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Find out a given word in a String using Regex in java [closed]

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Given a String,
Using regex in java or java code i have to find out Electors name from the given String :
"K ~\n" +
"m W swim\n" +
"sfiaruqsr\n" +
"wvnn 021m: r’ i\" W' _ ~\n" +
"_ 4'“; x ‘ D\n" +
"W ma ¢ “ii-a? “Rm qwfl\n" +
"Electors name ; Moleslwar Moreshva;\n" +
"Tuwal\n" +
"mam-a #02,qu _\n" +
"F\n" +
"Fm“ Name : Momma! Tuppal I\n" +
"‘ e\n" +
"Pam sq» : w! MALE '\n" +
"“WW/Dale m mm ; XX/xxnsae _‘"
Like for the given String the output should be : Moleslwar Moreshva Tuwal
Can't hardcode anything as the String response changes everytime.
Please help:)

There is no name end delimiter hence will have to assume the full name needs to be captured from the line where Electors name is present and the next line only. And once it is captured, replace unwanted character with a space.
Here is the java code that parses the name for you,
public static void main(String[] args) {
String s = "K ~\n" +
"m W swim\n" +
"sfiaruqsr\n" +
"wvnn 021m: r’ i\" W' _ ~\n" +
"_ 4'“; x ‘ D\n" +
"W ma ¢ “ii-a? “Rm qwfl\n" +
"Electors name ; Moleslwar Moreshva;\n" +
"Tuwal\n" +
"mam-a #02,qu _\n" +
"F\n" +
"Fm“ Name : Momma! Tuppal I\n" +
"‘ e\n" +
"Pam sq» : w! MALE '\n" +
"“WW/Dale m mm ; XX/xxnsae _‘";
Pattern p = Pattern.compile("Electors name\\s*;\\s*([\\w; ]+\\n\\w+)");
Matcher m = p.matcher(s);
if (m.find()) {
System.out.println(m.group(1).replaceAll(";\\n", " "));
} else {
System.out.println("Didn't match");
}
}
This prints following output,
Moleslwar Moreshva Tuwal

Evaluating "IF ELSE" shorthand as part of a toString() Implementation [closed]

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I have write the following toString() methode
public String toString() {
return "Product: "+ this.productName + ", Barcode: " + this.barCode
+ ", Expiration Date: " + this.expirationDate.toString() + ", Customer Price: "
+ this.customerPrice + ", Shops Price: " + this.shopsPrice
+ ", Instore Amount: " + this.inStoreAmount + ", Sale: "
+ (this.sale == null) ? "Not on sale" : + this.sale.toString();
}
But there is a problem with the way I use the if statement.
eclipse: "cannot covert from string to boolean"

You had an issue with the balancing of concatenation operator +. Also, I edited your method to be the following:
public String toString() {
return "Product: "+ this.productName + ", Barcode: " + this.barCode
+ ", Expiration Date: " + this.expirationDate.toString() + ", Customer Price: "
+ this.customerPrice + ", Shops Price: " + this.shopsPrice
+ ", Instore Amount: " + this.inStoreAmount + ", Sale: "+ ((this.sale == null) ? "Not on sale" : this.sale.toString());
}
When you are writing IF-ELSE short hand, try to keep everything in a (...) set of parentheses to keep track of things easily. I don't care what other professionals say about this, but if this helps you to understand things, so be it!

This is illegal syntax
(this.sale == null) ? "Not on sale" : + this.sale.toString()
and a compilation error should have alerted you to this problem.
Instead, use
((this.sale == null) ? "Not on sale" : this.sale.toString())
I've placed the entire ternary operator into parentheses to be clear.

Java NewLine "\n" not working in save to text file [closed]

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Here is my code:
public String display() {
return "\n......................\nFixed Employee:\n" +
"Name: " + super.fullName() +
"\nSalary: " + salary() +
" tk\n......................";
}
But when I'm invoking this method from main class, "\n" newLine not working. just showing one line output. Will you plz help to solve the problem?
Thanks

For saving in files use \r\n. \n as new lines is viable on printstreams but not writing to files.

You may need the system independent line separator as it might differ from one OS to another. Just replace the \n with the value of line separator:
I can be retrieve as you load any system property:
public String display() {
String separator = System.getProperty("line.separator"); // Load the system property using its key.
return "\n......................\nFixed Employee:\n"
+ "Name: "
+ super.fullName() +
"\nSalary: "
+ salary()
+ " tk\n......................"
.replace("\\n", separator); // replace the \n before returning your String
}
Or simply use System#lineSeparator method as #Deepanshu Bedi suggested:
public String display() {
String separator = System.lineSeparator(); // Consider it as a shortcut.
return "\n......................\nFixed Employee:\n"
+ "Name: "
+ super.fullName() +
"\nSalary: "
+ salary()
+ " tk\n......................"
.replace("\\n", separator); // replace the \n before returning your String
}

What are some better ways to print the following section? [closed]

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What are some better ways to print the following section? I think it will look better in a table format of some type with a heading. I want to adjust them to fixed lenghts, I think. Here is the code snippet I want to format to look better. The print elements are elements in a sql database.
System.out.println("\nAll records in your table:");
System.out.println("ID# Name GPA Status Mentor Level Thesis Advisor Company"); //table heading...
while (rs.next()) {
String output = " ";
output += rs.getString("studentID") + " "
+ rs.getString("firstName") + " "
+ rs.getString("lastName") + " "
+ rs.getString("gpa") + " "
+ rs.getString("status") + " "
+ rs.getString("mentor") + " "
+ rs.getString("level") + " "
+ rs.getString("thesisTitle") + " "
+ rs.getString("thesisAdvisor") + " "
+ rs.getString("company") + "\n";
System.out.printf("%s", output);
}

|You can|
|Do something|
|like this|
|studentname|
|firstname|
|middlename|
|.......|

"I want to adjust them to fixed lengths":
is that why your concatenating whitespaces in different lengths ?
You can try to concatenate "\t\t" instead.
In case it's HTML, using HTML Table will take care of it for you.

Find URL in String

hi im tring to find a URL in a string, i founded many topics about this using regex but i have a problem. Using this pattern:
String regex = "\\b(((ht|f)tp(s?)\\:\\/\\/|~\\/|\\/)|www.)" +
"(\\w+:\\w+#)?(([-\\w]+\\.)+(com|org|net|gov" +
"|mil|biz|info|mobi|name|aero|jobs|museum" +
"|travel|[a-z]{2}))(:[\\d]{1,5})?" +
"(((\\/([-\\w~!$+|.,=]|%[a-f\\d]{2})+)+|\\/)+|\\?|#)?" +
"((\\?([-\\w~!$+|.,*:]|%[a-f\\d{2}])+=?" +
"([-\\w~!$+|.,*:=]|%[a-f\\d]{2})*)" +
"(&(?:[-\\w~!$+|.,*:]|%[a-f\\d{2}])+=?" +
"([-\\w~!$+|.,*:=]|%[a-f\\d]{2})*)*)*" +
"(#([-\\w~!$+|.,*:=]|%[a-f\\d]{2})*)?\\b";
Its works pretty well in most of pages, but i have an issue with other. For example:
http://hello.com/hello world
returns
http://hello.com/hello
The problems is that space.
Anyone have a nice pattern that solve this?
Thanks.
EDIT:: this is my code
private ArrayList<String> pullLinks(String text) {
ArrayList<String> links = new ArrayList<String>();
String regex = "\\b(((ht|f)tp(s?)\\:\\/\\/|~\\/|\\/)|www.)" +
"(\\w+:\\w+#)?(([-\\w]+\\.)+(com|org|net|gov" +
"|mil|biz|info|mobi|name|aero|jobs|museum" +
"|travel|[a-z]{2}))(:[\\d]{1,5})?" +
"(((\\/([-\\w~!$+|.,=]|%[a-f\\d]{2})+)+|\\/)+|\\?|#)?" +
"((\\?([-\\w~!$+|.,*:]|%[a-f\\d{2}])+=?" +
"([-\\w~!$+|.,*:=]|%[a-f\\d]{2})*)" +
"(&(?:[-\\w~!$+|.,*:]|%[a-f\\d{2}])+=?" +
"([-\\w~!$+|.,*:=]|%[a-f\\d]{2})*)*)*" +
"(#([-\\w~!$+|.,*:=]|%[a-f\\d]{2})*)?\\b";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(text);
while(m.find()) {
String urlStr = m.group();
if (urlStr.startsWith("(") && urlStr.endsWith(")"))
{
urlStr = urlStr.substring(1, urlStr.length() - 1);
}
links.add(urlStr);
}
return links;
}

Spaces are not allowed in URLs (they need to be replaced by %20). See for instance the answer to this question:
Spaces in URLs?
If you allow URLs to include spaces anyway, then how would you interpret for instance http://www.google.com/ig is a nice webpage? Clearly the part after /ig should not be included!

Space is not a valid URL character.
Also, if you don't use whitespace as your terminator how are you going to find the end of the URL?
Your regex is also failing to account for other top level domains (like .int). I'm not actually sure why it is looking for specific TLDs at all as they are not required to form a valid URL.