Develop Reference

How to write a lambda expression when you are using a string array?

I want to use a lambda expression instead of a classic for.
String str = "Hello, Maria has 30 USD.";
String[] FORMAT = {"USD", "CAD"};
final String simbol = "$";
// This was the initial implementation.
// for (String s: FORMAT) {
// str = str.replaceAll(s + "\\s", "\\" + FORMAT);
// }
Arrays.stream(FORMAT).forEach(country -> {
str = str.replaceAll(country + "\\s", "\\" + simbol);
});
// and I tried to do like that, but I receiced an error
// "Variable used in lambda expression should be final or effectively final"
// but I don't want the str String to be final
For any string, I want to change the USD or CAD in $ simbol.
How can I changed this code to work ? Thanks in advance!

I see no problem with using a loop for this. That's how I'd likely do it.
You can do it with a stream using reduce:
str = Arrays.stream(FORMAT)
.reduce(
str,
(s, country) -> s.replaceAll(country + "\\s", Matcher.quoteReplacement(simbol)));
Or, easier:
str = str.replaceAll(
Arrays.stream(FORMAT).collect(joining("|", "(", ")")) + "\\s",
Matcher.quoteReplacement(simbol));

Consider using a traditional for loop, since you're changing a global variable:
for(String country: FORMAT) {
str = str.replaceAll(country + "\\s", "\\" + simbol);
}
Using Streams in this example will make things less readable.

How to replace a static string with a variable and convert to long?

I have the following code to generate sequential Mac Addresses. The code works well if I statically define the string when creating the hextint. However, if I change the string to a variable, as seen below, I get an error:
Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "6CDFFB6000000A
"
I'm sure I'm missing something simple here. Why is the string as a variable not converting? Any help is appreciated.
String macAddr = last_mac.getText();
// int qty = Integer.parseInt(label_qty.getText());
System.out.println("String to convert: " + macAddr);
// long hexint = Long.parseLong("6CDFFB60000000", 16);
long hexint = Long.parseLong(macAddr, 16);
System.out.println("String converted to Long:" +hexint);
System.out.println("10+ previous number:" + (10+hexint));
System.out.println("Convert back to hex: " + Long.toHexString(10+hexint).toUpperCase());
for(int i = 1; i < 10+1; i++){
System.out.println(i +" MAC: " + Long.toHexString(hexint + i).toUpperCase());
macAddr = Long.toHexString(hexint + i).toUpperCase();
}

Your text contains a trailing space. You should remove it by trim() method. Try the following code & it should work for you.
String macAddr = last_mac.getText().trim();

How to remove text between brackets in multiple lines

I have a big text files and I want to remove everything that is between
double curly brackets.
So given the text below:
String text = "This is {{\n" +
"{{the multiline\n" +
"text}} file }}\n" +
"what I\n" +
"{{ to {{be\n" +
"changed}}\n" +
"}} want.";
String cleanedText = Pattern.compile("(?<=\\{\\{).*?\\}\\}", Pattern.DOTALL).matcher(text).replaceAll("");
System.out.println(cleanedText);
I want the output to be:
This is what I want.
I have googled around and tried many different things but I couldn't find anything close to my case and as soon as I change it a little bit everything gets worse.
Thanks in advance

You can use this :
public static void main(String[] args) {
String text = "This is {{\n" +
"{{the multiline\n" +
"text}} file }}\n" +
"what I\n" +
"{{ to {{be\n" +
"changed}}\n" +
"}} want.";
String cleanedText = text.replaceAll("\\n", "");
while (cleanedText.contains("{{") && cleanedText.contains("}}")) {
cleanedText = cleanedText.replaceAll("\\{\\{[a-zA-Z\\s]*\\}\\}", "");
}
System.out.println(cleanedText);
}

A regular expression cannot express arbitrarily nested structures; i.e. any syntax that requires a recursive grammar to describe.
If you want to solve this using Java Pattern, you need to do it by repeated pattern matching. Here is one solution:
String res = input;
while (true) {
String tmp = res.replaceAll("\\{\\{[^}]*\\}\\}", "");
if (tmp.equals(res)) {
break;
}
res = tmp;
}
This is not very efficient ...
That can be transformed into an equivalent, but more concise form:
String res = input;
String tmp;
while (!(tmp = res.replaceAll("\\{\\{[^}]*\\}\\}", "")).equals(res)) {
res = tmp;
}
... but I prefer the first version because it is (IMO) a lot more readable.

