Android Return Line (\n) in String - android

In android, a String resource can be created in XML by using
<string name="name">data</string>
to create a new line, \n may be used. However, if the user personally inputs data into an EditText and includes a \n, the writing is saved to a String and when the writing is displayed again in a TextView the result would be something such as:
Line 1\nLine2
How come the \n doesn't apply to user written Strings?
The end goal for me is to be able to have the user type a \n into a String to make sure the String is displayed in multiple lines.
How would this be achieved? Thanks in advance!

Use this :
TextView view = your text view;
view.setSingleLine(false);
view.setText("first line" + "\n" + "second line" + "\n" + "third line");

How are you doing setText() in the textView.
Try this...
TextView view = your text view;
view.setSingleLine(false);
view.setText("first line\n"+"second line\n"+"third line");

with
System.getProperty("line.separator")
you get a String which results in a new line.
I would replace \n by this expression.

The text entered in an input field is used as an actual text. That means, when the user enters the Hello \n World and the code calls getText().toString(), what will actually be returned is: Hello \\n World that means, it's the actual slash symbol and not the "carriage return" symbol.
I don't believe android has any built-in API to give you the actual code typed by the user, that means, that you have to write that code yourself. If all you need is the \n as a carriage return it is simple:
getText().toString().replace("\\n", "\n");

Related

how to split text by "\n\n" Android?

In my string.xml I have a text which contains 2 paragraphs "\n\n". When I try to split it by String.split("\n\n")
or String.split("\\n\\n") it does not work. Knows anywhone the reason?
and here it is the sample data:
it displays well on the screen
val content = resources.getString(R.string.test).split("\n \n ")
strings.xml
<string name="test">first
\n
\n
second</string>
You should add gaps after \n. Use my pattern for spliting and it'll work.
Your text appears to actually have literal \n values in it. In that case, you may try splitting on \\n (which becomes \\\\n in Java syntax):
String yourText = "Part one \\n Part two";
String[] parts = yourText.split("\\\\n");
System.out.println(Arrays.toString(parts));
This prints:
[Part one , Part two]

JSON string to TextView newline

First of all, I have gone through questions similar to the problem I am facing and those solutions are not working for me.
I have a TextView field on my Android app which is supposed to display multiple paragraphs i.e multiple new lines. I am getting this string from a database present in my online server as a JSON.
The text contains \n in it and I am expecting it to create new lines once it is received by the app. But it displays the whole text without any breaks along with "\n" character.
Below is the text present in my database.
First line. \nSecond line. \nThird line.
JSON string received by me inside the app.
{
"server_response": [{
"news_expand": "First line. \\nSecond line. \\nThird line."
}]
}
Code to extract string from JSON. I have left out the code to get get JSONArray and JSONObject for simplicity.
na_expand = gna_jo.getString("news_expand");
String extracted from the JSON. Got this by printing the na_expand string.
First line. \nSecond line. \nThird line.
Code to display the text in the TextView. Note the below 'na_expand' is an SparseArray present in a different activity hence the 'get(position)' code.
art_expand.setText(na_expand.get(position));
Below is the text I get on the emulator.
First line. \nSecond line. \nThird line.
What am I doing wrong here?
I think you should replace \n with \n in your string before setting test to your textview same below
b= b.replaceAll("\\n","\n");
So I found a workaround to the problem. As I was not sure where the issue was happening with \n, I modified my text present in the database to have a symbol other than \n. For eg: ~
First line.~Second line.~Third line.
You can use a website like this - https://www.gillmeister-software.com/online-tools/text/remove-line-breaks.aspx to replace the line breaks with any symbol you want.
Next, I used the StringSplitter class to break the string received in JSON and then again join it together with \n.
String joined;
String expand_temp = na_expand.get(position);
TextUtils.StringSplitter splitter = new TextUtils.SimpleStringSplitter('~');
splitter.setString(expand_temp);
StringBuilder stringBuilder = new StringBuilder();
for (String s_temp : splitter) {
stringBuilder.append(s_temp + "\n");
}
joined = stringBuilder.toString().trim();
This worked! I used this string in setText.
art_expand.setText(joined);
Try below code
myTextView.setText(Html.fromHtml("yourString with additional html tags"));
It will resolve all the html tags accordingly and effects of the tags will be reflected as well.
NOte: For devices greater than Nougat use below code
myTextView.setText(Html.fromHtml("<h2>Title</h2><br><p>Description here</p>", Html.FROM_HTML_MODE_COMPACT));
Hope that helps
The \ character is an escape character in JSON. So, when you get \\n, it actually means \n, not the newline character, which should have been just \n. So what you see is an expected behaviour. The JSON you get should have ideally been:
{
"server_response": [{
"news_expand": "First line. \nSecond line. \nThird line."
}]
}
Get your server to respond properly, otherwise you'll have to strip the unnecessary \.
Do you haveandroid:singleLine="true" on your TextView? If yes it will ignore the \n and will place the text in a single line.
You can just add replaceAll("\\n","\n") when you set value to your art_expand EditText. It should be:
art_expand.setText(na_expand.get(position).replaceAll("\\n","\n"));

How to format a string for a textview android

I have a TextView and when I call .setText(string), it displays the string exactly as it is.
For instance, if the string contains " ' " , ' will be displayed.
How can I fix this?
Looks like your outside source already does the convertion from ' to ' since regular single quote will be displayed just fine in a TextView. What you can to do in this case is:
yourTextView.setText(Html.fromHtml(stringSource));
where stringSource is your string that comes from outside source. This will take care of more than just the single quote character.
The single quote ' in java needs to be escaped using a backslash (\). You can find other characters that need to be escaped here.

