創建某些JSON形式

我需要的形式是：創建某些JSON形式

{ 
    "suggestions": [ 
     { "value": "United Arab Emirates", "data": "AE" }, 
     { "value": "United Kingdom",  "data": "UK" }, 
     { "value": "United States",  "data": "US" } 
    ] 
}

我得到了這個PHP代碼：

header("Content-type: application/json"); 
require 'connect.inc.php'; 
$mysql = mysqli_connect("$db_host", "$db_username", "$db_pass", "$db_name"); 
$name = isset($_POST['query']) ? $_POST['query'] : ""; 


$sql = "SELECT MedicineName, MedicineID FROM medicinetypes WHERE MedicineName LIKE '%$name%'"; 

$res = mysqli_query($mysql, $sql); 
$res = mysqli_fetch_all($res, MYSQLI_ASSOC); 
$reply = array(); 
$reply['suggestions'][]=$res; 
echo json_encode($reply); 
mysqli_close($mysql);

是這樣的：

{"suggestions":[[{"MedicineName":"Amoxil 500","MedicineID":"1"},{"MedicineName":"Panadol","MedicineID":"2"},{"MedicineName":"Farmako3","MedicineID":"3"},{"MedicineName":"Beer","MedicineID":"4"},{"MedicineName":"Farmako4","MedicineID":"5"},{"MedicineName":"hjkki","MedicineID":"61"},{"MedicineName":"Amoxil 1","MedicineID":"577"},{"MedicineName":"Amoxil 502","MedicineID":"580"},{"MedicineName":"Amoxil 56","MedicineID":"582"},{"MedicineName":"Amoxil 600","MedicineID":"583"},{"MedicineName":"\u03b5\u03bb\u03b1 \u03bc\u03bf\u03c5 \u03bd\u03c4\u03b5","MedicineID":"586"}]]}

由於它不符合我需要的格式，因此導致了一個問題。我怎樣才能正確地做到這一點？

來源

2017-08-02 Constantine

下面是其他用戶發佈的解決方案： https://stackoverflow.com/questions/15678355/remove-extra-square-bracket-from-json – guillefix

@guillefix - 這是爲了解決JSON解析時的問題（假設它在JS中被解析）。問題在於詢問如何在源頭修復數據。 – Quentin

'MedicineName！==值'和'MedicineID！==數據' –

此：

$reply['suggestions'][]=$res;

獲取$reply['suggestions']
分配一個空數組，如果它一個不存在有
提出$res（已經陣列）作爲該陣列中的下一項

您只需要將$res設置爲$reply['suggestions']的值即可。

刪除[]。

$reply['suggestions']=$res;

來源

2017-08-02 10:49:55 Quentin

啊我真是太蠢！謝謝 – Constantine

嘗試改變$reply['suggestions'][]=$res;到$reply['suggestions']=$res;

來源

2017-08-02 10:51:08 Byte

創建某些JSON形式

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