如何在PHP中制作不同结构的JSON
我正在从 2 个表中检索数据：1) 主页横幅和 2) 热门交易。之后，我从中制作 REST API。但是我在 json 中获取数据是这样的：
[
{
"homuri": "/Home/Banner/activity.png"
},
{
"homuri": "/Home/Banner/fitness.png"
},
{
"hotdeal": "/Home/HotDeals/hotdeal1.png"
},
{
"hotdeal": "/Home/HotDeals/hotdeal2.png"
},
{
"hotdeal": "/Home/HotDeals/hotdeal3.png"
},
]
但我想要我的 json 结构是这样的：
[
"banner":{
"homuri": "/Home/Banner/activity.png",
"homuri": "/Home/Banner/fitness.png"
},
"hotdeals":{
"hotdeal": "/Home/HotDeals/hotdeal1.png",
"hotdeal": "/Home/HotDeals/hotdeal2.png",
"hotdeal": "/Home/HotDeals/hotdeal3.png",
},
]
如何做到这一点？以及如何在 HTML 中访问所有这些值？下面我附上我的代码：
<?php
include('dbconn.php');
$request=$_SERVER['REQUEST_METHOD'];
$data=array();
switch($request)
{
case 'GET':
response(getData());
}
function getData()
{
global $conn;
@$col=$_GET['col'];
$query=mysqli_query($conn,"select strHomeBannerUri as homuri from tblhomebannerdetails");
while($row=mysqli_fetch_assoc($query))
{
$data[]=array("homuri"=>$row['homuri']);
}
if($col=="pop")
{
$query=mysqli_query($conn,"select nHotrDealImageUri as hotdeal from tblhotdealimages");
while($row=mysqli_fetch_assoc($query))
{
$data[]=array("hotdeal"=>$row['hotdeal']);
}
}
return $data;
}
function response($data)
{
echo json_encode($data);
}
?>
智慧大石1回答
-
繁星coding
您希望 JSON 看起来的方式将不起作用 - 您不能在同一“嵌套级别”中多次使用一个键。function getData(){ global $conn; @$col=$_GET['col']; $query=mysqli_query($conn,"select strHomeBannerUri as homuri from tblhomebannerdetails"); while($row=mysqli_fetch_assoc($query)) { $data['homeuri'][]= $row['homuri']; } if($col=="pop") { $query=mysqli_query($conn,"select nHotrDealImageUri as hotdeal from tblhotdealimages"); while($row=mysqli_fetch_assoc($query)) { $data['hotdeals'][]= $row['hotdeal']; } } return $data; }可以解决这个问题 - 你的逻辑有一些问题，你应该检查自己如何在 PHP 中编写和修改数组。完成后，请检查 JSON 以及如何构建 JSON 对象/列表。上面的代码应该返回类似{ "banner":[ "/Home/Banner/activity.png", "/Home/Banner/fitness.png" ], "hotdeals":[ "/Home/HotDeals/hotdeal1.png", "/Home/HotDeals/hotdeal2.png", "/Home/HotDeals/hotdeal3.png" ], }
随时随地看视频慕课网APP
PHP