在 php 中创建动态 JSON 数组

Question

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 数据库获取值，并且我能够做到这一点：

$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")];

当只有一条记录时它有效，但如果有更多记录它就不起作用。有人能帮我吗？

Answer 1

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

Answer 2

那是因为您在 while 内使用 $response['story'] 来解决此问题：1-创建另一个名为 result 的数组，然后在 while 中使用它$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;你的代码将是这样的：   $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;