Develop Reference

Add new line character in remote config text

I'm doing Firebase RemoteConfig integration. In one of the scenarios, I need to break a text line, so I tried to use new line character (\n).
But this is not working, it is neither displaying as an extra character nor creating another line.

My solution is replace \n manually (assuming that in Firebase Console you put property for TITLE as "Title\nNewLine"):
FirebaseRemoteConfig.getInstance().getString(TITLE).replace("\\n", "\n")

Try using an uncommon character like two pipes || and then replacing every occurance of those with a newline after you do getString() in the code.

You can insert encoded text(with Base64) to Firebase panel.
After, decode the String from your Java class and use it.
Like
byte[] data = Base64.decode(base64, Base64.DEFAULT);
String text = new String(data, "UTF-8");

The trick (which actually works for all HTML tags supported on your target platform) is to wrap the String in a JSON Object on RemoteConfig, like so:
{
"text":"Your text with linebreaks...<br><br>...as well as <b>bold</b> and <i>italic</i> text.
}
On the target platform you then need to parse the JSON and convert it back to a simple string. On Android this looks like this:
// extract value from JSON
val text = JSONObject(remoteConfig.getString("remoteConfig_key")).getString("text")
// create Spanned and use it
view.text = HtmlCompat.fromHtml(text)

So what worked for me is to use "||" (or some other character combination you are confident will not be in the string) as the new line character. Then replace "||" with "\n". This string will then display properly for me.
For some reason sending "\n" in the string doesn't get recognized as expected but adding it manually on the receiving side seems to work.

To make the suggestion mentioned above, you can try this code(that can be generalized to "n" number of elements). Simply replace the sample text with yours with the same format and add the amount of elements
String text="#Elemento1#Elemento2#Elemento3#";
int cantElementos=3;
arrayElementosFinales= new String[cantElementos];
int posicionNum0=0;
int posicionNum1;
int posicionNum2;
for(int i=0;i<cantElementos;i++){
posicionNum1=text.indexOf("#",posicionNum0);
posicionNum2=text.indexOf("#", posicionNum1+1);
char [] m = new char[posicionNum2-posicionNum1-1];
text.getChars(posicionNum1+1, posicionNum2,m,0);
arrayElementosFinales[i]=String.valueOf(m);
posicionNum0=posicionNum2;
}

Use Cdata in the remote config in combination with "br" tag and HTML.fromHtml() .. for eg.
<![CDATA[ line 1<br/>line 2]]>

is it possible to cut text from string like this?

I want to put extra value from intent to other intent. But in other intent, app get all value. Example:
mAddress.setText(" from " + address);
String put_address = mAddress.getText().toString();
editIntent.putExtra("put_address", put_address);
is it possible to cut text "from" and get only address variable ???

you can split a string like
str = "From address#dd.com";
String modified = str.replace;
now splitstr contain your split strings
splitStr[1] contains "address#dd.com"
Can also use
str.substring(str.indexOf(" ")+1);

By the way, you can use jagapathi's answer. In his example he uses regular expression.
Regular expressions can help to parse, find, cut substrings using a particular pattern. In his code he splits string by any space character.
But, imho, the simplest solution is to create a substring using this code:
'put_address.substring(7);'

use one of these solutions:
String input = put_address.trim().substring(5);
*** note: 5 is index of real address first character;
String input = put_address..split(" ")[1];

why #character is missing while i split string using comma(,) in andorid?

1.here i want same string as on xml format but while i store into array it missed # from the string and added space
<string xmlns="http://tempuri.org/">7633,abcd#123</string>
loginDetailsArray = loginDetails.split(",");
intPersonID = loginDetailsArray[0];
strPassword = loginDetailsArray[1];
separeted using comma
resultstring=abcd 123//# is missing

You can try to replace # or any special characters with equivalent sign;
However you may not change password charactres.
Your SMS API might be not allowing passing Special characters.

You does not encode character '#'. Encode it ->
#amp;

IllegalArgumentException IN ANDROID URL

Using the below url I got an error:
java.lang.IllegalArgumentException: Illegal character in path at index 47: http://safetracker-threetinker.rhcloud.com/api/{userid}/locations?lat={latitude}&lng={longitude}.
URL:
URL=http://safetracker-threetinker.rhcloud.com/api/{userid}/locations?lat={latitude}&lng={longitude}
how to solve the error. I don't have good knowledge in URL encoding. please help me to find the solution.

The problem is actually it is looking for long/integer value and you are passing a { just put a $ so that it will be replaced by the actual value pointed by the variable
http://safetracker-threetinker.rhcloud.com/api/1/locations?lat=5&lng=5
your Address is like this
URL=http://safetracker-threetinker.rhcloud.com/api/{userid}/locations?lat={latitude}&lng={longitude}
it should be like this
URL=http://safetracker-threetinker.rhcloud.com/api/${userid}/locations?lat=${latitude}&lng={longitude}

We can not use some special characters in URL, so we have to replace these special characters with its encoded form.
Replace your URL with following URL
URL=http://safetracker-threetinker.rhcloud.com/api/%7Buserid%7D/locations?lat=%7Blatitude%7D&lng=%7Blongitude%7D
May this help you.

URLEncoder should be the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.
String q = "replace_with user_id/locations?lat=replace with latitude&lng=replace with longitude";
String url = "http://safetracker-threetinker.rhcloud.com/api/=" + URLEncoder.encode(q, "UTF-8");

android java URLDecoder problem

i have a String displayed on a WebView as "Siwy & Para Wino"
i fetch it from url , i got a string "Siwy%2B%2526%2BPara%2BWino". // be corrected
now i'm trying to use URLDecoder to solve this problem :
String decoded_result = URLDecoder.decode(url); // the url is "Siwy+%26+Para+Wino"
then i print it out , i still saw "Siwy+%26+Para+Wino"
Could anyone tell me why?

From the documentation (of URLDecoder):
This class is used to decode a string which is encoded in the application/x-www-form-urlencoded MIME content type.
We can look at the specification to see what a form-urlencoded MIME type is:
The form field names and values are escaped: space characters are replaced by '+', and then reserved characters are escaped as per [URL]; that is, non-alphanumeric characters are replaced by '%HH', a percent sign and two hexadecimal digits representing the ASCII code of the character. Line breaks, as in multi-line text field values, are represented as CR LF pairs, i.e. '%0D%0A'.
Since the specification calls for a percent sign followed by two hexadecimal digits for the ASCII code, the first time you call the decode(String s) method, it converts those into single characters, leaving the two additional characters 26 intact. The value %25 translates to % so the result after the first decoding is %26. Running decode one more time simply translates %26 back into &.
String decoded_result = URLDecoder.decode(URLDecoder.decode(url));
You can also use the Uri class if you have UTF-8-encoded strings:
Decodes '%'-escaped octets in the given string using the UTF-8 scheme.
Then use:
String decoded_result = Uri.decode(Uri.decode(url));

thanks for all answers , i solved it finally......
solution:
after i used URLDecoder.decode twice (oh my god) , i got what i want.
String temp = URLDecoder.decode( url); // url = "Siwy%2B%2526%2BPara%2BWino"
String result = URLDecoder.decode( temp ); // temp = "Siwy+%26+Para+Wino"
// result = "Swy & Para Wino". !!! oh good job.
but i still don't know why.. could someone tell me?