Encoding JSON with PHP from database table data

Question

I want to be able to encode JSON into a specific format so that it can be used with another script that I have created. I am able to encode it with PHP into JSON but need to be able to push these encoded JSON sets to a 'named' array.

At the moment I am getting this

{
    "ann_date":"14\/12\/2017",
    "ann_title":"Return Dates",
    "ann_content":"Have a good break. The course timetable resumes on Wednesday 3rd January 2018",
    "tutor":"John Smith"
}

From this code

class AnnData {
    public $ann_date = "";
    public $ann_title = "";
    public $ann_content = "";
    public $tutor = "";
}

while($row = mysqli_fetch_array($result)) {
    $ann_date = $row['ann_date'];
    $ann_title = $row['ann_title'];
    $ann_content = $row['ann_content'];
    $tutor = $row['tutor'];

    $annData = new AnnData();
    $annData->ann_date =  $ann_date;
    $annData->ann_title  = $ann_title;
    $annData->ann_content  = $ann_content;
    $annData->tutor  = $tutor;

    $annQ = json_encode($annData);

But need it to look like this

{
"announcements":[{
    "ann_date":"14\/12\/2017",
    "ann_title":"Return Dates",
    "ann_content":"Have a good break. The course timetable resumes on Wednesday 3rd January 2018",
   "tutor":"John Smith"}]
}

This thread will help you out: stackoverflow.com/questions/35099682/php-array-to-json-object Hope this helps! — Jenil Kanani
– Jenil Kanani, Commented Mar 13, 2018 at 11:09
$result['announcements'][] = $annData… You simply need to build the actual data structure you want… — deceze
– deceze ♦, Commented Mar 13, 2018 at 11:10
Try $result = array(); $result['announcements'] = $annData; $annQ = json_encode($result);, is it the result you want? — Mickaël Leger
– Mickaël Leger, Commented Mar 13, 2018 at 11:10

GrowingBrick · Accepted Answer · 2018-03-13 11:11:00Z

1

You can declare an associative array and then encode it:

$annData = new AnnData();
...
$out = array('announcements' => array($annData));
$outJson = json_encode($out);

answered Mar 13, 2018 at 11:11

GrowingBrick

7514 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

saunders07 Over a year ago

With this code I need to be able to add more JSON object to the array, at the moment it is creating an array for each of the JSON objects. Sorry bit of a novice with PHP.

GrowingBrick Over a year ago

$announcements = array(); for (...) { array_push($announcements, $object); } $outJson = json_encode(array('announcements' => $announcements));

jeroen · Accepted Answer · 2018-03-13 11:12:51Z

0

You can simplify this a lot by getting the correct columns from your database (only the 4 your need) and fetching all rows at once:

$annQ = json_encode([
    'announcements' => mysqli_fetch_all($result, MYSQLI_ASSOC),
]);

This replaces all the code you have posted.

An associative array will be stored in the string as an object so you don't need that class either if you are not going to use it anywhere else.

answered Mar 13, 2018 at 11:12

jeroen

91.8k22 gold badges117 silver badges132 bronze badges

5 Comments

YvesLeBorg Over a year ago

that would be true if the db schema has no other fields in the said table.

jeroen Over a year ago

@YvesLeBorg That's why I mentioned that the OP needs to get the correct columns from the database :-)

YvesLeBorg Over a year ago

duh ... sorry about the noise ;)

saunders07 Over a year ago

I'm getting back that fetch_all is undefined, suspect its because of PHP version, any other way of doing this?

jeroen Over a year ago

@saunders07 That sounds unlikely, it is available for php 5.3+. What version are you on?

Collectives™ on Stack Overflow

Encoding JSON with PHP from database table data

2 Answers 2

2 Comments

5 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related