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How do I convert string builder to string array in Android? [duplicate]

This question already has answers here:
StringBuilder to Array of Strings how to convert
(2 answers)
Closed 6 years ago.
I have a string builder in Android which is filled with data from a database, and I want to display that data using a list view. For that I want to covert string builder into an array of strings. Can somebody help me in this conversion, or suggest to me some other technique.

Here you can find Link
tempArray = sb.toString().split("ABCABC"); will split the string and return an array of strings for each line.

Try this :
String sbString = sb.ToString();
String[] ary = "abc".split("TABTAB");

How to convert a string mathematical expression into a BigInteger? [duplicate]

This question already has answers here:
How to evaluate a math expression given in string form?
(26 answers)
Closed 7 years ago.
A part of the goal in my app is to receive a mathematical expression (e.g. 1 + 1) by string, and convert it to a BigInteger.
My goal is to have an equivalent to:
BigInteger result = new BigInteger("1+1");
// This will throw an exception of invalid BigInteger
Also, ScriptEngineManager class isn't available for Android.
I still cannot find a way to achieve my goal.
Thanks a lot for helping!

You can try :
//If you have to parse only the result
String result = "2";
BigInteger.valueOf(Long.valueOf("2");
//If you have to parse the whole operation
String result = "1+1";
//create an algo to extract each 1 and determine operation
BigInteger bigIntUn = new BigInteger("1");
BigInteger bigIntAnotherUn = new BigInteger("1");
//you have add (+), min(-), multiply(*) (etc...)
bigIntUn.add(bigIntAnotherUn);

Comparing strings that use the substring method [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 8 years ago.
I hope somebody could give me an explaination why the below code wont work:
//Why doesnt this work
String l = myString.substring(cut, lengthLastBtn-1);
String c = myString.substring(cut, lengthLastBtn-1);
if(l==c){
Log.i(TAG, "Correct");
}
//End
//This work!
String l = "hi";
String c = "hi";
if(l==c){
Log.i(TAG, "Correct");
}
// End
// Or if i want the Vars as in the first code i have to use the if statement like this
if(l.contains(c)){
Log.i(TAG, "Correct");
}
//End
So, why cant a compare a string when i have used the substring method on it. I even see in the log for the strings that they are the same, or have the same text at least.

When you use the “==“ operator with String`s, it means a comparison between objects, not the value that objects hold.
In order to compare Strings values , you should use the built-in method equals. The result is true if the String object represents the same sequence of characters.
if(string1.equals(string2)) {
//Match
}

convert string array into integer array android [closed]

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Closed 9 years ago.
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I created a url string array.
String urls[]={"http://www.kllhjj.png","http://yui.kl.png"};
this url are not exactly correct. But in my code them are correct. Then I try to convert them into integer array just like following way.
int a[] = new int[urls.length];
for(int i=0; i<a.length; i++) {
try {
a[i] = Integer.parseInt(urls[i]);
}
catch(Exception e) { }
}
But here always show the a integer values as 0. why is it?
help me

Because Integer.parseInt(urls[i]); is throwing NumberFormatException and you are swallowing the Exception . The below code will not work in your case, but at least you will get to know the error:
try{
a[i]=Integer.parseInt(urls[i]);
}catch(Exception e){
e.printStackTrace();
throw new RunTimeException(e);
}
All the elements of a primitive int array are defaulted to 0. Hence you get the 0s .
You cannot parse string like "http://yui.kl.png" etc to int as they are not in numeric format . Read the documentation:
Throws:
NumberFormatException - if the string does not contain a parsable integer.

You are getting the NumberFormatException Exception hence not changing any values in a[] and because the default values of integer is 0 you are having 0 for every element in a[]

Urls are text string and you parsing text to int:
a[i]=Integer.parseInt(urls[i]);
which through an exception absorbed by:
catch(Exception e)
and default value for int is zero so u always getting it.

ParseInt is a function that search for integer in some string. it's like atoi if you're familiar with c language.
When you have a string that contains only letters it will return 0.
When you have a string that begins with letter and ends with number it also will return 0.
It will convert a string to integer if the string will contains a number characters like "0123".
I don't know what you're trying to do, but if you want that number will represent a url.
I think you can create an enum for the urls.

Android if statement does not fire? [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 8 years ago.
Why does my "if" statement not fire?
String something = "";
String category = json_data.getString("den");
Log.e("JSON", "category="+category);
if (category == "1"){
something = "Random something";
}
Even if in my logcat I can see JSON:"category=1", the "something" String does not take "Random something" value.
This must be some convention in java?
Please help.

use:
if (category.equals("1")){
something = "Random something";
}
or better way to avoid null pointer exception:
if ("1".equals(category)){
something = "Random something";
}
Also have a look at this link for detail : How do I compare strings in Java?

You need to do string comparison in Java using the .equals method on the String object. The comparison you are doing is only comparing the references not the actual string values.
Example:
if(category.equals("1")){
//do amazing stuff
}