So I have the following string:
String text = "\t\t\torder #168\n\t\t\tpaid\n\t\t\tview 4 items\n\t\t\tpicked up\n\t\t\tcomplete pickup\n\t\t\t2 stops";
How do I parse this string so that I always get the 2 in front of stops? I have tried the following, but it always returns 2 stops.
String substr = "complete pickup";
String numberOfStops = text.substring(text.indexOf(substr) + substr.length());
numberOfStops = numberOfStops.replaceAll("^\\s+","").replaceAll("\\s+$","");
The short way:
numberOfStops = numberOfStops.replaceAll("^\\s+","").replaceAll("\\s+$","").replace("stops","");
The flexible way is using Regex, and Pattern and Match classes. Let me know if you need it
I want use register in my application and i should send password and verifyCode with SMS to users phones.
But i should read verifyCode from message and set automatically number into verifyCode EditText.
My message format :
Hi, welcome to our service.
your password 12345
your verifyCode 54321
How can i do it? Please help me <3
Assuming that the number of digits are fixed in password and verify codes (Generally they are same as default values), We can extract digits from the string and then find substring which has verify code. This assumption is for simplicity.
String numberOnly= str.replaceAll("[^0-9]", "");
String verifyCode = numberOnly.substring(6);
Here String verifyCode = numberOnly.substring(6); is getting last 5 digits of the string which is your verification code. You can also write numberOnly.substring(6,10); to avoid confusions.
But this is prone to errors like StringIndexOutOfBoundsException, So whenever you want to get substring which is starting from index i till the end of the string, always write numberOnly.substring(i).
There are a lot ways to do this. You can use some complicated regex or use a simple spilt method.
Try this,
String str = "Hi, welcome to our service.\n"
+ "\n"
+ "your password \n"
+ "12345\n"
+ "\n"
+ "your verifyCode \n"
+ "54321";
// Solution #1
String[] parts = str.split("\n");
System.out.println(parts[3]);
System.out.println(parts[6]);
// Solution #2
String PAT = "(password|verifyCode)\\s+(\\d+)";
Pattern pats = Pattern.compile(PAT);
Matcher m = pats.matcher(str);
while (m.find()) {
String grp = m.group(2);
System.out.println(grp);
}
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")
I try to get only this part "9916-4203" in "Region Code:9916-4203 " in android. How can I do this?
I tried below code, I used substring method but it doesn't work:
firstNumber = Integer.parseInt(message.substring(11, 19));
If you know that string contains "Region Code:" couldn't you do a replace?
message = message.replace("Region Code:", "");
Assumed that you have only one phone number in your String, the following will remove any non-digit characters and parse the resulting number:
public static int getNumber(String num){
String tmp = "";
for(int i=0;i<num.length();i++){
if(Character.isDigit(num.charAt(i)))
tmp += num.charAt(i);
}
return Integer.parseInt(tmp);
}
Output in your case: 99164203
And as already mentioned, you won't be able to parse any String to Integer in case there are any non-digit characters
Im going to guess that what you want to extract is the full region code text minus the title. So maybe using regex would be a good simple fit for you?
String myString = "Region Code:9916-4203";
String match = "";
String pattern = "\:(.*)";
Pattern regEx = Pattern.compile(pattern);
Matcher m = regEx.matcher(myString);
// Find instance of pattern matches
Matcher m = regEx.matcher(myString);
if (m.find()) {
match = m.group(0);
}
Variable match will contain "9916-4203"
This should work for you.
Java code sourced from http://android-elements.blogspot.in/2011/04/regular-expressions-in-android.html
In Java the substring() method works with the first parameter being inclusive and the second parameter being exclusive. Meaning "Hello".substring(0, 2); will result in the string He.
In addition to excluding the parsing of something that isn't a number like #Opiatefuchs mentioned, your substring method should instead be message.substring(12, 21).
Hi I am trying to test the edit Text placing 2 different values but the second test is failing .reasons unknown...below is my code
TestCase1
public void testvalues1() {
// clearing the edit text
mTextView.clearComposingText();
TouchUtils.tapView(this, mTextView);
// sending input as 7
sendKeys("7");
String userInput1;
String expected = "158269.3778";
String parameterFrom1 = "0.0027";
String parameterTo1 = "61.04676";
// getting the input from the mTextView reference
userInput1 = mTextView.getText().toString();
String resultset = UnitCalculation.Converter(parameterFrom1,userInput1,parameterTo1);
assertEquals(resultset, expected);
}
In the above test case iam sending value 7 and output is as expected
TestCase2
public void testvalues2() {
// clearing the edit text
mTextView.clearComposingText();
TouchUtils.tapView(this, mTextView);
// sending input as 23
sendKeys("23");
String userInput1;
String expected = "150.5011";
String parameterFrom1 = "1.092607";
String parameterTo1 = "7.149502";
// getting the input from the mTextView reference
userInput1 = mTextView.getText().toString();
String resultset1 = UnitCalculation.Converter(parameterFrom1,userInput1,parameterTo1);
System.out.println("printing resilt set "+ resultset1);
assertEquals(resultset1, expected);
}
But the method is returning value 0 instead of 150.5011
Iam using the same methos to calculate, When i give User value hardcoded like this String userInput1="23"; it is working, but when is taking the value from edittext its is not working.
can i send multiple values to edit text on the same testfile??
sendKeys with more than one character is what messed it up. See this reference.
To sum up, sendKeys needs a String which contains space separated keys. Your sendKeys("23")is trying to find a key in the soft keyboard called 23, though there isn't. Try using this:
sendKeys("2 3");
As it will send these two key strokes individually, instead of trying to find a key named 23. That's why sending just a 7 worked, because 7 is the key name for "7".