Create dynamic JSON array in php

Question

I'd like to create a JSON dynamically like that:

{
"storie":[
{
"story_id":"111",
"username":"Username1",
"profile_photo":"boh.png",
"copertina":"ok.png",
"num_elements":"1",
"rating":"4.5"
},
{
"story_id":"222",
"username":"Username2",
"profile_photo":"hello.png",
"copertina":"hi.png",
"num_elements":"2",
"rating":"3.5"
}
]
}

I'm trying to do that getting values from MySQL database and I'm able to do that:

$response = array();
$sql = mysqli_query($conn, "SELECT * FROM storie WHERE userid IN (SELECT following FROM follow WHERE follower='$userid')");
while($row = mysqli_fetch_assoc($sql)){
  $usern = getuserinfo($row['userid'], "username", $conn);
  $prof_photo = getuserinfo($row['userid'], "profile_photo", $conn);
  $idsto=$row['storia_id'];
  $elem = mysqli_query($conn, "SELECT COUNT(*) AS da_vedere FROM `storie_images` WHERE storia_id='$idsto' AND imm_id NOT IN (SELECT imm_id FROM image_views WHERE viewer='$userid')");
  while($ok = mysqli_fetch_assoc($elem)){
    $num_elem = $ok['da_vedere'];
  }
  //here I put the line that add the array to the json array:
}

But the problem is this row, which should create a new array and put it into the json:

$response['storie'] = [array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5")];

It works when there's only a record, but it doesn't work if there are more records. Can someone help me?

Warning: You are wide open to SQL Injections and should use parameterized prepared statements instead of manually building your queries. They are provided by PDO or by MySQLi. Never trust any kind of input! Even when your queries are executed only by trusted users, you are still in risk of corrupting your data. Escaping is not enough! — Dharman
– Dharman ♦, Commented Aug 16, 2020 at 13:45

Egzon Hasi · Accepted Answer · 2020-08-16 13:28:28Z

2

simply replace:

$response['storie'] = [array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5")];

with

$response['storie'][] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");

answered Aug 16, 2020 at 13:28

Egzon Hasi

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Comments

anehme · Accepted Answer · 2020-08-16 13:27:14Z

That's because you are using $response['storie'] inside while , to fix this problem :

1- create another array called result for example , then use it in while

$response = array();
$result = array ()
while (....)
{
//here I put the line that add the array to the json array:
$result[] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
}

2- then use the response array outside the loop:

$response['storie'] = $result;

And your code will be like this :

   $response = array();
   $result = array ()

    $sql = mysqli_query($conn, "SELECT * FROM storie WHERE userid IN (SELECT following FROM follow WHERE follower='$userid')");
    while($row = mysqli_fetch_assoc($sql)){
    $usern = getuserinfo($row['userid'], "username", $conn);
    $prof_photo = getuserinfo($row['userid'], "profile_photo", $conn);
    $idsto=$row['storia_id'];
    $elem = mysqli_query($conn, "SELECT COUNT(*) AS da_vedere FROM `storie_images` WHERE storia_id='$idsto' AND imm_id NOT IN (SELECT imm_id FROM image_views WHERE viewer='$userid')");
    while($ok = mysqli_fetch_assoc($elem)){
    $num_elem = $ok['da_vedere'];
    }
    //here I put the line that add the array to the json array:
    $result[] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
    }

   $response['storie'] = $result;

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