I have a String:
String s="<p>Dear <span>{customerName}, your {accountName} is actived </span></p><p> </p><p><span>Congrats!.....</span></p>";
So I want to take CustomerName and accountName words and replace with customers details. Can anyone please tell me how can I replace. Here customerName and accountName are dynamically changing ..because those are columns in database sometimes different columns. So i want to find the words within the { and } and need to replace with column data.
Use the following code
s = s.replace("{customerName}", realCustomerName);
s = s.replace("{accountName}", realAccountNAme);
With String's replace function, the first argument is the string you want to replace, and the second argument is the string you want to insert.
Try:
s=s.replace('{customerName}',CustomerName ).replace('{accountName}',accountName);
where CustomerName and accountName will be the strings holding your customers details
If you simply want to replace the words, you could do the following:
String s="<p>Dear <span>{customerName}, your {accountName} is actived </span></p><p> </p><p><span>Congrats!.....</span></p>";
s.replace( "{customerName}", customer.getName() );
s.replace( "{accountName}", account.getName() );
Or, if you are building the string yourself and you can modify it, it might be better to do the following:
String s="<p>Dear <span>%1$s, your %1$s is actived </span></p><p> </p><p><span>Congrats!.....</span></p>";
// You may also just create a new String object...
s = String.format( s, customer.getName(), account.getName() );
Finally, I found the answer to replace the words using regular expressions. Here words b/w ~ need to replace and these words are not fixed and dynamically will be added to string from UI text Area.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegularEx {
/**
* #param args
*/
public static void main(String args[]) {
Pattern pattern = Pattern.compile("\\~.*?\\~");
StringBuilder s = new StringBuilder(
"~ABCD~~BBCc~All the best ~ABCD~~BBCc~~in~~Raja~ Such kind of people ~in~~Raja~~ABCD~~BBCc~~in~~Raja~rajasekhar~ABCD~~BBCc~~in~~Raja~ Bayanapalli ~Chinthalacheruvu~");
Matcher matcher = pattern.matcher(s);
// using Matcher find(), group(), start() and end() methods
String s1 =new String("~ABCD~~BBCc~All the best ~ABCD~~BBCc~~in~~Raja~ Such kind of people ~in~~Raja~~ABCD~~BBCc~~in~~Raja~rajasekhar~ABCD~~BBCc~~in~~Raja~ Bayanapalli ~Chinthalacheruvu~");
int i = 0;
while (matcher.find()) {
String grp = matcher.group();
int si = matcher.start();
int ei = matcher.end();
System.out.println("Found the text \"" + grp
+ "\" starting at " + si + " index and ending at index " + ei);
s1=s1.replaceAll(grp, "Raja");
//System.out.println("FinalString" + s1);
}
System.out.println("------------------------------------\nFinalString" + s1);
}
}
s = s.replace("{customerName}", "John Doe");
s = s.replace("{accountName}", "jdoe");
In my program I need to loop through a variety of dates. I am writing this program in java, and have a bit of experience with readers, but I do not know which reader would complete this task the best, or if another class would work better.
The dates would be input into a text file in the format as follows:
1/1/2013 to 1/7/2013
1/8/2013 to 1/15/2013
Or something of this manner. I would need to break each range of dates into 6 local variables for the loop, then change them for the next loop. The variables would be coded for example:
private static String startingMonth = "1";
private static String startingDay = "1";
private static String startingYear = "2013";
private static String endingMonth = "1";
private static String endingDay = "7";
private static String endingYear = "2013";
I imagine this could be done creating several delimiters to look for, but I do not know that this would be the easiest way. I have been looking at this post for help, but cant seem to find a relevant answer. What would be the best way to go about this?
There are several options.
You could use the scanner, and set the delimiter to include the slash. If you want the values as ints and not string, just use sc.nextInt()
Scanner sc = new Scanner(input).useDelimiter("\\s*|/");
// You can skip the loop to just read a single line.
while(sc.hasNext()) {
startingMonth = sc.next();
startingDay = sc.next();
startingYear = sc.next();
// skip "to"
sc.next()
endingMonth = sc.next();
endingDay = sc.next();
endingYear = sc.next();
}
You can use regex, as alfasin suggest, but this case is rather simple so you can just match the first and last space.
String str = "1/1/2013 to 1/7/2013";
String startDate = str.substring(0,str.indexOf(" "));
String endDate = str.substring(str.lastIndexOf(" ")+1);ยจ
// The rest is the same:
String[] start = startDate.split("/");
System.out.println(start[0] + "-" + start[1] + "-" + start[2]);
String[] end = endDate.split("/");
System.out.println(end[0] + "-" + end[1] + "-" + end[2]);
String str = "1/1/2013 to 1/7/2013";
Pattern pattern = Pattern.compile("(\\d+/\\d+/\\d+)");
Matcher matcher = pattern.matcher(str);
matcher.find();
String startDate = matcher.group();
matcher.find();
String endDate = matcher.group();
String[] start = startDate.split("/");
System.out.println(start[0] + "-" + start[1] + "-" + start[2]);
String[] end = endDate.split("/");
System.out.println(end[0] + "-" + end[1] + "-" + end[2]);
...
OUTPUT
1-1-2013
1-7-2013
I want to add Two java JSON String manually , so for this i need to remove "}" and replace it with comma "," of first JSON String and remove the first "{" of the second JSON String .
This is my program
import java.util.Map;
import org.codehaus.jackson.type.TypeReference;
public class Hi {
private static JsonHelper jsonHelper = JsonHelper.getInstance();
public static void main(String[] args) throws Exception {
Map<String, Tracker> allCusts = null;
String A = "{\"user5\":{\"Iden\":4,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
String B = "{\"user1\":{\"Iden\":4,\"Num\":1},\"user3\":{\"Iden\":6,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
String totalString = A + B;
if (null != totalString) {
allCusts = (Map<String, Tracker>) jsonHelper.toObject(
totalString, new TypeReference<Map<String, Tracker>>() {
});
}
System.out.println(allCusts);
}
}
When adding two Strings A + B
I want to remove the last character of "}" in A and replace it with "," and remove the FIrst character of "{" in B .
SO this should it look like .
String A = "{\"user5\":{\"Iden\":4,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1},";
String B = "\"user1\":{\"Iden\":4,\"Num\":1},\"user3\":{\"Iden\":6,\"Num\":1},\"user2\":{\"Iden\":5,\"Num\":1}}";
I have tried
String Astr = A.replace(A.substring(A.length()-1), ",");
String Bstr = B.replaceFirst("{", "");
String totalString = Astr + Bstr ;
With this i was getting
Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition
please suggest .
{ is a control character for Regular Expressions, and since replaceFirst takes a string representation of a Regular Expression as its first argument, you need to escape the { so it's not treated as a control character:
String Bstr = B.replaceFirst("\\{", "");
I would say that using the replace methods is really overkill here since you're just trying to chop a character off of either end of a string. This should work just as well:
String totalString = A.substring(0, A.length()-1) + "," + B.substring(1);
Of course, regex doesn't look like a very good tool for this. But the following seem to work:
String str = "{..{...}..}}";
str = str.replaceFirst("\\{", "");
str = str.replaceFirst("}$", ",");
System.out.println(str);
Output:
..{...}..},
Some issues in your first two statements. Add 0 as start index in substring method and leave with that. Put \\ as escape char in matching pattern and ut a , in second statement as replacement value.
String Astr = A.substring(0, A.length()-1);//truncate the ending `}`
String Bstr = B.replaceFirst("\\{", ",");//replaces first '{` with a ','
String totalString = Astr + Bstr ;
Please note: There are better ways, but I am just trying to correct your statements.
Looking for quick, simple way in Java to change this string
" hello there "
to something that looks like this
"hello there"
where I replace all those multiple spaces with a single space, except I also want the one or more spaces at the beginning of string to be gone.
Something like this gets me partly there
String mytext = " hello there ";
mytext = mytext.replaceAll("( )+", " ");
but not quite.
Try this:
String after = before.trim().replaceAll(" +", " ");
See also
String.trim()
Returns a copy of the string, with leading and trailing whitespace omitted.
regular-expressions.info/Repetition
No trim() regex
It's also possible to do this with just one replaceAll, but this is much less readable than the trim() solution. Nonetheless, it's provided here just to show what regex can do:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$|( )+", "$1")
);
}
There are 3 alternates:
^_+ : any sequence of spaces at the beginning of the string
Match and replace with $1, which captures the empty string
_+$ : any sequence of spaces at the end of the string
Match and replace with $1, which captures the empty string
(_)+ : any sequence of spaces that matches none of the above, meaning it's in the middle
Match and replace with $1, which captures a single space
See also
regular-expressions.info/Anchors
You just need a:
replaceAll("\\s{2,}", " ").trim();
where you match one or more spaces and replace them with a single space and then trim whitespaces at the beginning and end (you could actually invert by first trimming and then matching to make the regex quicker as someone pointed out).
To test this out quickly try:
System.out.println(new String(" hello there ").trim().replaceAll("\\s{2,}", " "));
and it will return:
"hello there"
Use the Apache commons StringUtils.normalizeSpace(String str) method. See docs here
This worked perfectly for me : sValue = sValue.trim().replaceAll("\\s+", " ");
trim() method removes the leading and trailing spaces and using replaceAll("regex", "string to replace") method with regex "\s+" matches more than one space and will replace it with a single space
myText = myText.trim().replaceAll("\\s+"," ");
The following code will compact any whitespace between words and remove any at the string's beginning and end
String input = "\n\n\n a string with many spaces, \n"+
" a \t tab and a newline\n\n";
String output = input.trim().replaceAll("\\s+", " ");
System.out.println(output);
This will output a string with many spaces, a tab and a newline
Note that any non-printable characters including spaces, tabs and newlines will be compacted or removed
For more information see the respective documentation:
String#trim() method
String#replaceAll(String regex, String replacement) method
For information about Java's regular expression implementation see the documentation of the Pattern class
"[ ]{2,}"
This will match more than one space.
String mytext = " hello there ";
//without trim -> " hello there"
//with trim -> "hello there"
mytext = mytext.trim().replaceAll("[ ]{2,}", " ");
System.out.println(mytext);
OUTPUT:
hello there
To eliminate spaces at the beginning and at the end of the String, use String#trim() method. And then use your mytext.replaceAll("( )+", " ").
You can first use String.trim(), and then apply the regex replace command on the result.
Try this one.
Sample Code
String str = " hello there ";
System.out.println(str.replaceAll("( +)"," ").trim());
OUTPUT
hello there
First it will replace all the spaces with single space. Than we have to supposed to do trim String because Starting of the String and End of the String it will replace the all space with single space if String has spaces at Starting of the String and End of the String So we need to trim them. Than you get your desired String.
String blogName = "how to do in java . com";
String nameWithProperSpacing = blogName.replaceAll("\\\s+", " ");
trim()
Removes only the leading & trailing spaces.
From Java Doc,
"Returns a string whose value is this string, with any leading and trailing whitespace removed."
System.out.println(" D ev Dum my ".trim());
"D ev Dum my"
replace(), replaceAll()
Replaces all the empty strings in the word,
System.out.println(" D ev Dum my ".replace(" ",""));
System.out.println(" D ev Dum my ".replaceAll(" ",""));
System.out.println(" D ev Dum my ".replaceAll("\\s+",""));
Output:
"DevDummy"
"DevDummy"
"DevDummy"
Note: "\s+" is the regular expression similar to the empty space character.
Reference : https://www.codedjava.com/2018/06/replace-all-spaces-in-string-trim.html
In Kotlin it would look like this
val input = "\n\n\n a string with many spaces, \n"
val cleanedInput = input.trim().replace(Regex("(\\s)+"), " ")
A lot of correct answers been provided so far and I see lot of upvotes. However, the mentioned ways will work but not really optimized or not really readable.
I recently came across the solution which every developer will like.
String nameWithProperSpacing = StringUtils.normalizeSpace( stringWithLotOfSpaces );
You are done.
This is readable solution.
You could use lookarounds also.
test.replaceAll("^ +| +$|(?<= ) ", "");
OR
test.replaceAll("^ +| +$| (?= )", "")
<space>(?= ) matches a space character which is followed by another space character. So in consecutive spaces, it would match all the spaces except the last because it isn't followed by a space character. This leaving you a single space for consecutive spaces after the removal operation.
Example:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$| (?= )", "")
);
}
See String.replaceAll.
Use the regex "\s" and replace with " ".
Then use String.trim.
String str = " hello world"
reduce spaces first
str = str.trim().replaceAll(" +", " ");
capitalize the first letter and lowercase everything else
str = str.substring(0,1).toUpperCase() +str.substring(1,str.length()).toLowerCase();
you should do it like this
String mytext = " hello there ";
mytext = mytext.replaceAll("( +)", " ");
put + inside round brackets.
String str = " this is string ";
str = str.replaceAll("\\s+", " ").trim();
This worked for me
scan= filter(scan, " [\\s]+", " ");
scan= sac.trim();
where filter is following function and scan is the input string:
public String filter(String scan, String regex, String replace) {
StringBuffer sb = new StringBuffer();
Pattern pt = Pattern.compile(regex);
Matcher m = pt.matcher(scan);
while (m.find()) {
m.appendReplacement(sb, replace);
}
m.appendTail(sb);
return sb.toString();
}
The simplest method for removing white space anywhere in the string.
public String removeWhiteSpaces(String returnString){
returnString = returnString.trim().replaceAll("^ +| +$|( )+", " ");
return returnString;
}
check this...
public static void main(String[] args) {
String s = "A B C D E F G\tH I\rJ\nK\tL";
System.out.println("Current : "+s);
System.out.println("Single Space : "+singleSpace(s));
System.out.println("Space count : "+spaceCount(s));
System.out.format("Replace all = %s", s.replaceAll("\\s+", ""));
// Example where it uses the most.
String s = "My name is yashwanth . M";
String s2 = "My nameis yashwanth.M";
System.out.println("Normal : "+s.equals(s2));
System.out.println("Replace : "+s.replaceAll("\\s+", "").equals(s2.replaceAll("\\s+", "")));
}
If String contains only single-space then replace() will not-replace,
If spaces are more than one, Then replace() action performs and removes spacess.
public static String singleSpace(String str){
return str.replaceAll(" +| +|\t|\r|\n","");
}
To count the number of spaces in a String.
public static String spaceCount(String str){
int i = 0;
while(str.indexOf(" ") > -1){
//str = str.replaceFirst(" ", ""+(i++));
str = str.replaceFirst(Pattern.quote(" "), ""+(i++));
}
return str;
}
Pattern.quote("?") returns literal pattern String.
My method before I found the second answer using regex as a better solution. Maybe someone needs this code.
private String replaceMultipleSpacesFromString(String s){
if(s.length() == 0 ) return "";
int timesSpace = 0;
String res = "";
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(c == ' '){
timesSpace++;
if(timesSpace < 2)
res += c;
}else{
res += c;
timesSpace = 0;
}
}
return res.trim();
}
Stream version, filters spaces and tabs.
Stream.of(str.split("[ \\t]")).filter(s -> s.length() > 0).collect(Collectors.joining(" "))
I know replaceAll method is much easier but I wanted to post this as well.
