json_decode returns an empty result.

Question

mysqli

SELECT ... ,concat('[[', GROUP_CONCAT(`id`, ',', '\"',`name`, '\"' SEPARATOR '],['),']]') as npc

A string of the following type arrives from the base (json):

 var_dump($row['npc']);//→ string(62) "[[679,"<Привратник> Хранитель Храма"]]"

I want to convert this string to an array, but I can't

 $row['npc'] = json_decode($row['npc']);// пусто

If to copy a line from the browser and to substitute in json_decode

 $row['npc'] = json_decode('[[679,"<Привратник> Хранитель Храма"]]');

then convert it turns out. What kind of nonsense I do not understand: (

Look at the work with the database turn PDO. It can create ready-made arrays based on the expanded data. And no group_concat needed. ru.stackoverflow.com/questions/504422/… - Mike

Ipatiev Ipatiev 16.9k 3 17 44 · Answer 1 · 2016-08-03T16:08:06

If json_decode returned void, then an error occurred during decoding. Which one should look through json_last_error and correct accordingly. Most likely the problem is trivial with the encoding.

But in general, of course, generating Jason with his hands (!) In a SQL query (!!) is a tin of tin. It is not clear why you can not get the data and then encode them normally

You can of course of two implode () parse a two-dimensional array, I just wanted to make it easier for you to get by with just one json_decode :)

json_decode returns an empty result.

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