我正在开发一个IOS应用,对于API,我正在将请求发送到应该使用PHP返回JSON数据的URL。
[{"Child Care":"After Scool,Breakfast Club\n"},{"Child Care":"Breakfast Club"}]
但我想变得像
[{
"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);
?>
只需将其JSON
正确放置在您的阵列中即可。
$emparray = array();
while($row = mysqli_fetch_assoc($result)) {
$emparray['Child Care'][] = $row['Child Care'];
}
header('Content-Type: application/json');
echo json_encode($emparray);
要从评论中回答您的第二个问题:
$emparray = array();
while($row = mysqli_fetch_assoc($result)) {
foreach($row as $key => $val) {
$emparray[$key][] = $val;
}
}
header('Content-Type: application/json');
echo json_encode($emparray);
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我来说两句