我需要在php中的json响应方面的帮助

Question

这里是我的代码：

<?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);
}
?>

，在这之后，我得到了这个格式的响应，

{“错误”：“错误”：“service_name”}{“错误”}{“错误”：“service_name”：“米歇尔沙龙”}{“错误”：假，“service_name”：“米歇尔沙龙”}{错误“错误”：“service_name”：“迈克沙龙”}{“错误”}{“错误”：假，"service_name":"Etta沙龙“}

，我只想让这个响应像这个

{“错误”：“假”，“service_name”：“迈克沙龙”}，{“错误”：假，“service_name”：“米歇尔沙龙”}，{“错误”：假，“service_name”：“米歇尔沙龙”}，{“错误”：假，"service_name":"Mike“}，{”错误“：false，”service_name“：”service_name“

请帮助我为json找一份合适的回复表格。谢谢

Answer 1

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的结果都编码出来，并将其编码到循环的外面。

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 );

Answer 2

不要对单个结果进行json_encode()，而是将它们放入一个数组中，最后是json_encode()，即：

<?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);