Convert a string into mathematical function of x :f(x)? [closed] - android

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Hello I'm a beginner in Android programming.
I'm working on a graphic calculator but i still have the problem of converting the function to math
For example :
y=cos(x^2)-ln(x)
should look like
y=Math.cos(x*x) - Math.log(x)
And than we plot it.
Thank You

I assume you want to convert a String to a float or double?
String input = "3.14";
float x = Float.parseFloat(input);
This code converts the String "3.14" into a float. You can use this float in the function you described in the question.
EDIT:
If you want to convert the full function String into Java code, you should limit the input to buttons. For example, add a button named "Cos", and remember in the code the user wants to use the Cosinus function. Then the user enters a value, and presses "Enter". The code knows it should use the cosinus function, and x is the value entered in an EditText. That value should be parsed using Float.parseFloat().

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Regular expression for song name in android [closed]

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In my application I want to display the song's name. Now it is showing the entire song path, like this
String song_name = /storage/Music/Lean on.mp3
but I want to display only the song's name, that means
String song_name = Lean on.mp3
Can any one help me with the regular expressions used for extracting the name of song from the file path?
You can use subString function with combination of LastInedexOf function.
please find below example
String path=":/storage/Music/Lean on.mp3";
String filename=path.substring(path.lastIndexOf("/")+1);

search text in a website programmatically [closed]

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i want my app to search a specific website(from an url) for a specific text and give me the 3 chars after that text or to search for a pattern with a placeholder and give me the first matching string. Is that possible? There is no need to display the website.
Once you download the page using something like HTTPUrlConnection, you can use a regex with your search term
Pattern p = Pattern.compile("specific text(\w\w\w)");
Matcher m = p.matcher(site_text);
boolean b = m.matches();
The three \w will be captured in a group for you to use if there's a match.

how do i user pattern matches like search engine [closed]

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I want to generate Search Suggestions that matches the input in the AutoCompleteTextView.
It will check for the matches for the first word I type then all the related words should be populated.
After I type next character then it will match combination of that word and populate the result .
I do not understand how to do pattern matches.
For example: Suppose, there are three country names.
INDIA, IRAQ, IRAN.
When I type I then I want get All Country names Starting with I to be displayed.
When I type IR then result should be IRAN,IRAQ. ETC.
Your requirement will be satisfied in AutoCompleteTextView itself.
Threshold is used to set from the length where auto complete show it's suggestion:
For ex:
String[] languages={"INDIA", "IRAQ","IRAN"};
AutoCompleteTextView text(AutoCompleteTextView) findViewById(R.id.autoCompleteTextView1);
ArrayAdapter adapter = new ArrayAdapter(this,android.R.layout.simple_list_item_1,languages);
text.setAdapter(adapter);
text.setThreshold(1);

How do I return an int from EditText and than show that number in Toast? [closed]

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I am a beginner at programming Android, please help with this simple project.
I want an integer to show up in toast, which I write in textBox. I do not know what to put in the highlighted part here.
Sorry for the link, but I cannot post the picture
You need to use the getText() method to get the text the user entered.
Change the myEdit declaration to:
final EditText myEdit = ... ;
and then use:
Toast.makeText(MainActivity.this, myEdit.getText().toString(), ...

adding string as integer [closed]

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in my application i have got 0.00 in edit text.next when i enter 2or3 or any digit its getting concat with 0.00 and getting 0.003.
if(item.equals("0.00")){
item=item+string;}
i changed the code.but this time too its getting an error
if(item.equals("0.00")){
int a=Integer.parseInt(item.toString());
int b=Integer.parseInt(string.toString());
int c=a+b;
item=String.valueOf(c);}
If you are using decimal places then you will want to use a data type that supports those, such as double. Try this:
double d = Double.parseDouble(item.toString());
NOTE: If both your variables are of string type, then you won't need to use toString()
You want O.OO but still using Integer. Try using FLOAT.
Also, FLOAT is NEVER a perfect value. So you might have .XX added.
You have an error, because "0.00" has decimal places, so it is parsed as a floating point number, not integer.
Parse it as a double:
String item = "0.00";
double result = Double.parseDouble(item) + 0.03; // result is 0.03

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