How to insert a new line in strings in Android

I'm creating an android application and within it there is a button that will send some information in an email, and I don't want to have everything all in one paragraph.
Here's what my app is doing for the putExtra for the email's body:
I am the first part of the info being emailed. I am the second part. I am the third part.
Here's what I want it to do:
I am the first part of the info being emailed.
I am the second part.
I am the third part.
How would I put a new line into a string or with the putExtra method to accomplish that?
Try:
String str = "my string \n my other string";
When printed you will get:
my string
my other string
Try using System.getProperty("line.separator") to get a new line.
I would personally prefer using "\n". This just puts a line break in Linux or Android.
For example,
String str = "I am the first part of the info being emailed.\nI am the second part.\n\nI am the third part.";
Output
I am the first part of the info being emailed.
I am the second part.
I am the third part.
A more generalized way would be to use,
System.getProperty("line.separator")
For example,
String str = "I am the first part of the info being emailed." + System.getProperty("line.separator") + "I am the second part." + System.getProperty("line.separator") + System.getProperty("line.separator") + "I am the third part.";
brings the same output as above. Here, the static getProperty() method of the System class can be used to get the "line.seperator" for the particular OS.
But this is not necessary at all, as the OS here is fixed, that is, Android. So, calling a method every time is a heavy and unnecessary operation.
Moreover, this also increases your code length and makes it look kind of messy. A "\n" is sweet and simple.
I use <br> in a CDATA tag.
As an example, my strings.xml file contains an item like this:
<item><![CDATA[<b>My name is John</b><br>Nice to meet you]]></item>
and prints
My name is John
Nice to meet you
If you want to add line break at runtime into a String from same string you are deriving the value then this Kotlin code works for me:
str = "<br>"+str?.replace("," , "</br><br>")+"</br>"
value = HtmlCompat.fromHtml(${skill_et_1}",Html.FROM_HTML_MODE_LEGACY)
tv.text = value

How to add a line break in an Android TextView?