public static String removeExtraSpace(String input) {
input= input.trim();
ArrayList <String> x= new ArrayList<>(Arrays.asList(input.split("")));
for(int i=0; i<x.size()-1;i++) {
if(x.get(i).equals(" ") && x.get(i+1).equals(" ")) {
x.remove(i);
i--;
}
}
String word="";
for(String each: x)
word+=each;
return word;
}
String myText = " Hello World ";
myText = myText.trim().replace(/ +(?= )/g,'');
// Output: "Hello World"
string.replaceAll("\s+", " ");
If you already use Guava (v. 19+) in your project you may want to use this:
CharMatcher.whitespace().trimAndCollapseFrom(input, ' ');
or, if you need to remove exactly SPACE symbol ( or U+0020, see more whitespaces) use:
CharMatcher.anyOf(" ").trimAndCollapseFrom(input, ' ');
public class RemoveExtraSpacesEfficient {
public static void main(String[] args) {
String s = "my name is mr space ";
char[] charArray = s.toCharArray();
char prev = s.charAt(0);
for (int i = 0; i < charArray.length; i++) {
char cur = charArray[i];
if (cur == ' ' && prev == ' ') {
} else {
System.out.print(cur);
}
prev = cur;
}
}
}
The above solution is the algorithm with the complexity of O(n) without using any java function.
Please use below code
package com.myjava.string;
import java.util.StringTokenizer;
public class MyStrRemoveMultSpaces {
public static void main(String a[]){
String str = "String With Multiple Spaces";
StringTokenizer st = new StringTokenizer(str, " ");
StringBuffer sb = new StringBuffer();
while(st.hasMoreElements()){
sb.append(st.nextElement()).append(" ");
}
System.out.println(sb.toString().trim());
}
}
I want to make strings like "a b c" to "prefix_a prefix_b prefix_c"
how to do that in java?
You can use the String method: replaceAll(String regex,String replacement)
String s = "a xyz c";
s = s.replaceAll("(\\w+)", "prefix_$1");
System.out.println(s);
You may need to tweek the regexp to meet your exact requirements.
Assuming a split character of a space (" "), the String can be split using the split method, then each new String can have the prefix_ appended, then concatenated back to a String:
String[] tokens = "a b c".split(" ");
String result = "";
for (String token : tokens) {
result += ("prefix_" + token + " ");
}
System.out.println(result);
Output:
prefix_a prefix_b prefix_c
Using a StringBuilder would improve performance if necessary:
String[] tokens = "a b c".split(" ");
StringBuilder result = new StringBuilder();
for (String token : tokens) {
result.append("prefix_");
result.append(token);
result.append(" ");
}
result.deleteCharAt(result.length() - 1);
System.out.println(result.toString());
The only catch with the first sample is that there will be an extraneous space at the end of the last token.
hope I'm not mis-reading the question. Are you just looking for straight up concatenation?
String someString = "a";
String yourPrefix = "prefix_"; // or whatever
String result = yourPrefix + someString;
System.out.println(result);
would show you
prefix_a
You can use StringTokenizer to enumerate over your string, with a "space" delimiter, and in your loop you can add your prefix onto the current element in your enumeration. Bottom line: See StringTokenizer in the javadocs.
You could also do it with regex and a word boundary ("\b"), but this seems brittle.
Another possibility is using String.split to convert your string into an array of strings, and then loop over your array of "a", "b", and "c" and prefix your array elements with the prefix of your choice.
You can split a string using regular expressions and put it back together with a loop over the resulting array:
public class Test {
public static void main (String args[]) {
String s = "a b c";
String[] s2 = s.split("\\s+");
String s3 = "";
if (s2.length > 0)
s3 = "pattern_" + s2[0];
for (int i = 1; i < s2.length; i++) {
s3 = s3 + " pattern_" + s2[i];
}
System.out.println (s3);
}
}
This is C# but should easily translate to Java (but it's not a very smart solution).
String input = "a b c";
String output (" " + input).Replace(" ", "prefix_")
UPDATE
The first solution has no spaces in the output. This solution requires a place holder symbol (#) not occuring in the input.
String output = ("#" + input.Replace(" ", " #")).Replace("#", "prefix_");
It's probably more efficient to use a StringBuilder.
String input = "a b c";
String[] items = input.Split(new[] {' '}, StringSplitOptions.RemoveEmptyEntries);
StringBuilder sb = new StringBuilder();
foreach (String item in items)
{
sb.Append("prefix_");
sb.Append(item);
sb.Append(" ");
}
sb.Length--;
String output = sb.ToString();