JavaScript-Developer-I合格問題、JavaScript-Developer-I資格復習テキスト

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JavaScript-Developer-I認定試験はたいへん難しい試験ですね。しかし、難しい試験といっても、試験を申し込んで受験する人が多くいます。なぜかと言うと、もちろんJavaScript-Developer-I認定試験がとても大切な試験ですから。IT職員の皆さんにとって、この試験のJavaScript-Developer-I認証資格を持っていないならちょっと大変ですね。この認証資格はあなたの仕事にたくさんのメリットを与えられ、あなたの昇進にも助けになることができます。とにかく、これは皆さんのキャリアに大きな影響をもたらせる試験です。こんなに重要な試験ですから、あなたも受験したいでしょう。

Salesforce JavaScript-Developer-I 認定試験の出題範囲：

トピック出題範囲
トピック 1
  • Debugging and Error Handling: This section will measure the ability of JavaScript programmer to properly handle errors and debug JavaScript code. You will be expected to apply debugging techniques, such as using the console and breakpoints, to resolve issues. Proper error handling is crucial, and the exam will evaluate your skills in managing and correcting errors efficiently within your code.
トピック 2
  • Objects, Functions, and Classes: In this section, the focus is on applying object-oriented principles, functions, and classes. As a Salesforce JavaScript programmer, you will need to locate optimal function implementations and understand how to structure objects and classes for business solutions. You will also be tested on your ability to use JavaScript modules and decorators, while analyzing variable scope and execution flow.
トピック 3
  • Browser and Events: The topic will challenge your ability to work with events, event handlers, and the DOM. You will need to demonstrate how to manage event propagation and utilize browser-specific APIs. Additionally, proficiency in using browser developer tools to investigate code behavior will be tested, making it crucial for you to understand browser interactions thoroughly.

Salesforce JavaScript-Developer-I認定は、開発者がSalesforceプラットフォームでJavaScriptのスキルと専門知識を紹介する絶好の機会です。この認定は、JavaScriptを使用してカスタムアプリケーションを開発した経験があり、Salesforce開発のスキルと知識を検証したい専門家向けに設計されています。

>> JavaScript-Developer-I合格問題 <<

JavaScript-Developer-I資格復習テキスト、JavaScript-Developer-I試験番号

JavaScript-Developer-I学習教材があれば、あなたは自分の夢を叶えます。JavaScript-Developer-I学習教材はすごく人気があります。全世界のお客様からいい評価をもらいました。なんといっても、自分はJavaScript-Developer-I学習教材を利用したら、その資料のよさを感じることができます。大切なのは、JavaScript-Developer-I学習教材の合格率が高いので、多くのお客様はJavaScript-Developer-I認定試験資格証明書を取得したということです。

Salesforce Certified JavaScript Developer (JS-Dev-101) 認定 JavaScript-Developer-I 試験問題 (Q86-Q91):

質問 # 86
Corrected code:
let a = " * " ;
let b = " ** " ;
// x = 3;
console.log(a);
What is displayed when the code executes?

正解：C

解説：
The correct answer is B because variable a is declared and initialized before it is printed.
let a = " * " ;
This creates a block-scoped variable named a and assigns it the string value:
" * "
The line:
// x = 3;
is a comment. JavaScript ignores comments during execution, so this line has no effect.
Then this line runs:
console.log(a);
Since a already exists and contains " * " , the console displays:
* Option A is incorrect because a is defined.
Option C is incorrect because a has an assigned value, so it is not undefined.
Option D is incorrect because a was never assigned null.
Therefore, the verified answer is B .


質問 # 87
Refer to the code below:
let inArray = [ [1, 2], [3, 4, 5] ];
Which two statements result in the array [1, 2, 3, 4, 5]?
(With corrected typing errors: usArray # inArray, .. # ....)

正解：A、B

解説：
We start with the array:
let inArray = [ [1, 2], [3, 4, 5] ];
This is an array of two inner arrays:
* First element: [1, 2]
* Second element: [3, 4, 5]
The desired result is to transform this into a single, flat array:
[1, 2, 3, 4, 5]
This is accomplished by using Array.prototype.concat to concatenate the inner arrays into one new array, optionally combined with the spread syntax (...) or Function.prototype.apply.
* Option A: [].concat(...inArray);
Relevant concepts:
* Spread syntax (...inArray) expands an iterable into separate arguments.
* Array.prototype.concat joins arrays or values into a new array. When you pass arrays as arguments to concat, it flattens them one level into the result.
Step-by-step behavior:
* inArray is [ [1, 2], [3, 4, 5] ].
* ...inArray expands into [1, 2] and [3, 4, 5] as separate arguments.
* So [].concat(...inArray) is equivalent to:
* [].concat([1, 2] , [3, 4, 5]);
* concat processes each argument:
* For [1, 2], it adds 1 and 2 into the result array.
* For [3, 4, 5], it adds 3, 4, and 5 into the result array.
The final outcome is:
[1, 2, 3, 4, 5]
Therefore, Option A correctly produces the desired array.
* Option B: [].concat.apply(inArray, [] );
Corrected name from usArray to inArray.
Relevant concepts:
* Function.prototype.apply(fnThis, argsArray) invokes a function with a specific this value and a list of arguments passed as an array.
Here:
* thisArg is inArray.
* argsArray is [] (no actual arguments passed).
So the call:
[].concat.apply(inArray, [] );
is equivalent to:
Array.prototype.concat.apply(inArray, []);
// i.e. inArray.concat();
Since there are no extra arguments, inArray.concat() simply returns a shallow copy of inArray:
[ [1, 2], [3, 4, 5] ]
This remains an array of arrays and is not flattened into [1, 2, 3, 4, 5]. Therefore, Option B does not produce the required result.
* Option C: [].concat::...inArray();
Corrected name from usArray to inArray and .. to ....
This expression uses syntax that is not part of standard JavaScript:
* :: (double colon) was proposed in early drafts as a bind operator but is not part of the official ECMAScript standard.
* The combination concat::...inArray() is invalid in normal JavaScript engines and cannot be used as a valid way to flatten arrays.
* In addition, inArray() would imply calling inArray as a function, but it is an array, which would cause a runtime error.
Hence, Option C is syntactically or semantically invalid in standard JavaScript and does not provide the required result [1, 2, 3, 4, 5].
* Option D: [].concat.apply({}, inArray);
Corrected name from usArray to inArray.
Relevant concepts:
* Again, Function.prototype.apply is used to call concat with a specific this value and arguments provided as an array.
* concat treats its this value as an array-like object but ultimately returns a new array containing concatenated elements.
In this expression:
[].concat.apply({}, inArray);
* thisArg is {} (an empty object).
* argsArray is inArray, which is [ [1, 2], [3, 4, 5] ].
Using apply, this is interpreted as calling concat like:
Array.prototype.concat.call({}, [1, 2], [3, 4, 5] );
concat then:
* Starts from the array-like this (here {} is treated as an empty base).
* Takes the first argument [1, 2] and appends its elements 1 and 2 to the result.
* Takes the second argument [3, 4, 5] and appends its elements 3, 4, and 5 to the result.
The resulting new array is:
[1, 2, 3, 4, 5]
Therefore, Option D also correctly produces the required array.
* Final evaluation of all options:
* Option A: Uses spread syntax with concat and correctly flattens one level: [1, 2, 3, 4, 5].
* Option B: Equivalent to inArray.concat(), leaving it as [[1, 2], [3, 4, 5]] . Does not flatten the structure.
* Option C: Uses non-standard and invalid syntax; not a valid or correct JavaScript solution.
* Option D: Uses apply with concat, passing the inner arrays as arguments and flattening one level to [1,
2, 3, 4, 5].
Thus, the two correct statements are:
The answer: A, D
References of JavaScript knowledge documents or Study Guide (concept names only):
* Array.prototype.concat (concatenating and one-level flattening behavior)
* Spread syntax for arrays (...array)
* Function.prototype.apply (calling a function with a given this value and arguments list)
* Array flattening by one level using concat with array arguments
* Distinction between nested arrays and flat arrays in JavaScript


質問 # 88
Refer to the code below:
Let searchString = ' Look for this ';
Which two options remove the whitespace from the beginning of searchString? Choose 2 answers