我正在嘗試使用 PHP 生成 JSON 以專門采用所需的形狀。下面是我的 PHP 代碼片段：
if($con){
$sql="select * from people";
$result=mysqli_query($con,$sql);
if($result){
header("Content-Type: JSON");
$i=0;
while($row = mysqli_fetch_assoc($result)){
$response[$i]['gender']=$row['gender'];
$response[$i]['first']=$row['first'];
$i ;
}
echo json_encode($response, JSON_PRETTY_PRINT);
}
}
這是我當前的 JSON 回應
[
{
"gender":"male",
"first":"Angela"
},
{
"gender":"female",
"first":"Krista"
}
]
這是我想要的回應：
{
"inputs": [
{
"values": {
"gender": "male",
"first": "Angela"
}
},
{
"values": {
"gender": "female",
"first": "Krista"
}
}
]
}
uj5u.com熱心網友回復：
幾乎，添加“值”，然后將您的專案添加$response到專案陣列中，json 的內容型別也是 application/json，您應該將 $response 預定義為一個陣列，否則在使用它時會收到未定義的警告和空值json_encode 如果查詢為空。
...
header("Content-Type: application/json");
$response = [];
while($row = mysqli_fetch_assoc($result)){
$response[]['values'] = [
'gender' => $row['gender'],
'first' => $row['first']
];
}
echo json_encode(['inputs' => $response], JSON_PRETTY_PRINT);
...
uj5u.com熱心網友回復：
這條路：
if ($con) {
$sql = "select * from people";
$result = mysqli_query($con,$sql);
if ($result) {
header("Content-Type: JSON");
$response = [
"inputs" => []
];
while ($row = mysqli_fetch_assoc($result)){
$response["inputs"][] = [
"values" => [
'gender' => $row['gender'],
'first' => $row['first']
]
];
}
echo json_encode($response, JSON_PRETTY_PRINT);
}
}
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