Erstellen verschachtelten Json-Objekt mit PHP mysql

Question

Ich habe zwei Tabellen, Tabelle 1 hat 2 Felder (Frage_pk, Frage_Name) und Tabelle 2 hat 4 Felder (ans_pk, Optionen, Frage_fk und Right_answer). möchte ich Json wie die folgende Struktur

{ 
    "type": "quiz", 
    "name": "Brand Colors", 
    "description": "Can you identify these brands by the background color?", 
    "questions": [ 
     { 
      "name": "Can you identify this color?", 
      "description": "#ea4c89", 
      "answers": [ 
       { 
        "name": "Dribbble", 
        "description": "dribbble.png", 
        "weight": 1 
       }, 
       { 
        "name": "Amazon", 
        "description": "amazon.png", 
        "weight": 0 
       }, 
       { 
        "name": "Apple", 
        "description": "apple.png", 
        "weight": 0 
       } 
      ] 
     }, 
     { 
      "name": "Can you identify this color?", 
      "description": "#3d9ae8", 
      "answers": [ 
       { 
        "name": "Youtube", 
        "description": "youtube.png", 
        "weight": 0 
       }, 
       { 
        "name": "Dropbox", 
        "description": "dropbox.png", 
        "weight": 1 
       }, 
       { 
        "name": "Wordpress", 
        "description": "wordpress.png", 
        "weight": 0 
       } 
      ] 
     }, 
     { 
      "name": "Can you identify this color?", 
      "description": "#c4302b", 
      "answers": [ 
       { 
        "name": "Youtube", 
        "description": "youtube.png", 
        "weight": 1 
       }, 
       { 
        "name": "Twitter", 
        "description": "twitter.png", 
        "weight": 0 
       }, 
       { 
        "name": "Vimeo", 
        "description": "vimeo.png", 
        "weight": 0 
       } 
      ] 
     } 

    ] 
} 

MY PHP CODE

<?php 
include '../config/config.php'; 
if(isset($_GET['sub_cat_id'])) 
{ 
     $sub_cat_id = $_GET['sub_cat_id']; 
     $result = mysql_query("select * from $questions where sub_cat='$sub_cat_id' order by level_fk asc"); 
     $json_response = array(); //Create an array 
     $i=1; 
         while ($row = mysql_fetch_array($result)) 
         { 
         $row_array['qus_pk'] = $row['qus_pk'];   
         $row_array['question'] = $row['question']; 
         $qus_pk = $row['qus_pk']; 


         $option_qry = mysql_query("select * from $qus_ans where qus_pk=$qus_pk"); 
         while ($opt_fet = mysql_fetch_array($option_qry)) 
         { 

         $row_array['options'] = $opt_fet['options']; 
         $row_array['right_ans'] = $opt_fet['right_ans']; 
         array_push($json_response,$row_array); //push the values in the array 
          }      


         $i++; 
         } 
     echo json_encode($json_response); 
} 

?> 

Und ich mein Ergebnis zu schaffen, bin immer json Antwort wie die folgende

[ 
    { 
     "qus_pk": "1", 
     "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?", 
     "options": "45", 
     "right_ans": "0" 
    }, 
    { 
     "qus_pk": "1", 
     "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?", 
     "options": "40", 
     "right_ans": "0" 
    }, 
    { 
     "qus_pk": "1", 
     "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?", 
     "options": "35", 
     "right_ans": "1" 
    }, 
    { 
     "qus_pk": "1", 
     "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?", 
     "options": "50", 
     "right_ans": "0" 
    }, 
    { 
     "qus_pk": "2", 
     "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?", 
     "options": "4 times", 
     "right_ans": "0" 
    }, 
    { 
     "qus_pk": "2", 
     "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?", 
     "options": "1 times", 
     "right_ans": "0" 
    }, 
    { 
     "qus_pk": "2", 
     "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?", 
     "options": "3 times", 
     "right_ans": "1" 
    }, 
    { 
     "qus_pk": "2", 
     "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?", 
     "options": "5 times", 
     "right_ans": "0" 
    } 
] 

In meinem respose jeder Zeit wird die Frage wiederholt, so wie zu vermeiden und , wenn ich die erste JSON-Struktur erreichen will, in meinem PHP-Code was & wo muss ich Änderungen vornehmen ?. Wenn jemand weiß, hilf mir.

