Question

您好我的问题是我想要提供的请求是以下数组

["notes":{"email_id":"123","title":"John","notes":"15"},{"email_id":"15","title":"raj","notes":"hello"}],

但我的数组格式不同

{"email_id":"10","notes": ["hi,hello,how"],"title":"hello"}

喜欢这个请求只接受请帮帮我...... !!

<?php 

include('db.php');
$input = file_get_contents('php://input');
$input = json_decode($input);
$json=array();
$email= $input->email_id;
$title = $input->title;
$notes= $input->notes;
$noteid= $input->note_id;
//echo json_encode($email);
if($email != '')
{
    foreach($email as $key=>$value){    
        if($noteid == '')
           $qry= mysqli_query($conn,"INSERT INTO `notes`( `email_id`, `title`, `notes`) VALUES ('$value','$title[$key]','$notes[$key]')");
        else
           $qry= mysqli_query($conn,"update `notes` set `email_id` = '$value', `title` = '$title[$key]', `notes` = '$notes[$key]' where id = '$noteid' ");
    }

    if($qry){
        $json = array("response" => "success", "status"=>1 ,"msg" => (($noteid == '') ? "insert ": "Update ")."done!");
    }
    else{
        $json = array("response" => "failed", "status"=>0 ,"msg" => (($noteid == '') ? "insert ": "Update ")."failed!");
    }
}
else
   $json = array("response" => "failed", "status"=>2 ,"msg" => "Request Not Reached!");

echo json_encode($json);
?>

上面是我的代码在上面的代码抄袭请求格式中作为多个数组，但我没有得到我在一个对象中有不同的发送请求如何解决这个问题请使用原始数据作为邮递员请求发送提前感谢你.. !!

Answer 1

如果问题是你在期待一个json对象时收到一个字符串，那么就这样做：

int value = 5;
NSMutableArray * anArray  = [[NSMutableArray alloc] init];
[anArray addObject: @(value)];

Answer 2

您的json格式不正确

public class ApplicationUser : IdentityUser
{
   public string Name {get;set;}  //?? missing?

   public ICollection<UserTask> OwnedUserTasks { get; set; }
   public ICollection<UserTask> ExecutedUserTasks { get; set; }
}

正确的格式是

["notes":{"email_id":"123","title":"John","notes":"15"},{"email_id":"15","title":"raj","notes":"hello"}]

你可以使用json_decode（）函数

在php中解码json

`

我想使用PHP将输入作为多个数组格式发送

2 个答案: