[Solved] How to do offline maps in android? [closed]

You have the option of using OpenStreet Map via osmdroid. osmdroid is a (almost) full/free replacement for Android’s MapView (v1 API) class. It also includes a modular tile provider system with support for numerous online and offline tile sources and overlay support with built-in overlays for plotting icons, tracking location, and drawing shapes. 4 solved … Read more

[Solved] ” Missing ; before statement ” In a long code [closed]

I have updated the function… check now? navigator.geolocation.getCurrentPosition(function(position){ var long = position.coords.longitude; var lat = position.coords.latitude; var theDateC = new Date(); var D = (367*theDateC.getFullYear())-(parseInt((7/4)*(theDateC.getFullYear+parseInt((theDateC.getMonth()+9)/12))))+parseInt(275*(theDateC.getMonth()/9))+theDateC.getDate()-730531.5; var L = 280.461+0.9856474*D; var M = 357.528+0.9856003*D; var Lambda = L +1.915*Math.sin(M)+0.02*Math.sin(2*M); var Obliquity = 23.439-0.0000004*D; var Alpha = Math.atan (Math.cos(Obliquity)*Math.tan(Lambda)); Alpha = Alpha – (360 * parseInt(Alpha/360)); Alpha … Read more

[Solved] How to see who access my ip address from web service via wifi [closed]

to find informations about visitor request , please check you xampp apache log file . here is the code you can add in your index.php to deny access for some adress ip : <?php $deny = array(“111.111.111”, “222.222.222”, “333.333.333”); if (in_array ($_SERVER[‘REMOTE_ADDR’], $deny)) { header(“location: http://www.google.com/”); exit(); } ?> add the adress ip in $deny … Read more

[Solved] Adding data dynamically from one json object to another

You’re pretty much going to need a server-side process if you want to save your changes. You can load the JSON via ajax: $.ajax({ url: “/path/to/friends.json”, dataType: “json”, success: function(data) { // Here, `data` will be the object resulting from deserializing the JSON // Store `data` somewhere useful, perhaps you might have a `friends` // … Read more

[Solved] Error “no such element: Unable to locate element”

The email field is inside the frame. Before access any element in the frame, you have to switch. please try following code. public static WebElement Email_Field(WebDriver driver) throws InterruptedException { WebElement element; (new WebDriverWait(driver, 30)).until(ExpectedConditions.frameToBeAvailableAndSwitchToIt(“alibaba-login-box”)); element = driver.findElement(By.xpath(“//input[@id=’fm-login-id’]”)); while (!isDisplayed(element)) { Thread.sleep(3000); System.out.println(“Element is not visible yet”); } return element; } solved Error “no such … Read more

[Solved] how can parse this xml file? [closed]

try This try { InputStream mInputStream = getAssets().open(“XmlData.xml”); XmlPullParserFactory xmlFactoryObject = XmlPullParserFactory.newInstance(); XmlPullParser myparser = xmlFactoryObject.newPullParser(); myparser.setInput(mInputStream, null); int event = myparser.getEventType(); while (event != XmlPullParser.END_DOCUMENT) { String name=myparser.getName(); switch (event){ case XmlPullParser.START_TAG: break; case XmlPullParser.END_TAG: if(name.equals(“page”)){ Log.e(“TAG”,””+ myparser.getAttributeValue(null,”id”)); Log.e(“Page content “,””+ myparser.getAttributeValue(null,”text”)); } break; } event = myparser.next(); } } catch (Exception e) { … Read more

[Solved] Can Java delimit input itself, without explicit delimiters?

Assuming you’re using scanner, yes, it could. The scanner operates on the notion that a regexp serves as delimiter: Each match of the regex delimits, and whatever the regexp matches is tossed out (because nobody ‘cares’ about reading the spaces or the commas or whatever). The scanner then gives you stuff in between the delimiters. … Read more

[Solved] Asking the user for a URL to receive a JSON

Create a constructor in your async Task private class JSONTask extends AsyncTask<String, String, String> { String url; public JSONTask(String url){ this.url=url; } use the url string in place of params[0] And wherever you call your async task do it like this new JSONTask(textView.getText()).execute() This should solve it. Else you can directly use the do in … Read more

[Solved] Java Switch Statement for operators

Try out this : package com.sujit; import java.util.Scanner; public class UserInput { public static void main(String[] args) { Scanner input = new Scanner(System.in); boolean flag = true; do { System.out.println(“Enter 1st number”); int num1 = input.nextInt(); System.out.println(“Enter 2nd number”); int num2 = input.nextInt(); System.out.println(“select one operator :\n 1)+\n2)-\n3)*\n4)/\n5)Exit(Enter E)\n”); System.out.println(“Enter your choice :”); char choice … Read more

[Solved] My android application is getting stopped

Go to the your Manifest file and do this in the activity where you want to show the Toolbar <activity android:name=”.YourActivity” android:theme=”@style/AppTheme.NoActionBar”><!– ADD THIS LINE –> Then in styles.xml add the following: <style name=”AppTheme.NoActionBar”> <item name=”windowActionBar”>false</item> <item name=”windowNoTitle”>true</item> </style> solved My android application is getting stopped

[Solved] Choosing language at startup

If I understand you correctly you wish to display that activity only when the user runs the app for the first time. Well, here’s what you can do: 1) Get a handle to a SharedPreference. This is to store if the user has already selected the language or not. SharedPreferences sharedPref = getActivity().getPreferences(Context.MODE_PRIVATE); 2) Create … Read more

[Solved] How to remove duplicate from string/mixed list

I think this is what you are trying to achieve: x = [] while True: data = input() if data.lower() == “done”: break if data not in x: x.append(data) Note the use of while True and break to avoid having two input calls. Alternatively, use a set: x = set() while True: data = input() … Read more