在 php 中创建动态 JSON 数组答案

【问题标题】：Create dynamic JSON array in php在 php 中创建动态 JSON 数组
【发布时间】：2020-12-05 18:59:25
【问题描述】：

我想像这样动态创建一个 JSON：

{
"storie":[
{
"story_id":"111",
"username":"Username1",
"profile_photo":"boh.png",
"copertina":"ok.png",
"num_elements":"1",
"rating":"4.5"
},
{
"story_id":"222",
"username":"Username2",
"profile_photo":"hello.png",
"copertina":"hi.png",
"num_elements":"2",
"rating":"3.5"
}
]
}

我正在尝试从 MySQL 数据库中获取值，并且我能够做到：

$response = array();
$sql = mysqli_query($conn, "SELECT * FROM storie WHERE userid IN (SELECT following FROM follow WHERE follower='$userid')");
while($row = mysqli_fetch_assoc($sql)){
  $usern = getuserinfo($row['userid'], "username", $conn);
  $prof_photo = getuserinfo($row['userid'], "profile_photo", $conn);
  $idsto=$row['storia_id'];
  $elem = mysqli_query($conn, "SELECT COUNT(*) AS da_vedere FROM `storie_images` WHERE storia_id='$idsto' AND imm_id NOT IN (SELECT imm_id FROM image_views WHERE viewer='$userid')");
  while($ok = mysqli_fetch_assoc($elem)){
    $num_elem = $ok['da_vedere'];
  }
  //here I put the line that add the array to the json array:
}

但问题是这一行，它应该创建一个新数组并将其放入json中：

$response['storie'] = [array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5")];

只有一条记录时有效，多条记录时无效。有人可以帮我吗？

【问题讨论】：

警告：您对SQL Injections 持开放态度，应该使用参数化的prepared statements，而不是手动构建查询。它们由PDO 或MySQLi 提供。永远不要相信任何形式的输入！即使您的查询仅由受信任的用户执行，you are still in risk of corrupting your data。 Escaping is not enough!

标签： php mysql arrays json

【解决方案1】：

只需替换：

$response['storie'] = [array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5")];

与

$response['storie'][] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");

【讨论】：

【解决方案2】：

那是因为你在 while 中使用了 $response['storie'] 来解决这个问题：

1- 例如，创建另一个名为 result 的数组，然后在 while 中使用它

$response = array();
$result = array ()
while (....)
{
//here I put the line that add the array to the json array:
$result[] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
}

2-然后在循环外使用响应数组：

$response['storie'] = $result;

你的代码会是这样的：

   $response = array();
   $result = array ()

    $sql = mysqli_query($conn, "SELECT * FROM storie WHERE userid IN (SELECT following FROM follow WHERE follower='$userid')");
    while($row = mysqli_fetch_assoc($sql)){
    $usern = getuserinfo($row['userid'], "username", $conn);
    $prof_photo = getuserinfo($row['userid'], "profile_photo", $conn);
    $idsto=$row['storia_id'];
    $elem = mysqli_query($conn, "SELECT COUNT(*) AS da_vedere FROM `storie_images` WHERE storia_id='$idsto' AND imm_id NOT IN (SELECT imm_id FROM image_views WHERE viewer='$userid')");
    while($ok = mysqli_fetch_assoc($elem)){
    $num_elem = $ok['da_vedere'];
    }
    //here I put the line that add the array to the json array:
    $result[] = array("story_id" => $idsto, "username" => $usern, "profile_photo" => $prof_photo, "copertina" => $row['copertina'], "num_elements" => $num_elem, "rating" => "4.5");
    }

   $response['storie'] = $result;

【讨论】：