授權的Salesforce Salesforce Certified JavaScript Developer (JS-Dev-101)中的最佳JavaScript-Developer-I題庫資料和領導者資格考試

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JavaScript-Developer-I最新考題，JavaScript-Developer-I考試大綱

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要成為 Salesforce 認證的 JavaScript 開發人員 I，候選人必須通過 JavaScript-Developer-I 考試。此考試由 60 道多項選擇題組成，必須在 105 分鐘內完成。考試旨在測試候選人在 Salesforce 平台上的 JavaScript 開發知識，以及他們設計、開發和維護自定義應用程序的能力。候選人必須達到 68% 或更高的及格分數才能獲得認證。一旦獲得認證，開發人員將證明他們在 Salesforce 平台上的 JavaScript 開發專業知識，使他們對潛在雇主和客戶更具價值。

最新的 Salesforce Developer JavaScript-Developer-I 免費考試真題 (Q128-Q133):

問題 #128
Refer to the following array:
Let arr1 = [ 1, 2, 3, 4, 5 ];

Which two lines of code result in a second array, arr2 being created such that arr2 is not a reference to arr1?

A. Let arr2 = arr1;
B. Let arr2 = Array.from(arr1);
C. Let arr2 = arr1.sort();
D. Let arr2 = arr1.slice(0, 5);

答案：B,D

問題 #129
A developer writes the code below to return a message to a user attempting to register a new username. If the username is available, a variable named nag is declared and assigned a value on line 03.

What is the value of msg when getAvailableabilityMessage ("newUserName") is executed and get Availability ("newUserName") returns true?

A. "msg is not defined"
B. undefined
C. "newUserName"
D. "User-name available"

答案：D

問題 #130
static delay = async delay = > {
return new Promise(resolve = > {
setTimeout(resolve, delay);
});
};
static asyncCall = async () = > {
await delay(1000);
console.log(1);
};
console.log(2);
asyncCall();
console.log(3);
Assume delay and asyncCall are in scope as functions.
What is logged to the console?

A. 2 1 3
B. 1 2 3
C. 1 3 2
D. 2 3 1

答案：D

解題說明：
The correct answer is D , because JavaScript executes synchronous code first, while the code after await runs later after the Promise resolves.
The execution order is:
console.log(2);
This runs first, so JavaScript logs:
2
Then this line runs:
asyncCall();
The asyncCall() function starts executing. Inside it, JavaScript reaches:
await delay(1000);
The delay(1000) function returns a Promise that resolves after 1000 milliseconds:
return new Promise(resolve = > {
setTimeout(resolve, delay);
});
Because await pauses the rest of the asyncCall() function, this line does not run immediately:
console.log(1);
Instead, JavaScript continues executing the remaining synchronous code outside the async function:
console.log(3);
So JavaScript logs:
3
After approximately 1000 milliseconds, the Promise returned by delay(1000) resolves. Then the paused async function continues and runs:
console.log(1);
So JavaScript logs:
1
Final console output:
2
3
1
Important JavaScript concepts involved:
An async function always returns a Promise.
The await keyword pauses execution only inside the async function where it appears.
Code outside the async function continues running normally.
setTimeout() schedules its callback to run later, after the current synchronous code has finished.
Therefore, the verified answer is D. 2 3 1 .

問題 #131
Refer to the code below:
new Promise((resolve, reject) => {
const fraction = Math.random();
if( fraction >0.5) reject("fraction > 0.5, " + fraction);
resolve(fraction);
})
.then(() =>console.log("resolved"))
.catch((error) => console.error(error))
.finally(() => console.log(" when am I called?"));

When does Promise.finally on line 08 get called?

A. When resolved or rejected
B. WHen resolved
C. When resolved and settled
D. When rejected

答案：A

問題 #132
The developer wants to test the array shown:
const arr = Array(5).fill(0)
Which two tests are the most accurate for this array ?
Choose 2 answers:

A. console.assert(arr[0] === 0 && arr[ arr.length] === 0);
B. console.assert (arr.length >0);
C. console.assert( arr.length === 5 );
D. arr.forEach(elem => console.assert(elem === 0)) ;

答案：C,D

問題 #133
......

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