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U can go to : https://developer.android.com
and download necessary software needed to develop android apps
(Free of charge, of course)
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I have developing a app.in this app i want to point a location in google map.please help me i want code develop in eclipse
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Help with what? You need to give clearer details of what your code is doing and where the problem lies.
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See here.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
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I got a code written in java on my Android phone as java program only. That works fine, but its not an app. Just a program. I then created an android app with the same code. But that wont work. It has a textbox to input numbers into an arraylist, then I have a HashSet which takes the data from the arraylist. Then finds duplicates via collections.frequency but that part doesn't work if the data is input by user via textbox. It puts out: 24 shows that many times: 1. However, when I disable the userinput. And create the numbers in the code like so: list.add("24"); list.add("33"), and so forth. It then picks a number randomly and put it out like so: 24 shows that many times: 2
regardless if other numbers in there are also duplicates. Its picks a number randomly. Unlike in eclipse or even on the phone as just java no Android involed it puts out any number which is a duplicate like so: 24 shows that many times:2. 33 shows that many times: 4. 39 shows that many times: 3. It puts out any number which is a duplicate. Not so in android. It pick one number only and that randomly.
Any help is appreciated.
Thanks
Newbie88
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You need to show the code that is failing and explain where the error occurs.
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K, here is the android code:
ArrayList<string> = new ArrayList<>();
String item = inputText.getText().toString();
ll.add(item);
HashSet<string> set = new HashSet<>(ll);
for (String temp : set)
{answerText.setText(temp + "shows that many times : " + Collections.frequency(ll, temp));
//the output as follows:
33 44 33 44 shows that many times: 1
Basically not finding any duplicates.
However, if I disable the userinput textbox and replace it with manual input like so:
ll.add("33");
ll.add("44");
ll.add("33");
ll.add("44");
ll.add("24");
ll.add("24");
the output will be this: 44 shows that many times: 2
Well, that's finding duplicates. But only for one number. Why? Why not putting out all duplicate numbers? Like it does in Java as a pure Java code without android involved?
That code is this:
List<string> list = new ArrayList<string>();
Scanner stdin = new Scanner(System.in);
System.out.println("Enter the amount of numbers you want to input: Input numbers separated
by a space.");
int n = stdin.nextInt();
for (int i = 0; i < n; i++)
{
list.add(stdin.next());
}
System.out.println("\nCount all with frequency");
Set<string> uniqueSet = new HashSet<string>(list);
for (String temp : uniqueSet) {
System.out.println(temp + " shows that many times : " + Collections.frequency(list, temp));
//Enter the amount of numbers you want to input
12 //hit the return key
22 33 44 22 33 44 22 33 44 22 33 44
//the output is like so:
Count all with frequency
33 shows that many times: 4
44 shows tham many times: 4
22 shows that many times: 4
That's how I want it in android as well. With userinput via textbox if possible. Unfortunately the scanner input is not working in android.
Please can you help me?
Thanks
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I put your Android code into an activity, like:
public class MainActivity extends Activity
{
EditText inputText = null;
ArrayList<String> ll = null;
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
ll = new ArrayList<String>();
inputText = (EditText) findViewById(R.id.inputText);
Button add = (Button) findViewById(R.id.add);
add.setOnClickListener(new OnClickListener()
{
@Override
public void onClick(View v)
{
String item = inputText.getText().toString();
ll.add(item);
inputText.setText("");
}
});
Button results = (Button) findViewById(R.id.results);
results.setOnClickListener(new OnClickListener()
{
@Override
public void onClick(View v)
{
android.util.Log.d("Test", "List has " + ll.size() + " items.");
HashSet<String> set = new HashSet<String>(ll);
for (String temp : set)
android.util.Log.d("Test", temp + " was found " + Collections.frequency(ll, temp) + " times.");
}
});
}
}
Entering the numbers:
33
44
33
44
24
24
produced the following output:
List has 6 items.
44 was found 2 times.
24 was found 2 times.
33 was found 2 times.
So, I'm not sure what else you have in your Android code that is causing it not to work.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
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Hello, can anyone tell me how can i export my DB information in a file. I tried to save file from file Explorer in Eclipse DDMS.
but it must be in a File restriction and that's why i cannot saving the file 
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Does the database belong to your app?
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
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I googled on the keywords Android and widgets, hoping to find clear definition of a widget. But all the definitions are vague and could be applied to any apps.
My feeling is that a widget is a small app that runs in the background at all times, as a separate thread. Usually, it displays something or waits for input at all times. As opposed to an app that has to be started/terminated explicitly when the need arises. If this is the definition of a widget then it is the oldest concept in multi-tasking operating system where such concept used to be called a background process. If this is so, then why it is said that widgets are unique to Android. I bet iPhone has them too but under a different name. What is so unique about widgets?
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A widget is a visual element (e.g., textview, button, spinner). See here for more.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
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Hello,
Widget in Android is an application compoment to pin on phone screen.
Its provide fast communication between user and app.
It also enables fast data display from the application because user can see it on the phone screen without opening the app.
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Hello everybody;
I installs a digital certificate from globalsign in my smartphone.I wants to use it for signing a PDF documents but i don't know how to exploit it in my android application to do my purpose .There is someone who know how to do it and thanks.
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Have you seen this?
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
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Thanks for the answer ,I want to develop an android application which use this certficate in order to sign a PDF document . There is someone who know how to do that !
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I don't know now how to introduce it in android application by using java code
Especially when i want to call it i find no way to do it please help me because it's urgent.
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Hi..
I have to develop an android app for a given website,in which we write and read data from website, I searched on too many sites , but did not get anything other than Webview , I have to implement app without using webview... can any one help me to do this plzz??
thanx in advance
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It rather depends on what help you need. We cannot do your work for you, but will try to help resolve problems with code that you have written. Your question lacks any real detail that could help us to help you, so I can only suggest you look at some of the articles at http://www.codeproject.com/KB/android/#Android+Tutorial+Contest[^] as a first step.
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Hello,
Could you be more precised?
What exactly do you need?
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how do we provide security for an web app connecting to server in android
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Need help I have tried everything, I have installed IntelliJ reinstalled (5 times) Eclipse finally got it working.
Took advice of ppl here and also build path changed Workspace and I still get R.cannot be resolved in my Eclipse Juno I checked my main.xml corrected my Android Manifest XML and i am still stuck with this error can someone pls help me?
I will make a deal, I have a library of over 100 plus eBooks, on Java, Android SQL,JavaScript Scala, PHP C# you name it, if someone helps me resolve this issue and the code works I will send you any ebook you would like.
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Have a look in your project's generated files directory tree, (mine is C:\Users\My Documents\Android\BaseTest\gen\com\example\basetest) for a file named R.java. If you don't have one then your project is not getting built correctly. Try a manual build and look at the messages that get produced.
BTW please don't offer 'incentives' to get people to help you; it looks cheap.
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