Develop Reference

How do you print out a string exactly as it is?

I had an issue with my code because my file path somehow ended up with a "\n" at the end of the path which caused issues when trying to use the file, as it would not be able to find that file.
For debugging purposes, how can I print out a string INCLUDING things like \b \n \r etc.?
E.g.
System.out.println(file.getAbsolutePath).withSpecials()
which will print to console:
C:/folder/filename.extension\n

You could try using this code, which escapes a string. This takes care of all escapes except \u, which should display fine anyway.
public static String escape(String str) {
str = str.replace("\b", "\\b");
str = str.replace("\t", "\\t");
str = str.replace("\n", "\\n");
str = str.replace("\r", "\\r");
str = str.replace("\f", "\\f");
str = str.replace("\'", "\\'");
str = str.replace("\\", "\\\\");
return str;
}
This function can be used as follows:
System.out.println(escape("123\n\rabc"));

public class Main {
public static void main(String arg[]) {
String str = "bla\r\n";
System.out.print(str); // prints "bla" and breaks line
System.out.print(Main.withEndings(str)); // prints "bla\r\n"
// Breaks a line
System.out.println();
// Every char is a number, Java uses by default UTF-16 char encoding
char end = '\n';
System.out.println("Char code: " + (int)end); // prints "Char code: 10"
}
public static String withEndings(String str) {
// Replace the character '\n' to a string with 2 characters the '\' and the 'n',
// the same to '\r'.
return str.replace("\n", "\\n").replace("\r", "\\r");
}
}

You can print the \n by doing string.replace("\n", "\\\\n");
So to print it out do: System.out.println(file.getAbsolutePath().replace("\n", "\\\\n"));

java how to escape accented character in string

For example
{"orderNumber":"S301020000","customerFirstName":"ke ČECHA ","customerLastName":"张科","orderStatus":"PENDING_FULFILLMENT_REQUEST","orderSubmittedDate":"May 13, 2015 1:41:28 PM"}
how to get the accented character like "Č" in above json string and escape it in java
Just give some context of this question, please check this question from me
Ajax unescape response text from java servlet not working properly
Sorry for my English :)

You should escape all characters that are greater than 0x7F. You can loop through the String's characters using the .charAt(index) method. For each character ch that needs escaping, replace it with:
String hexDigits = Integer.toHexString(ch).toUpperCase();
String escapedCh = "\\u" + "0000".substring(hexDigits.length) + hexDigits;
I don't think you will need to unescape them in JavaScript because JavaScript supports escaped characters in string literals, so you should be able to work with the string the way it is returned by the server. I'm guessing you will be using JSON.parse() to convert the returned JSON string into a JavaScript object, like this.
Here's a complete function:
public static escapeJavaScript(String source)
{
StringBuilder result = new StringBuilder();
for (int i = 0; i < source.length(); i++)
{
char ch = source.charAt(i);
if (ch > 0x7F)
{
String hexDigits = Integer.toHexString(ch).toUpperCase();
String escapedCh = "\\u" + "0000".substring(hexDigits.length) + hexDigits;
result.append(escapedCh);
}
else
{
result.append(ch);
}
}
return result.toString();
}

Get certain substring from String java

I can have this string as below :
String s = "chapterId=c_1&sectionId=s_24666&isHL=1&cssFileName=haynes";
or
String s = "chapterId=c_1&sectionId=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&sectionId=s_24666";
//OR
//String s = "chapterId=c_1&sectionId=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&sectionId=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&sectionId=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&sectionId=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&sectionId=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&sectionId=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&sectionId=s_24666&isHL=1&cssFileName=haynes";
OUPUT :
24666

SubString replacement in text

I want to replace all the occurrences of a string in a text except the first one.
for eg:
input: Example [2] This is a sample text. This is a sample text. This is a sample text.
replaced word: sample (sImple)
output: Example [2] This is a sample text. This is a sImple text. This is a sImple text.
In string functions what I see is replace, replaceAll, replaceFirst.
How should I handle this case.
Thanks in advance.

You can use this regex to search:
((?:\bsample\b|(?<!^)\G).*?)\bsample\b
And this for replcement:
$1simple
RegEx Demo
Java Code:
String r = input.replaceAll("((?:\\bsample\\b|(?<!^)\\G).*?)\\bsample\\b", "$1simple");

replaceAll and replace will replace all substrings (difference between them is that replaceAll uses regular expression as argument, while replace uses literals).
replaceFirst will replace only first substring which will match pattern you want to find.
What you can do is
use indexOf(String str, int fromIndex) method to determine indexes of first and second sample word,
then substring(int beginIndex) on index of second sample to get part of string from which you want to let replacing possible
and call your replace method on this part
when replacement is done you can concatenate part which shouldn't be changed (before index of second sample word) and part with replaced values
Other solution would be using appendReplacement and appendTail form Matcher class and use replacing value after you find second sample word. Code for it can look like
String yourString = "Example [2] This is a sample text. This is a sample text. This is a sample text.";
Pattern p = Pattern.compile("sample", Pattern.LITERAL);
Matcher m = p.matcher(yourString);
StringBuffer sb = new StringBuffer();
boolean firstWordAlreadyFound = false;
while (m.find()) {
if (firstWordAlreadyFound) {
m.appendReplacement(sb, "sImple");
} else {
m.appendReplacement(sb, m.group());
firstWordAlreadyFound = true;
}
}
m.appendTail(sb);
String result = sb.toString();
System.out.println(result);
Output:
Example [2] This is a sample text. This is a sImple text. This is a sImple text.

