Deal of The Day! Hurry Up, Grab the Special Discount - Save 25% - Ends In 00:00:00 Coupon code: SAVE25
Welcome to Pass4Success

- Free Preparation Discussions

Zend 200-710 Exam - Topic 2 Question 37 Discussion

Actual exam question for Zend's 200-710 exam
Question #: 37
Topic #: 2
[All 200-710 Questions]

Given a PHP value, which sample shows how to convert the value to JSON?

Show Suggested Answer Hide Answer
Suggested Answer: B

Contribute your Thoughts:

0/2000 characters
Rebeca
5 months ago
Surprised that people still confuse these methods!
upvoted 0 times
...
Adelina
5 months ago
I thought D was a thing, but it seems off.
upvoted 0 times
...
Earleen
5 months ago
Wait, are B and C even valid?
upvoted 0 times
...
Ashlyn
5 months ago
Definitely agree, json_encode is the standard.
upvoted 0 times
...
Lezlie
5 months ago
A is the correct way to convert to JSON!
upvoted 0 times
...
Jamey
6 months ago
I have a vague memory of option D, but I don't think PHP has a __toJson() method for objects.
upvoted 0 times
...
Marta
6 months ago
Option C seems off to me; I don't recall using the new Json() syntax in PHP.
upvoted 0 times
...
Carline
6 months ago
I'm not entirely sure, but I feel like Json::encode in option B is related to a framework I studied.
upvoted 0 times
...
Tamar
6 months ago
I think option A looks familiar; I remember using json_encode in my practice questions.
upvoted 0 times
...
Louvenia
6 months ago
Wait, I'm a little confused. How do we determine the number of APs needed? Is it just a simple calculation based on the client count, or are there other factors to consider? I want to make sure I understand this properly before selecting an answer.
upvoted 0 times
...
Daniela
6 months ago
Hmm, I'm not entirely sure about this one. I think I need to double-check the GoldenGate documentation to make sure I fully understand the differences between the available options. I want to make sure I select the right approach for detecting update and delete conflicts.
upvoted 0 times
...
Maxima
6 months ago
This is a tricky one, but I'm pretty sure the answer is A. The other options seem to be describing physical characteristics, not leadership types. I'll mark A and move on.
upvoted 0 times
...
Salena
6 months ago
I feel like I recall something about nodes and decentralization, but I'm struggling to remember exactly how they all fit together.
upvoted 0 times
...
Simona
6 months ago
Aha, I see what they're getting at. If an iRule references an event that doesn't occur, that could cause issues with the client's connection. I'll go with that option.
upvoted 0 times
...
Rickie
11 months ago
I wonder if option D involves turning the value into a Transformer first, and then converting it to JSON. Probably best to keep it simple with json_encode().
upvoted 0 times
Arlene
9 months ago
B) $string = Json::encode($value);
upvoted 0 times
...
Therese
9 months ago
Yes, that's the correct way to convert a PHP value to JSON.
upvoted 0 times
...
Viola
10 months ago
A) $string = json_encode($value);
upvoted 0 times
...
...
Trina
11 months ago
If I see 'Json' with a capital 'J', I automatically assume it's not the standard PHP function. json_encode() all the way!
upvoted 0 times
...
Lynda
11 months ago
Converting the value to an object and then calling a custom __toJson() method? That's just weird. json_encode() is the way to go.
upvoted 0 times
Jeannetta
10 months ago
B) $string = Json::encode($value);
upvoted 0 times
...
Delbert
10 months ago
Yeah, json_encode() is the standard way to convert PHP values to JSON.
upvoted 0 times
...
Laurel
10 months ago
B) $string = Json::encode($value);
upvoted 0 times
...
Lili
10 months ago
A) $string = json_encode($value);
upvoted 0 times
...
Brynn
10 months ago
Yeah, json_encode() is definitely the way to go.
upvoted 0 times
...
Ashlee
10 months ago
A) $string = json_encode($value);
upvoted 0 times
...
...
Natalie
11 months ago
Creating a new Json object and then converting it to a string? That seems overly complicated. I'll stick with the simple json_encode() method.
upvoted 0 times
Louisa
9 months ago
B) $string = Json::encode($value);
upvoted 0 times
...
Wilda
9 months ago
A) $string = json_encode($value);
upvoted 0 times
...
Berry
10 months ago
B) $string = Json::encode($value);
upvoted 0 times
...
Angella
10 months ago
A) $string = json_encode($value);
upvoted 0 times
...
Kristeen
10 months ago
I agree, json_encode() is much simpler.
upvoted 0 times
...
Simona
10 months ago
A) $string = json_encode($value);
upvoted 0 times
...
...
Brice
11 months ago
Option B looks like it might be a custom function, but I'm not familiar with it. I'll go with the built-in json_encode() function.
upvoted 0 times
Whitley
11 months ago
Yeah, json_encode() is a built-in function for converting PHP values to JSON.
upvoted 0 times
...
Clarinda
11 months ago
I agree, json_encode() is the way to go.
upvoted 0 times
...
...
Dylan
12 months ago
The correct answer is A. json_encode() is the standard PHP function to convert a value to JSON.
upvoted 0 times
Adrianna
10 months ago
Yes, that's correct. json_encode() is the way to go.
upvoted 0 times
...
Nichelle
11 months ago
A) $string = json_encode($value);
upvoted 0 times
...
...
Kattie
12 months ago
I'm not sure, but I think B) $string = Json::encode($value); could also be a valid option.
upvoted 0 times
...
Chan
12 months ago
I agree with Tonette, json_encode is the right function to convert PHP value to JSON.
upvoted 0 times
...
Tonette
12 months ago
I think the correct answer is A) $string = json_encode($value);
upvoted 0 times
...

Save Cancel