I am trying to add a line break in the TextView.
I tried suggested \n but that does nothing. Here is how I set my texts.
TextView txtSubTitle = (TextView)findViewById(r.id.txtSubTitle);
txtSubTitle.setText(Html.fromHtml(getResources().getString(R.string.sample_string)));
This is my String: <string name="sample_string">some test line 1 \n some test line 2</string>
It should show like so:
some test line 1
some test line 2
But it shows like so: some test line 1 some test line 2.
Am I missing something?
\n works for me, like this:
<TextView android:text="First line\nNext line"
ok figured it out:
<string name="sample_string"><![CDATA[some test line 1 <br />some test line 2]]></string>
so wrap in CDATA is necessary and breaks added inside as html tags
Android version 1.6 does not recognize \r\n.
Instead, use: System.getProperty("line.separator")
String s = "Line 1"
+ System.getProperty("line.separator")
+ "Line 2"
+ System.getProperty("line.separator");
Linebreaks (\n) only work if you put your string resource value in quotes like this:
<string name="sample_string">"some test line 1 \n some test line 2"</string>
It won't do linebreaks if you put it without quotes like this:
<string name="sample_string">some test line 1 \n some test line 2</string>
yes, it's that easy.
Tried all the above, did some research of my own resulting in the following solution for rendering linefeed escape chars:
string = string.replace("\\\n", System.getProperty("line.separator"));
Using the replace method you need to filter escaped linefeeds (e.g. '\\n')
Only then each instance of line feed '\n' escape chars gets rendered into the actual linefeed
For this example I used a Google Apps Scripting noSQL database (ScriptDb) with JSON formatted data.
Cheers :D
There are two ways around this.
If you use your string as a raw string, you need to use the newline
character. If you use it as html, e.g. by parsing it with Html.fromString,
the second variant is better.
1) Newline character \n
<string name="sample">This\nis a sample</string>
2) Html newline tag <br> or <br />
<string name="sample">This<br>is a sample</string>
This worked for me
android:text="First \n Second"
This worked for me, maybe someone will find out this helpful:
TextView textField = (TextView) findViewById(R.id.textview1);
textField.setText("First line of text" + System.getProperty("line.separator") + "Linija 2");
If you're using XML to declare your TextView use android:singleLine = "false" or in Java, use txtSubTitle.setSingleLine(false);
Used Android Studio 0.8.9. The only way worked for me is using \n.
Neither wrapping with CDATA nor <br> or <br /> worked.
I use the following:
YOUR_TEXTVIEW.setText("Got some text \n another line");
very easy : use "\n"
String aString1 = "abcd";
String aString2 = "1234";
mSomeTextView.setText(aString1 + "\n" + aString2);
\n corresponds to ASCII char 0xA, which is 'LF' or line feed
\r corresponds to ASCII char 0xD, which is 'CR' or carriage return
this dates back from the very first typewriters, where you could choose to do only a line feed (and type just a line lower), or a line feed + carriage return (which also moves to the beginning of a line)
on Android / java the \n corresponds to a carriage return + line feed, as you would otherwise just 'overwrite' the same line
As I know in the previous version of android studio uses separate lines " \n " code. But new one (4.1.2) uses "<br/" to separate lines. For example -
Old one:
<string name="string_name">Sample text 1 \n Sample text 2 </string>
New one:
<string name="string_name">Sample text 1 <br/> Sample text 2 </string>
Also you can add "<br/>" instead of \n.
It's HTML escaped code for <br/>
And then you can add text to TexView:
articleTextView.setText(Html.fromHtml(textForTextView));
Try to double-check your localizations.
Possible, you trying to edit one file (localization), but actually program using another, just like in my case. The default system language is russian, while I trying to edit english localization.
In my case, working solution is to use "\n" as line separator:
<string name="string_one">line one.
\nline two;
\nline three.</string>
You could also use the String-Editor of Android Studio, it automatically generates line brakes and stuff like that...
As Html.fromHtml deprecated I simply I used this code to get String2 in next line.
textView.setText(fromHtml("String1 <br/> String2"));
.
#SuppressWarnings("deprecation")
public static Spanned fromHtml(String html){
Spanned result;
if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.N) {
result = Html.fromHtml(html,Html.FROM_HTML_MODE_LEGACY);
} else {
result = Html.fromHtml(html);
}
return result;
}
The most easy way to do it is to go to values/strings (in your resource folder)
Declare a string there:
<string name="example_string">Line 1\Line2\Line n</string>
And in your specific xml file just call the string like
<TextView
android:id="#+id/textView"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="#string/example_string" />
I found another method:
Is necessary to add the "android:maxWidth="40dp"" attribute.
Of course, it may not work perfectly, but it gives a line break.
\n was not working for me. I was able to fix the issue by changing the xml to text and building the textview text property like below.
android:text="Line 1
Line 2
Line 3
DoubleSpace"
Hopefully This helps those who have said that \n did not work for them.
I'm reading my text from a file, so I took a slightly different approach, since adding \n to the file resulted in \n appearing in the text.
final TextView textView = (TextView) findViewById(R.id.warm_up_view);
StringBuilder sb = new StringBuilder();
Scanner scanner = new Scanner(getResources().openRawResource(R.raw.warm_up_file));
while (scanner.hasNextLine()) {
sb.append(scanner.nextLine());
sb.append("\n");
}
textView.setText(sb.toString());
In my case, I solved this problem by adding the following:
android:inputType="textMultiLine"
Maybe you are able to put the lf into the text, but it doesn't display? Make sure you have enough height for the control. For example:
Correct:
android:layout_height="wrap_content"
May be wrong:
android:layout_height="10dp"
I feel like a more complete answer is needed to describe how this works more thoroughly.
Firstly, if you need advanced formatting, check the manual on how to use HTML in string resources.
Then you can use <br/>, etc. However, this requires setting the text using code.
If it's just plain text, there are many ways to escape a newline character (LF) in static string resources.
Enclosing the string in double quotes
The cleanest way is to enclose the string in double quotes.
This will make it so whitespace is interpreted exactly as it appears, not collapsed.
Then you can simply use newline normally in this method (don't use indentation).
<string name="str1">"Line 1.
Line 2.
Line 3."</string>
Note that some characters require special escaping in this mode (such as \").
The escape sequences below also work in quoted mode.
When using a single-line in XML to represent multi-line strings
The most elegant way to escape the newline in XML is with its code point (10 or 0xA in hex) by using its XML/HTML entity
or
. This is the XML way to escape any character.
However, this seems to work only in quoted mode.
Another method is to simply use \n, though it negatively affects legibility, in my opinion (since it's not a special escape sequence in XML, Android Studio doesn't highlight it).
<string name="str1">"Line 1.
Line 2.
Line 3."</string>
<string name="str1">"Line 1.\nLine 2.\nLine 3."</string>
<string name="str1">Line 1.\nLine 2.\nLine 3.</string>
Do not include a newline or any whitespace after any of these escape sequences, since that will be interpreted as extra space.
I would recommend querying the line.separator property, and using that whenever you want to add a line break.
Here is some sample code:
TextView calloutContent = new TextView(getApplicationContext());
calloutContent.setTextColor(Color.BLACK);
calloutContent.setSingleLine(false);
calloutContent.setLines(2);
calloutContent.setText(" line 1" + System.getProperty ("line.separator")+" line2" );

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