Here is a naive approach:
public static String replaceAllButFirst(String text, String toReplace, String replacement) {
String[] parts = text.split(toReplace, 2);
if(parts.length == 2) { //Found at least one match
return parts[0] + toReplace + parts[1].replaceAll(toReplace, replacement);
} else { //no match found giving original text
return text;
}
}
public static void main(String[] args) {
String x = "This is a sample test. This is a sample test. This is a sample test";
System.out.println(replaceAllButFirst(x, "sample", "simple"));
}
Which will give:
This is a sample test. This is a simple test. This is a simple test

Try with substring and indexOf methods to break it in two string then replace in second string and finally append both the strings back
sample code:
String str = "Example [2] This is a sample text. This is a sample text. This is a sample text.";
String findWhat = "sample";
int index = str.indexOf(findWhat) + findWhat.length();
String temp = str.substring(0, index + 1); // first string
str = str.substring(index + 1); // second string
//replace in second string and combine back
str = temp + str.replace(findWhat, "simple"); // final string
System.out.println(str);
combine all in few statements:
int index = str.indexOf(findWhat) + findWhat.length();
str = str.substring(0, index + 1) + str.substring(index + 1).replace(findWhat, "simple");

There is no built-in function that does exactly what you want, either in the String or StringBuilder classes. You'll need to write your own. Here's a quickie:
private string ReplaceText(string originalText, string textToReplace, string replacementText)
{
string tempText;
int firstIndex, lastIndex;
tempText = originalText;
firstIndex = originalText.IndexOf(textToReplace);
lastIndex = tempText.LastIndexOf(textToReplace);
while (firstIndex >= 0 && lastIndex > firstIndex)
{
tempText = tempText.Substring(0,lastIndex) + replacementText + tempText.Substring(lastIndex + textToReplace.Length);
lastIndex = tempText.LastIndexOf(textToReplace);
}
return tempText;
}

Another option:
(?<=\bsample\b)(.*?)\bsample\b
And replacement:
$1yourstring
Java Code:
String s=input.replaceAll("(?<=\\bsample\\b)(.*?)\\bsample\\b", "$1yourString");

Insert a character before and after all letters in a string in Java

I want to insert a % character before after every letter in a string, but using StringBuilder to make it fast.
For example, if a string is 'AA' then it would be '%A%A%'. If it is 'XYZ' then it would be '%X%Y%Z%'

String foo = "VWXYZ";
foo = "%" + foo.replaceAll("(.)","$1%");
System.out.println(foo);
Output:
%V%W%X%Y%Z%
You don't need a StringBuilder. The compiler will take care of that simple concatenation prior to the regex for you by using one.
Edit in response to comment below:
replaceAll() uses a Regular Expression (regex).
The regex (.) says "match any character, and give me a reference to it" . is a wildcard for any character, the parenthesis create the backreference. The $1 in the second argument says "Use backreference #1 from the match".
replaceAll() keeps running this expression over the whole string replacing each character with itself followed by a percent sign, building a new String which it then returns to you.

Try something like this:
String test = "ABC";
StringBuilder builder = new StringBuilder("");
builder.append("%");
for (char achar : test.toCharArray()) {
builder.append(achar);
builder.append("%");
}
System.out.println(builder.toString());

public static String escape(String s) {
StringBuilder buf = new StringBuilder();
boolean wasLetter = false;
for (char c: s.toCharArray()) {
boolean isLetter = Character.isLetter(c);
if (isLetter && !wasLetter) {
buf.append('%');
}
buf.append(c);
if (isLetter) {
buf.append('%');
}
wasLetter = isLetter;
}
return buf.toString();
}

StringBuilder sb = new StringBuilder("AAAAAAA");
for(int i = sb.length(); i >= 0; i--)
{
sb.insert(i, '%');
}

You may see this.
String s="AAAA";
StringBuilder builder = new StringBuilder();
char[] ch=s.toCharArray();
for(int i=0;i<ch.length;i++)
{
builder.append("%"+ch[i]);
}
builder.append("%");
System.out.println(builder.toString());
Output
%A%A%A%A%

I agree with #Brian Roach to add character to before and after but if you want to add any specific character then do like this
String source = "hello good old world";
StringBuffer res = new StringBuffer();
String[] strArr = tagList.split(" ");
for (String str : strArr) {
char[] stringArray = str.trim().toCharArray();
stringArray[0] = stringArray[0];
str = new String(stringArray);
//here you need to specify your first and last character which you want to set
res.append("#"+ str + "$").append(" ");
}
System.out.println("Result: " + res.toString().trim());
Output :- #hello$ #good$ #old$ #world$