Get JSON variable into webpage

Question

I have started to learn JSON with PHP. First I wan't to learn, how to create some easy API which will get some data from database from main server and use it in other webpage. I have found some solutions which will get these data into web browser console and not just specific data I need for that web but all data from database.

For example:

DATABASE

id: 1 name: test status: active

id: 2 name: test2 status: disable

INTO TEST.COM WEB

It get just record whos name is TEST and use its status.

$status = RECORD FROM DATABASE;
if ($status == 'disable') { 
  echo 'THIS WEBPAGE STATUS IS DISABLE';
} else { 
  echo 'THIS WEBPAGE STATUS IS ACTIVE';
}

Could some one help me with this, please? It looks very east from what I have seen, but just not for me, because I am new with this JSON.

Thanks a lot

What exactly do you need help with? I.e. are you getting the wrong result? What do you think it ought to be? — Erick G. Hagstrom
– Erick G. Hagstrom, Commented Jul 20, 2015 at 15:44
In API response what exactly you need to show and you want to manage the response using javascript? — Mitul
– Mitul, Commented Jul 20, 2015 at 15:55
You should take a look at PHP json_encode() and jQuery AJAX, then provide a more specific question on what you need help with. — Siphon
– Siphon, Commented Jul 20, 2015 at 16:36

João Reis · Accepted Answer · 2015-07-20 18:10:31Z

1

Simply use json_enconde() as @Siphon said.

If that doesnt work try to do this:

<?php
   echo "<pre>";
   print_r($status);
   echo "</pre>";
?>

Check what that says, if something like this shows up:

$data = RECORD FROM DATABASE;



 Array $data(
        ['status'] => 'enabled',
    );

if that shows up then your if statment is wrong and you need to do this:

$data= RECORD FROM DATABASE;
if ($data['status'] == 'disable') { 
  echo 'THIS WEBPAGE STATUS IS DISABLE';
} else { 
  echo 'THIS WEBPAGE STATUS IS ACTIVE';
}

answered Jul 20, 2015 at 18:10

João Reis

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2 Comments

brass Over a year ago

Thanks for that. But my problem isnt about making php code part, my problem is about making json code... I will try to write more about that.. I would like help with making that part which is on online server X where are json code and some php code, like DB with websites (Y, Z) and its status... and into these websites (Y, Z) I check if into server X website status is disable or active. something like this youtube.com/watch?v=F5pXxS0y4bg but just with results for one website not all and dont show these results into consule but use it in website functions.

João Reis Over a year ago

If you give me your email i can help you.

meda · Accepted Answer · 2015-07-20 18:56:31Z

1

Your question is broad but here are some pointers:

you will need a query

SELECT * FROM `table` WHERE name = ?

Then you encode the data (array) to PDO using json_encode()

Example in PDO:

$data = $stmt->fetchAll();
$json = json_encode($data);

answered Jul 20, 2015 at 18:56

meda

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