Question

我正在开发一个IOS应用程序，对于API我将我的请求发送到一个应该用PHP返回JSON数据的URL 我很喜欢

[{"Child Care":"After Scool,Breakfast Club\n"},{"Child Care":"Breakfast Club"}]

但我希望得到像

[{

"Childcare":[

"All of Childcare",

"After school",

"Breakfast Club"
]}

我的代码是

    <?php
session_start();
$connection=mysqli_connect('localhost','root','','testing') or die(mysqli_error());

    $sql="select `Child Care` from Activity_type ";
    $result = mysqli_query($connection,$sql) or die("Error in Selecting " . mysqli_error($connection));
    $emparray=array();
    while($row =mysqli_fetch_assoc($result)){
        array_push($emparray,$row);
        }   
        header('Content-Type: application/json');
        echo json_encode($emparray); 

    ?>

Answer 1

只需将其正确放入JSON阵列即可。

$emparray = array();
while($row = mysqli_fetch_assoc($result)) {
    $emparray['Child Care'][] = $row['Child Care'];
}

header('Content-Type: application/json');
echo json_encode($emparray);

从评论中回答你的第二个问题：

$emparray = array();
while($row = mysqli_fetch_assoc($result)) {
    foreach($row as $key => $val) {
        $emparray[$key][] = $val;
    }
}

header('Content-Type: application/json');
echo json_encode($emparray);

Answer 2

这将提供您想要的格式。让我知道这是否适合您。

while($row =mysqli_fetch_assoc($result)){
    array_push($emparray,$row['Child Care']);
}   
header('Content-Type: application/json');
echo json_encode(array("Childcare" => $emparray));

Answer 3

尝试类似的事情，

$sql="select `Child Care` from Activity_type ";
$result = mysqli_query($connection,$sql) or die("Error in Selecting " . mysqli_error($connection));
$emparray=array();

while($row =mysqli_fetch_assoc($result)){
    $cares = explode(',' $row['Child Care']);
    foreach($cares as $care){
      $emparray[] = $care;
  }
}   

header('Content-Type: application/json');    
echo json_encode($emparray);

使用php explode功能

如何使用PHP以正确的格式获取JSON数据

3 个答案: