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I want to print a Forward Slash, but i am getting a warning , how to solve [duplicate]

I am setting text using setText() by following way.
prodNameView.setText("" + name);
prodOriginalPriceView.setText("" + String.format(getString(R.string.string_product_rate_with_ruppe_sign), "" + new BigDecimal(price).setScale(2, RoundingMode.UP)));
In that First one is simple use and Second one is setting text with formatting text.
Android Studio is so much interesting, I used Menu Analyze -> Code Cleanup and i got suggestion on above two lines like.
Do not concatenate text displayed with setText. Use resource string
with placeholders. less... (Ctrl+F1)
When calling TextView#setText:
Never call Number#toString() to format numbers; it will not handle fraction separators and locale-specific digits properly. Consider
using String#format with proper format specifications (%d or %f)
instead.
Do not pass a string literal (e.g. "Hello") to display text. Hardcoded text can not be properly translated to other languages.
Consider using Android resource strings instead.
Do not build messages by concatenating text chunks. Such messages can not be properly translated.
What I can do for this? Anyone can help explain what the thing is and what should I do?

Resource has the get overloaded version of getString which takes a varargs of type Object: getString(int, java.lang.Object...). If you setup correctly your string in strings.xml, with the correct place holders, you can use this version to retrieve the formatted version of your final String. E.g.
<string name="welcome_messages">Hello, %1$s! You have %2$d new messages.</string>
using getString(R.string.welcome_message, "Test", 0);
android will return a String with
"Hello Test! you have 0 new messages"
About setText("" + name);
Your first Example, prodNameView.setText("" + name); doesn't make any sense to me. The TextView is able to handle null values. If name is null, no text will be drawn.

Don't get confused with %1$s and %2$d in the accepted answer.Here is a few extra information.
The format specifiers can be of the following syntax:
%[argument_index$]format_specifier
The optional argument_index is specified as a number ending with a “$” after the “%” and selects the specified argument in the argument list. The first argument is referenced by "1$", the second by "2$", etc.
The required format specifier is a character indicating how the argument should be formatted. The set of valid conversions for a given argument depends on the argument's data type.
Example
We will create the following formatted string where the gray parts are inserted programmatically.
Hello Test! you have 0 new messages
Your string resource:
< string name="welcome_messages">Hello, %1$s! You have %2$d new
messages< /string >
Do the string substitution as given below:
getString(R.string.welcome_message, "Test", 0);
Note:
%1$s will be substituted by the string "Test"
%2$d will be substituted by the string "0"

I ran into the same lint error message and solved it this way.
Initially my code was:
private void displayQuantity(int quantity) {
TextView quantityTextView = (TextView) findViewById(R.id.quantity_text_view);
quantityTextView.setText("" + quantity);
}
I got the following error
Do not concatenate text displayed with setText. Use resource string with placeholders.
So, I added this to strings.xml
<string name="blank">%d</string>
Which is my initial "" + a placeholder for my number(quantity).
Note: My quantity variable was previously defined and is what I wanted to append to the string. My code as a result was
private void displayQuantity(int quantity) {
TextView quantityTextView = (TextView) findViewById(R.id.quantity_text_view);
quantityTextView.setText(getString(R.string.blank, quantity));
}
After this, my error went away. The behavior in the app did not change and my quantity continued to display as I wanted it to now without a lint error.

Do not concatenate text inside your setText() method, Concatenate what ever you want in a String and put that String value inside your setText() method.
ex: correct way
int min = 120;
int sec = 200;
int hrs = 2;
String minutes = String.format("%02d", mins);
String seconds = String.format("%02d", secs);
String newTime = hrs+":"+minutes+":"+seconds;
text.setText(minutes);
Do not concatenate inside setText() like
text.setText(hrs+":"+String.format("%02d", mins)+":"+String.format("%02d", secs));

You should check this thread and use a placeholder like his one (not tested)
<string name="string_product_rate_with_ruppe_sign">Price : %1$d</string>
String text = String.format(getString(R.string.string_product_rate_with_ruppe_sign),new BigDecimal(price).setScale(2, RoundingMode.UP));
prodOriginalPriceView.setText(text);

Don't Mad, It's too Simple.
String firstname = firstname.getText().toString();
String result = "hi "+ firstname +" Welcome Here";
mytextview.setText(result);

the problem is because you are appending "" at the beginning of every string.
lint will scan arguments being passed to setText and will generate warnings, in your case following warning is relevant:
Do not build messages by
concatenating text chunks. Such messages can not be properly
translated.
as you are concatenating every string with "".
remove this concatenation as the arguments you are passing are already text. Also, you can use .toString() if at all required anywhere else instead of concatenating your string with ""

I fixed it by using String.format
befor :
textViewAddress.setText("Address"+address+"\n"+"nCountry"+"\n"+"City"+"city"+"\n"+"State"+"state")
after :
textViewAddress.setText(
String.format("Address:%s\nCountry:%s\nCity:%s\nState:%s", address, country, city, state));

You can use this , it works for me
title.setText(MessageFormat.format("{0} {1}", itemList.get(position).getOppName(), itemList.get(position).getBatchNum()));

If you don't need to support i18n, you can disable this lint check in Android Studio
File -> Settings -> Editor -> Inspections -> Android -> Lint -> TextView Internationalization(uncheck this)

prodNameView.setText("" + name); //this produce lint error
val nameStr="" + name;//workaround for quick warning fix require rebuild
prodNameView.setText(nameStr);

I know I am super late for answering this but I think you can store the data in a varible first then you can provide the variable name. eg:-
// Java syntax
String a = ("" + name);
String b = "" + String.format(getString(R.string.string_product_rate_with_ruppe_sign);
String c = "" + new BigDecimal(price).setScale(2, RoundingMode.UP));
prodNameView.setText(a);
prodOriginalPriceView.setText(b, c);

if it is textView you can use like that : myTextView.text = ("Hello World")
in editText you can use myTextView.setText("Hello World")

Extra java input validation for strings

I want to make this so that short inputs can still be detected, such as "Londo" and "Lon", but want to keep it small and use it without basically copying and pasting the code, any tips? thank you.
if (Menu.answer1.equals("London"))
{
if (location.equals("London")) {
System.out.print(location + " ");
System.out.print(date + " ");
System.out.print(degrees + "C ");
System.out.print(wind + "MPH ");
System.out.print(winddirection + " ");
System.out.print(weather + " ");
System.out.println("");
}

You can use startsWith()
String city = "London";
if (city.startsWith("Lon")) {
// do something
}

Also if you need to check some substring, you can use contains method:
Menu.answer1 = "London";
Menu.answer1.contains("ondo"); // true

If you want to check against a fixed set of alternatives, you may use a list of valid inputs using contains:
List<String> londonNames = Arrays.asList("London", "Londo", "Lon");
if (londonNames.contains(Menu.answer1)) {
...
}

You can use (case-insensitive) regex to do the same, e.g.:
(?)Lon[a-z]{0,3} where
(?) = case insensitivity
Lon = Initial 3 characters
[a-z]{0,3} = any number of alphabets between 0 and 3
Here's an example:
String regex = "(?)Lon[a-z]{0,3}";
System.out.println("London".matches(regex));
System.out.println("Lond".matches(regex));
System.out.println("Lon".matches(regex));

If the underlying problem is that the user can enter one of several names, and you want to allow abbreviations, then a fairly standard approach is to have a table of acceptable names.
Given the user input, loop through the table testing "does the table entry start with the string typed by the user?" (like one of the previous answers here). If yes, then you have a potential match.
Keep looking. If you get a second match then the user input was ambiguous and should be rejected.
As a bonus, you can collect all names that match, and then use them in an error message. ("Pick one of London, Lonfoo, Lonbar").
This approach has the advantage (compared to a long chain of if-then-else logic) of not requiring you to write more code when all you want to do is have more data.
It automatically allows the shortest unique abbreviation, and will adjust when a once-unique abbreviation is no longer unique because of newly-added names.

Regex to parse multiline data

I have a following data from a file and I would like to see if I can do a regex parsing here
Name (First Name) City Zip
John (retired) 10007
Mark Baltimore 21268
....
....
Avg Salary
70000 100%
Its not a big file and the entire data from the file is available in a String object with a new line characters (\n) (String data = "data from the file")
I am trying to get name, city, zip and then the salary, percentage details
data inside () considered part of Name field.
For Name field space is considered valid and there are no space for other fields.
'Avg Salary' is available only at the end of the file
Will it be easy to do this via regex parsing in Java?

If the text file is space-aligned, you can (and probably should) extract the fields based on the number of characters. So, you take the first n characters in each line as first name, the next m characters as City, and so on.
This is one code to extract using the above method, by calculating the field length of the fields automatically, assuming we know the header.
String data = "data from the file";
// This is just to ensure we have enough space in the array
int numNewLines = data.length()-data.replace("\n","").length();
String[][] result = new String[numNewLines][3];
String[] lines = data.split("\n");
int avgSalary = 0;
int secondFieldStart = lines[0].indexOf("City");
int thirdFieldStart = lines[0].indexOf("Zip");
for(int i=1; i<lines.length; i++){
String line = lines[i].trim();
if(line.equals("Avg Salary")){
avgSalary = Integer.parseInt(lines[i+1].substring(0,secondFieldStart).trim());
break;
}
result[i-1][0] = line.substring(0,secondFieldStart).trim(); // First Name
result[i-1][1] = line.substring(secondFieldStart,thirdFieldStart).trim(); // City
result[i-1][2] = line.substring(thirdFieldStart).trim(); // Zip
}
Using regex will be possible, but it will be more complicated. And regex won't be able to differentiate person's name and city's name anyway:
Consider this case:
John Long-name Joe New York 21003
How would you know the name is John Long-name Joe instead of John Long-name Joe New if you don't know that the length of the first field is at most 20 characters? (note that length of John Long-name Joe is 19 characters, leaving one space between it and New in New York)
Of course if your fields are separated by other characters (like tab character \t), you can split each line based on that. And it's easy to modify the code above to accommodate that =)
Since the solution I proposed above is simpler, I guess you might want to try it instead =)

Getting Week for a given number from a String of the form `SunMonTueWedThuFriSat`

Ok, I am stumped here. I really looked into this and cant seem to find specifically what I am trying to do. I am two weeks into Java in school and new to programming so please pardon any errors I may make.
I am working on a string manipulation program and among other tasks in it I am to create a sting called names that holds SunMonTueWedThuFriSat, now I need to figure out how to grab and display in JOptionPane the 3 characters in that string that go with the numbers I am supposed to associate with them (0=Sun, 1=Mon, etc.) So if the user inputs 2 it should display Tue. I have a basic understanding on how to display those characters in a string, but for the life of me i cant seem to figure out how to associate those numbers with those characters. Every time I try to work something I keep getting errors and frustration.
Thanks for all your help!!!

If the days of the week has to be in a single string, you can parse the string for each 3 character substring, and store then in an ArrayList. Then, the provided number will match the index of the Day of the Week in the List.

You should be able to just use substring, since all the abbreviations are three characters.
String s = "SunMonTueWedThuFriSat";
Prompt the user to enter a number (0-6):
String in = JOptionPane.showInputDialog("Please input a number (0-6)");
Now convert it to a number
int choice = Integer.parseInt(in);
And select the right portion of the string:
String day = s.substring(choice*3, choice*3+3);
Now just show this day in a JOptionPane:
JOptionPane.showMessageDialog(null, "Information", "You chose: " + day,
JOptionPane.INFORMATION_MESSAGE));

Check HashMap.
For example:
HashMap<int, String> aWeekMap = new HashMap();
aWeekMap.put(0, "Mon");
aWeekMap.put(1, "Tue");
System.out.println(aWeekMap.get(0)); // prints Mon.
In memory you can visualize as following:

If you can use an array of string, it will be easier:
String[] names = {"Sun","Mon","Tue","Wed","Thu","Fri","Sat"};
int input = 2;
System.out.println("The day is " + names[input]);

use the follwing code
private ArrayList arr=new ArrayList();
arr.Add(0,names.subString(0,2);
arr.Add(0,names.subString(3,5);
arr.Add(0,names.subString(6,8);
arr.Add(0,names.subString(9,11);
arr.Add(0,names.subString(12,14);
arr.Add(0,names.subString(15,17);
arr.Add(0,names.subString(18,20);
and
JOptionPane.showInputDialog(null, "Please choose a name", "Example 1",
JOptionPane.QUESTION_MESSAGE, null, arr);

string index out of range exception

I'm looking for help correcting an exception error for 'string index out of range'. My code is supposed to take two strings as input from the user(string1 and string2) and create new strings that are parts of the originals.
So far I have the following:
modString1 = string1.substring(string1.length() -3, string1.length());
modString2 = string2.substring(0,3);
The above code is supposed to take the last 3 characters of string1 and the first 3 characters of string2. The problem I am having comes when the user inputs a string that is shorter than 3 characters.
I'm wondering if there is a way to check the input and add a character (x for example) if the string is too short?
For example, if the user enters 'A' for the first string it will change the string to 'xxA' and if 'A' is entered for the second string it will change that to 'Axx'?

Put an if statement before your code, checking the length of the string before you process it.
For example:
if(string1.length() < 3) {
// Add characters to the string
}

I'm wondering if there is a way to check the input and add a character (x for example) if the string is too short?
What you are looking for is called padding.
It can be done in a number of ways. The simplest is probably to use an external library such as Apache's StringUtils. You could also write a padding method yourself using a StringBuilder.
Related:
How can I pad a String in Java?

put the validation like below and add the string.
For ex.
if(string1.length()<3){
String op = 'xx';
string1 += op;
}