Answer 1

Hallo versuchen diese,

<?php 
include '../config/config.php'; 
if(isset($_GET['sub_cat_id'])) 
{ 
    $sub_cat_id = $_GET['sub_cat_id']; 
    $result = mysql_query("SELECT * FROM $questions WHERE sub_cat='$sub_cat_id' ORDER BY level_fk ASC"); 
    $json_response = array(); //Create an array 
    while ($row = mysql_fetch_array($result)) 
    { 
     $row_array = array(); 
     $row_array['qus_pk'] = $row['qus_pk'];   
     $row_array['question'] = $row['question']; 
     $row_array['answers'] = array(); 
     $qus_pk = $row['qus_pk']; 

     $option_qry = mysql_query("SELECT * FROM $qus_ans WHERE qus_pk=$qus_pk"); 
     while ($opt_fet = mysql_fetch_array($option_qry)) 
     { 
      $row_array['answers'][] = array(
       'options' => $opt_fet['options'], 
       'right_ans' => $opt_fet['right_ans'], 
      ); 

     } 
     array_push($json_response, $row_array); //push the values in the array 
    } 
    echo json_encode($json_response); 
} 
?>  

Answer 2

Ich denke, dass dieser Code einfacher ist und es Mysqli verwendet durch die Art und Weise, um herauszufinden, ...

Dieses auf meine eigene Datenstruktur basiert, ich bin in die Mitte von etwas und ich habe keine Zeit atm es um die Frage anzupassen, sondern sollte einfach, wie es auf andere Strukturen anzupassen, um herauszufinden:

$usersList_array =array(); 
$user_array = array(); 
$note_array = array(); 

$fetch_users = mysqli_query($mysqli, "SELECT 
     ID, 
     Surname, 
     Name 
    FROM tb_Users 
    WHERE Name LIKE 'G%' 
    ORDER BY ID") or die(mysqli_error($mysqli)); 
while ($row_users = mysqli_fetch_assoc($fetch_users)) { 
    $user_array['id'] = $row_users['ID']; 
    $user_array['surnameName'] = $row_users['Surname'].' '.$row_users['Name']; 
    $user_array['notes'] = array(); 

    $fetch_notes = mysqli_query($mysqli, "SELECT 
     id, 
     dateIns, 
     type, 
     content 
    FROM tb_Notes 
    WHERE fk_RefTable = 'tb_Users' AND 
     fk_RefID = ".$row_users['ID']."" 
    ) or die(mysqli_error($mysqli)); 
    while ($row_notes = mysqli_fetch_assoc($fetch_notes)) { 
     $note_array['id']=$row_notes['id']; 
     $note_array['dateIns']=$row_notes['dateIns']; 
     $note_array['type']=$row_notes['type']; 
     $note_array['content']=$row_notes['content']; 
     array_push($user_array['notes'],$note_array); 
    } 

    array_push($usersList_array,$user_array); 
} 

$jsonData = json_encode($usersList_array, JSON_PRETTY_PRINT); 


echo $jsonData; 

Resultierende JSON:

[ 
{ 
    "id": "1", 
    "surnameName": "Xyz Giorgio", 
    "notes": [ 
     { 
      "id": "1", 
      "dateIns": "2016-05-01 03:10:45", 
      "type": "warning", 
      "content": "warning test" 
     }, 
     { 
      "id": "2", 
      "dateIns": "2016-05-18 20:51:32", 
      "type": "error", 
      "content": "error test" 
     }, 
     { 
      "id": "3", 
      "dateIns": "2016-05-18 20:53:00", 
      "type": "info", 
      "content": "info test" 
     } 
    ] 
}, 
{ 
    "id": "2", 
    "cognomeNome": "Xyz Georg", 
    "notes": [ 
     { 
      "id": "4", 
      "dateIns": "2016-05-20 14:38:20", 
      "type": "warning", 
      "content": "georg warning" 
     }, 
     { 
      "id": "5", 
      "dateIns": "2016-05-20 14:38:20", 
      "type": "info", 
      "content": "georg info" 
     } 
    ] 
} 
]