Conditional Validation for parameter algorithm in isHash function

[KUSHAGRARAJTIWARI]

1. Added utility function to check presence of given key in given object
2. Conditional validation for parameter algorithm in isHash function

Issue: #2295

Checklist

• ✅ PR contains only changes related; no stray files, etc.
• ✅ README updated (where applicable)
• ✅ Tests written (where applicable)

[codecov]

Codecov Report

All modified and coverable lines are covered by tests ✅

Project coverage is 99.95%. Comparing base (b958bd7) to head (2ade081).
Report is 17 commits behind head on master.

Additional details and impacted files

@@           Coverage Diff           @@
##           master    #2296   +/-   ##
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  Coverage   99.95%   99.95%           
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  Files         107      108    +1     
  Lines        2454     2458    +4     
  Branches      619      622    +3     
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+ Hits         2453     2457    +4     
  Partials        1        1           

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[Daniyal-Qureshi]

Suggested change

	if (object[key] === undefined) {
	if(object[key]) {
	return true;
	}
	throw new Error(errorMessage);

[KUSHAGRARAJTIWARI]

@Daniyal-Qureshi
Thanks for your suggestion. But, if(object[key]) will result false if object[key] is any falsy value like 0, "" (empty string), null, etc. This will defeat the purpose of the function validateKey.

I used a better way of checking the key of any object. Please provide your suggestions on that code.

[Daniyal-Qureshi]

@KUSHAGRARAJTIWARI Previously, you were checking the value of the given key,I suggested the better way to check if the value is undefined, I think you need to check whether is key is present in the object or not.

[KUSHAGRARAJTIWARI]

Yes, you are correct. I changed the code and am now using the hasOwnProperty() method. This method sends true if the key is present in the object irrespective of the corresponding value. Using Javascript utility functions is good for code.

After your comment, I recollected that there is a function hasOwnProperty() to check if any key is present in the object or not. Thanks for your comment.