PHP- putting database results in array and attach a variable

Question

I'm pulling records from a database which i'm able to display the data in JSON format the way i want and it works perfectly. The problem here its there's a token i generate on the page which i want to add to the payload that the user gets and this's what i'm getting

{"user_details":[{
                "id":"1",
                "fname":"xxx",
                "lname":"xxx Mensah",
                "phone_number":"0000000",
                "email":"[email protected]",
                "username":"xx",
                "password":"xx",
                "user_type":"xx"
                }],
"token":"xxxxx"}

but i want to get the results like this

[{"user_details":{
                "id":"1",
                "fname":"xxx",
                "lname":"xxx Mensah",
                "phone_number":"0000000",
                "email":"[email protected]",
                "username":"xx",
                "password":"xx",
                "user_type":"xx"
               },
"token":{"xxxxx"}
]

PHP

$token = xxxxx
$row = $conn->query("SELECT * from users where username='".$username."' and password='".$password."'");
$row->setFetchMode(PDO::FETCH_ASSOC);
$userdetails = $row->fetchAll(PDO::FETCH_OBJ);

$results = array(
    'user_details' => $userdetails,
    'token' => $token
);
echo json_encode(($results));

Your script is wide open to SQL Injection Attack Even if you are escaping inputs, its not safe! Use prepared parameterized statements in either the MYSQLI_ or PDO API's — RiggsFolly
– RiggsFolly, Commented Apr 1, 2019 at 22:35

gview · Accepted Answer · 2019-04-01 23:23:39Z

1

It seems that expected json is not valid (the token part), but you can have something similar that looks like this :

[{"user_details":{"id":"1","fname":"xxx","lname":"xxx Mensah","phone_number":"0000000","email":"[email protected]","username":"xx","password":"xx","user_type":"xx"}},{"token":"xxxxx"}]

Code :

$token = xxxxx;

$row = $conn->query("SELECT * from users where username='".$username."' and password='".$password."'");
$row->setFetchMode(PDO::FETCH_ASSOC);
$userdetails = $row->fetch(PDO::FETCH_OBJ);

$results = array(
    array(
        'user_details' => $userdetails
    ), 
    array(
        'token' => $token
    )
);
echo json_encode($results);

Notice that the result is wrapped inside another array and the token is inside another array.

The fetchAll is replaced by a fetch since you need only one row not all the rows, right ?

This example can help you to know how your data should be structured in order to get the desired result :

<?php

$array = array(
  array (
    'user_details' => 
      array (
        'id' => '1',
        'fname' => 'xxx',
        'lname' => 'xxx Mensah',
        'phone_number' => '0000000',
        'email' => '[email protected]',
        'username' => 'xx',
        'password' => 'xx',
        'user_type' => 'xx',
      ),
  ),
  array(
      'token' => 'xxxxx'
  )
);

echo json_encode($array);

edited Apr 1, 2019 at 23:23

gview

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answered Apr 1, 2019 at 21:49

Abdelkarim EL AMEL

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1 Comment

gview Over a year ago

This is really close, so I'm upvoting. Key is identification that code needs to do fetch() vs fetchall().

gview · Accepted Answer · 2019-04-01 23:25:31Z

1

The [] brackets in a JSON encoded string representation of an object, whilst the {} brackets represents an array. So your user details are showing an object which is an array of key, value pairs. Then you have an array which wraps everything up, the token and user details.

I'd be inclined to try something along the lines of:

$userdetails = $row->fetch(PDO::FETCH_OBJ);
$object = (object) ['userdetails' => $userdetails, 'token' => $token];
echo json_encode($results);

I haven't tested it out, but should be close enough.

edited Apr 1, 2019 at 23:25

gview

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answered Apr 1, 2019 at 22:11

Jim Grant

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4 Comments

gview Over a year ago

It's the opposite of your statement. [] is array, {} is an object.

Jim Grant Over a year ago

I knew replying to questions so late at night was a mistake. :)

gview Over a year ago

Happens to the best of us. I figured as much given that your code reflects the basic issue. With that said, the underlying problem is that he gets an array with an object in it, and that will serialize in the form he doesn't like regardless.

gview Over a year ago

Add the fetch() change and I think your answer gets him closest to the desired result.

Goe Roe Ku · Accepted Answer · 2019-04-01 21:56:26Z

-1

did you make a typo? please try this one :

echo json_encode ([$results]);

answered Apr 1, 2019 at 21:56

Goe Roe Ku

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