I am not an expert in regular expression, so I just write a loop which does this for you. If you don't have/want to use a regEx, then it could be helpful for you;)
public static void main(String args[]) {
String text = "This is {{\n" +
"{{the multiline\n" +
"text}} file }}\n" +
"what I\n" +
"{{ to {{be\n" +
"changed}}\n" +
"}} want.";
int openBrackets = 0;
String output = "";
char[] input = text.toCharArray();
for(int i=0;i<input.length;i++){
if(input[i] == '{'){
openBrackets++;
continue;
}
if(input[i] == '}'){
openBrackets--;
continue;
}
if(openBrackets==0){
output += input[i];
}
}
System.out.println(output);
}

My suggestion is to remove anything between curly brackets, starting at the innermost pair:
String text = "This is {{\n" +
"{{the multiline\n" +
"text}} file }}\n" +
"what I\n" +
"{{ to {{be\n" +
"changed}}\n" +
"}} want.";
Pattern p = Pattern.compile("\\{\\{[^{}]+?}}", Pattern.MULTILINE);
while (p.matcher(text).find()) {
text = p.matcher(text).replaceAll("");
}
resulting in the output
This is
what I
want.
This might fail when having single curly brackets or unpaired pair of brackets, but could be good enough for your case.

Insert a space after every given character - java

I need to insert a space after every given character in a string.
For example "abc.def..."
Needs to become "abc. def. . . "
So in this case the given character is the dot.
My search on google brought no answer to that question
I really should go and get some serious regex knowledge.
EDIT : ----------------------------------------------------------
String test = "0:;1:;";
test.replaceAll( "\\:", ": " );
System.out.println(test);
// output: 0:;1:;
// so didnt do anything
SOLUTION: -------------------------------------------------------
String test = "0:;1:;";
**test =** test.replaceAll( "\\:", ": " );
System.out.println(test);

You could use String.replaceAll():
String input = "abc.def...";
String result = input.replaceAll( "\\.", ". " );
// result will be "abc. def. . . "
Edit:
String test = "0:;1:;";
result = test.replaceAll( ":", ": " );
// result will be "0: ;1: ;" (test is still unmodified)
Edit:
As said in other answers, String.replace() is all you need for this simple substitution. Only if it's a regular expression (like you said in your question), you have to use String.replaceAll().

You can use replace.
text = text.replace(".", ". ");
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replace%28java.lang.CharSequence,%20java.lang.CharSequence%29

If you want a simple brute force technique. The following code will do it.
String input = "abc.def...";
StringBuilder output = new StringBuilder();
for(int i = 0; i < input.length; i++){
char c = input.getCharAt(i);
output.append(c);
output.append(" ");
}
return output.toString();

Java add chars to a string

I have two strings in a java program, which I want to mix in a certain way to form two new strings. To do this I have to pick up some constituent chars from each string and add them to form the new strings. I have a code like this(this.eka and this.toka are the original strings):
String muutettu1 = new String();
String muutettu2 = new String();
muutettu1 += this.toka.charAt(0) + this.toka.charAt(1) + this.eka.substring(2);
muutettu2 += this.eka.charAt(0) + this.eka.charAt(1) + this.toka.substring(2);
System.out.println(muutettu1 + " " + muutettu2);
I'm getting numbers for the .charAt(x) parts, so how do I convert the chars to string?

StringBuilder builder = new StringBuilder();
builder
.append(this.toka.charAt(0))
.append(this.toka.charAt(1))
.append(this.toka.charAt(2))
.append(' ')
.append(this.eka.charAt(0))
.append(this.eka.charAt(1))
.append(this.eka.charAt(2));
System.out.println (builder.toString());

Just use always use substring() instead of charAt()
In this particular case, the values are mutable, consequently, we can use the built in String class method substring() to solve this problem (#see the example below):
Example specific to the OP's use case:
muutettu1 += toka.substring(0,1) + toka.substring(1,2) + eka.substring(2);
muutettu2 += eka.substring(0,1) + eka.substring(1,2) + toka.substring(2);
Concept Example, (i.e Example showing the generalized approach to take when attempting to solve a problem using this concept)
muutettu1 += toka.substring(x,x+1) + toka.substring(y,y+1) + eka.substring(z);
muutettu2 += eka.substring(x,x+1) + eka.substring(y,y+1) + toka.substring(z);
"...Where x,y,z are the variables holding the positions from where to extract."

The obvious conversion method is Character.toString.
A better solution is:
String muutettu1 = toka.substring(0,2) + eka.substring(2);
String muutettu2 = eka.substring(0,2) + toka.substring(2);
You should create a method for this operation as it is redundant.
The string object instatiantiation new String() is unnecessary. When you append something to an empty string the result will be the appended content.

You can also convert an integer into a String representation in two ways: 1) String.valueOf(a) with a denoting an integer 2) Integer.toString(a)

This thing can adding a chars to the end of a string
StringBuilder strBind = new StringBuilder("Abcd");
strBind.append('E');
System.out.println("string = " + str);
//Output => AbcdE
str.append('f');
//Output => AbcdEf