Why does json_encode put square brackets around my json object? - php

Question

I have a function in PHP which is below:
public function displayMenuType() {
$stmt = $this->conn->query("SELECT * FROM `Type`") or die($this->conn->error);
$json = array();
if($stmt) {
while($result = mysqli_fetch_all($stmt,MYSQLI_ASSOC)) {
$json = json_encode($result, JSON_PRETTY_PRINT);
}
echo $json;
} else {
return false;
}
}
The problem is, the output isn't like I'd expect to see from a json object, it's all on one line and for some reason inside an array/square brackets. Where have I went wrong? The output is below:
[ { "ID": "1", "Type": "Classic Starters", "Description": "" }, { "ID": "2", "Type": "Special Starters", "Description": "" }, { "ID": "3", "Type": "Tapas Menu", "Description": "" }, { "ID": "4", "Type": "Speciality Curry Dishes", "Description": "" }, { "ID": "5", "Type": "Healthier Options", "Description": "Using coconut oil or coconut milk, no ghee, no sugars, we present our healthier range of curries." }, { "ID": "6", "Type": "Traditional Curries", "Description": "" }, { "ID": "7", "Type": "Tandoori & Grilled", "Description": "" }, { "ID": "8", "Type": "Biryanis", "Description": "" }, { "ID": "9", "Type": "Stir Fry's", "Description": "Our new range of Stir fry\u2019s, all made with chicken tikka, served with salad & mint sauce." }, { "ID": "10", "Type": "Nanbabs", "Description": "A choice of Stirfry\u2019s made with chicken tikka, served on a fresh nan bread with mint sauce." }, { "ID": "11", "Type": "Vegetarian Side Dishes", "Description": "" }, { "ID": "12", "Type": "Rice & Chips", "Description": "" }, { "ID": "13", "Type": "Rice Box", "Description": "" }, { "ID": "14", "Type": "Breads", "Description": "" }, { "ID": "15", "Type": "Dips & Drinks", "Description": "" } ]

Answer 1

The square brackets mean a JSON array, which you want, as you want a json array of rows, I'm assuming.
Try this:
if($stmt) {
$array = array();
while($result = mysqli_fetch_all($stmt,MYSQLI_ASSOC)) {
$array[] = $result;
}
$json = json_encode($array, JSON_PRETTY_PRINT);
header('Content-Type: application/json');
echo $json;
exit;
}

Answer 2

According to its documentation mysqli_fetch_all() returns a set of rows (an array). Because this array has numeric integer consecutive keys, it is encoded as a Javascript array by json_encode(). The outer square brackets in the generated JSON denote this array.
Each item of the array returned by mysqli_fetch_all() is an array itself but, because of the MYSQLI_ASSOC argument, they are associative arrays, i.e. they have string keys. Such arrays are encoded as Javascript objects by json_encode(). They are enclosed in curly braces in the JSON. F.e. { "ID": "1", "Type": "Classic Starters", "Description": "" }
The argument JSON_PRETTY_PRINT tells json_encode() to produce a nicely formatted JSON that can be easily read by humans. But everything is in vain if you echo the value it returns in HTML context and then look at it in the browser. The browser doesn't care about the newlines and spaces; it renders any sequence of spaces and newlines as a single white space.
Check the source code of the page in the browser and you'll see the JSON pretty formatted.

Answer 3

Well the obvious error is that you are Over Writing $json each time round the loop.
Then also build an complete array of all the results and then once complete convert it to a JSON String.
public function displayMenuType() {
$stmt = $this->conn->query("SELECT * FROM `Type`")
or die($this->conn->error);
$json = array();
if($stmt) {
while($result = mysqli_fetch_all($stmt,MYSQLI_ASSOC)) {
$json[] = $result;
}
echo json_encode($json);
} else {
return json_encode(['result'=>false]);
}
}

Answer 4

mysqli_fetch_all returns all of the results from your query, which will be represented as an array of associative arrays. The square brackets are array syntax, so the square brackets are properly present if the goal is to show a JSON entity. Without the square brackets, a JSON parser word know how to interpret the commas between each object.
There are a few things you can do to improve your code, like appending to your array instead of potentially overwriting it each loop.
$json = array();
if ($stmt) {
while ($row = mysqli_fetch_assoc($stmt)) {
$json[] = $row;
}
echo json_encode($json);
} else {
return false;
}