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What is Dynamic XPath in Selenium?

Dynamic XPath is also called as custom XPath and it is one way to locate element uniquely.  

Dynamic XPath is used to locate exact attribute or decrease the number of matching nodes/result from a webpage and following XPath expressions can be used for the same:

  • Basic XPath
  • Using ‘OR’ & ‘AND’
  • Using Contains()
  • Using Starts-With
  • Using Text()
  • Using Index
  • Using Chained XPath
  • XPath Axes
  • Following
  • Following Sibling
  • Preceding
  • Preceding Sibling
  • Child
  • Parent
  • Descendants
  • Ancestors

Example

.//*[@class='product']//h4[contains(text(),'Text')]//ancestor::div[@class='table-good']
/li[@class="b"]//select[@name="s"]//option[contains (text(), 'title')]/preceding-sibling::option
//*[text()=’Text']//ancestor::div[@class='table-goods']

Given an array of ints length 3, return the sum of all the elements.

Check below code:

package com.test.arrays;

public class ThreeElementsSum 
{
	public static void main(String args[])
	{
		int sum = ThreeElementsSum.sum3(new int[] {1,2,3});
		System.out.println(sum);
	}
	
	public static int sum3(int[] nums)
	{
		return nums[0] + nums[1] + nums[2];
	}
}

Output

6

Given two strings, a and b, return the result of putting them together in the order abba, e.g. "Hi" and "Bye" returns "HiByeByeHi".

Check below code:

package com.test.strings;

public class abba 
{
	public static void main(String args[])
	{
		System.out.println(abba.makeAbba("Hi","Bye"));
		System.out.println(abba.makeAbba("Programs","Buzz"));
	}

	public static String makeAbba(String a, String b)
	{
		return(a+b+b+a);
	}
}

Output

HiByeByeHi
ProgramsBuzzBuzzPrograms

Modify and return the given map as follows: if the key "a" has a value, set the key "b" to have that value, and set the key "a" to have the value "". Basically "b" is a bully, taking the value and replacing it with the empty string.

Check below code:

package com.test.collections;

import java.util.HashMap;
import java.util.Map;

public class MapBully 
{
	public static void main(String args[])
	{
		MapBully mb = new MapBully();
		HashMap<String, String> hm = new HashMap<String, String>();
		hm.put("a","candy");
		hm.put("b","dirt");
		System.out.println(mb.mapBully(hm));
	}

	public Map<String, String> mapBully(HashMap<String, String> map) 
	{
        if(map.containsKey("a"))
        {
        	map.put("b", map.get("a"));
        	map.put("a", "");
        }
        return map;
    }
}

Output

{a=, b=candy}

Given 2 arrays of ints, a and b, return true if they have the same first element or they have the same last element. Both arrays will be length 1 or more.

Check Below Code:

package com.test.arrays;

public class CommonArrayEnd 
{
	public static void main(String args[])
	{
		CommonArrayEnd ca = new CommonArrayEnd();
		System.out.println(ca.commonEnd(new int[] {1, 2, 6}, new int[] {1, 2, 6}));
		System.out.println(ca.commonEnd(new int[] {1, 2, 3}, new int[] {7, 3, 2}));
		System.out.println(ca.commonEnd(new int[] {1, 2, 3}, new int[] {1}));
	}
	
	public boolean commonEnd(int[] a, int[] b) 
	{
		return (a[0] == b[0] || a[a.length-1] == b[b.length-1]);
	}
}

Output

true
false
true

Given a string name, e.g. "Bob", return a greeting of the form "Hello Bob!".

Check below code:

package com.test.strings;

public class HelloName 
{
	public static void main(String args[])
	{
		HelloName hn = new HelloName();
		System.out.println(hn.helloName("Ram"));
		System.out.println(hn.helloName("Sham"));
		System.out.println(hn.helloName("X"));
	}
	
	public String helloName(String name) 
	{
		 return "Hello " + name + "!";
	}
}

Output

Hello Ram!
Hello Sham!
Hello X!

Given an array of ints, return true if the array is length 1 or more, and the first element and the last element are equal.

Check below code:

package com.test.arrays;

public class FirstLastEqual 
{
	public static void main(String args[])
	{
		FirstLastEqual fl = new FirstLastEqual();
		System.out.println(fl.sameFirstLast(new int[] {1, 2, 1}));
		System.out.println(fl.sameFirstLast(new int[] {1, 2, 3, 1}));
		System.out.println(fl.sameFirstLast(new int[] {1, 2, 3}));
	}
	
	public boolean sameFirstLast(int[] nums) 
	{
		if(nums.length >= 1 && nums[0] == nums[nums.length-1])
			return true;
		else
			return false;	  
	}
}

Output

true
true
false

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