Question

这是我的代码：

<?php
while ($row = mysqli_fetch_assoc($searching_user))
{
    $salon_name = ucfirst($row['service_name']);
    $salon_id = ucfirst($row['id']);
    $salon_address = ucwords($row['address']);
    $salon_area = ucwords($row['area']);
    $salon_city = ucwords($row['city']);
    $salon_specialty = ucwords($row['specialty']);
    $img = $row['image_url'];
    $response["error"] = FALSE;
    $response["service_name"] = $salon_name;
    echo json_encode($response);
}
?>

在此之后我收到了这种格式的回复

{ “错误”：假， “服务名”：“麦克沙龙“} {”错误“：false，”service_name“：”米歇尔沙龙“} {“error”：false，“service_name”：“Michel salon”} {“error”：false，“service_name”：“Mike Salon”} {“error”：false，“service_name”：“Etta Salon”}

我只想要这样的回复

[{“error”：false，“service_name”：“迈克沙龙 “}，{” 错误 “：假，” 服务名 “：” 米歇尔沙龙 “}，{” 错误 “：假，” 服务名 “：” 米歇尔 salon“}，{”error“：false，”service_name“：”Mike Salon“}， {“error”：false，“service_name”：“Etta Salon”}]

请帮助我为json获取一份正确的回复表格。感谢

Answer 1

不要json_encode（）单个结果，而是把它们放到一个数组中，最后json_encode（）表示：

<?php
$response = [];
while ($row=mysqli_fetch_assoc($searching_user)) {
  $salon_name = ucfirst($row['service_name']); 
  $salon_id = ucfirst($row['id']);
  $salon_address = ucwords($row['address']);
  $salon_area = ucwords($row['area']);
  $salon_city = ucwords($row['city']);
  $salon_specialty = ucwords($row['specialty']);
  $img = $row['image_url'];

  $response[] = [
    'error' => FALSE,
    'service_name' => $salon_name,
    // you may want to add more attributes here...
  ];
}
echo json_encode($response);

我个人建议缩短这一点：

<?php
$response = [];
while ($row=mysqli_fetch_assoc($searching_user)) {
  $response[] = [
    'error'           => FALSE,
    'service_name'    => ucfirst($row['service_name']),
    'salon_id'        => $row['id'],
    'salon_address'   => ucwords($row['address']),
    'salon_area'      => ucwords($row['area']),
    'salon_city'      => ucwords($row['city']),
    'salon_specialty' => ucwords($row['specialty']),
    'img'             => $row['image_url'],
  ];
}
echo json_encode($response);

Answer 2

您正在尝试对单个结果进行编码，尝试创建一个数组将所有结果并将其编码到循环旁边。

while ($row=mysqli_fetch_assoc($searching_user)) {
    $salon_name = ucfirst($row['service_name']);
     $salon_id = ucfirst($row['id']);
     $salon_address = ucwords($row['address']);
     $salon_area = ucwords($row['area']);
     $salon_city = ucwords($row['city']);
     $salon_specialty = ucwords($row['specialty']);
     $img = $row['image_url'];
     $response["error"] = FALSE;
     $response["service_name"]=$salon_name;
     // Added this line
     $responses[] = $response;
}
//Encode all results
 echo json_encode($responses );

我在php中需要json响应方面的帮助

2 个答案: