the string makes no sense. You will need to specify how you have determined that CADA is the string to extract.
E.g. is it because it is the middle 4 characters? Is it because you are stripping off 3 characters each side? Are you just looking for the String "CADA"? Is it characters 3,7 of the String? Is it the first 4 of the last 7 characters of a String? Is it because it contains 2 vowels and 2 consanants? I could go on..
String regex = "CADA";
Pattern p = Pattern.compile(regex, Pattern.MULTILINE);
Matcher m = p.matcher(originalText);
while (m.find()) {
String outputThis = m.group(1);
}
Use this tool http://www.regexplanet.com/advanced/java/index.html
Probably, you don't take in account the fact of java.lang.String immutability. That's why you need to assign the result of substringing to a new variable.

Parsing String Java

I got a String:
["4fd1cf1783353a15415","4ffecf87fcc40d110a965626"]
or
["4fd5f684815345","4fd6ef3e60a676854651","4fd83c33c19164512153"]
And I'd like to store every id (eg. 4fd5f684815345, 4fd6ef3e60a676854651, 4fd83c33c19164512153...) in a independant String (or ArrayList).
How to parse it, because the String can be dynamic (1,2,3 values or more)?
OR JSON Array Parsing:
My JSON
"idCards":[
"4fc52f95egvt418541515",
"4fd1d05454151541545115"
],
A part of my code:
msg3 = (JSONArray) myobject.get("idCards");
System.out.println(msg3.toJSONString());
The result:
[4fc52f95egvt418541515","4fd1d05454151541545115"]
I'd like this 2 values in 2 differents String.
Many thanks for your help!
It would appear to be that this could be a JSON String. In which case, you may make use of a Java JSON Library to help you parse that into Java native objects.
http://www.json.org/
http://code.google.com/p/google-gson/
String data = "[\"4fd5f684815345\",\"4fd6ef3e60a676854651\",\"4fd83c33c19164512153\"]";
// parse JSON String to JSON Array
JsonArray array = (JsonArray) (new JsonParser()).parse(data);
// build a Java ArrayList
List<String> stringList = new ArrayList<String>();
// for each item in JsonArray, add to Java ArrayList
for (int i = 0; i < array.size(); i++) {
stringList.add((array.get(i)).getAsString());
}
I fully agree with the JSON answers, but if this is a one-off hack, you could just do this:
String input = "[\"4fd5f684815345\",\"4fd6ef3e60a676854651\",\"4fd83c33c19164512153\"]";
input = input.replace("[", "");
input = input.replace("]", "");
input = input.replace("\"", "");
String[] parts = input.split(",");
I make a number of assumptions here:
Assume no spaces before and after the delimiting [, ], ,
Assume no , and " character in the Strings you want to extract
input.substring(1, input.length() - 1).replaceAll("\\\"", "").split(",");
Or if you don't want to mess with regular expression (replaceAll function works with regular expression), then you can use replace method:
input.substring(1, input.length() - 1).replace("\"", "").split(",");
Due to the assumptions above, this answer is very brittle. Consider using JSON parser if the data is really JSON.
String str = "[\"4fd5f684815345\",\"4fd6ef3e60a676854651\",\"4fd83c33c19164512153\"]";
String strArry[] = null;
if(str.trim().length() > 0){
str = str.substring(1 , str.length()-1).replaceAll("\\\"", "");
strArry = str.split(",");
}
If s is the input string, it can just be as simple as
String[] idArray = s.replaceAll("[\\[\\]\"]", "").split(",");
it would be more secure (because ',' may be a decimal separator) to split with ("\",\""), and not remove trailing " in replaceAll, here subtring do not parse all the string :
final String str = "[\"4fd5f684815345\",\"4fd6ef3e60a676854651\",\"4fd83c33c19164512153\"]";
final String strArray[] = str.substring(2, str.length() - 2).split("\",\"");
final ArrayList<String> al = new ArrayList<String>();
for (final String string : strArray) {
al.add(string);
System.out.println(string);
}
System.out.println(al);
for (final String string : strArray) {
System.out.println(string);
}
Output :
4fd5f684815345
4fd6ef3e60a676854651
4fd83c33c19164512153
[4fd5f684815345, 4fd6ef3e60a676854651, 4fd83c33c19164512153]

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