我有两张桌子;一个用于图片，另一个用于他们的评论（l
picture_table (pic_id,pic_owner,pic_insert_date)
picture_comment_table (pic_id,pic_commentor_name,pic_comments,pic_comment_date)
如何创建一个PHP代码，在所有图片的特定图片对象中创建一个注释数组，如下面的代码？
[
{
"pic_id": "picture100",
"pic_owner": "Stanley Kin",
"pic_insert_date": "23-02-2017",
[
{
"pic_commentor_name": "Kim Love",
"pic_comments": "You look beautifull"
"pic_comment_date": "31-01-2017"
},
{
"pic_commentor_name": "John doe Love",
"pic_comments": "Marvelous you are"
"pic_comment_date": "13-01-2017"
}
]
},
{
"pic_id": "picture101",
"pic_owner": "Linky Nam",
"pic_insert_date": "21-04-2017",
[
{
"pic_commentor_name": "Lira Love",
"pic_comments": "You look Gorgeous"
"pic_comment_date": "31-04-2017"
},
{
"pic_commentor_name": "Bram Javan",
"pic_comments": "love the pics more....."
"pic_comment_date": "11-01-2017"
}
]
}
]
答案 0 :(得分：0)
您不必知道从mysql表创建json的列。 我在下面创建了简单的代码，将表格带到json。 注意： 解析时通过'comments'键获取图片的注释; 假设conn变量已被定义。
public function getPictureInfos(){
$pictures = $this->getNewDataFromTable("picture_table");
for($i = 0; $i < count($pictures); $i++) {
$pictures[$i]['comments'] = $this->getNewDataFromTable("picture_comment_table");
}
return json_encode($pictures);
}
public function getNewDataFromTable($tableName){
$sql = "SELECT * FROM ".$tableName;
$result=$this->conn->query($sql);
$data = array();
$count=0;
if($result)
while($row = $result->fetch_assoc()) {
$data[$count]= $row;
$count++;
}
return $data;
}
编辑：使用“评论”键接收评论的含义。
举一个你的json字符串的例子：
[
{
"pic_id": "picture100",
"pic_owner": "Stanley Kin",
"pic_insert_date": "23-02-2017",
[
{
"pic_commentor_name": "Kim Love",
"pic_comments": "You look beautifull"
"pic_comment_date": "31-01-2017"
},
{
"pic_commentor_name": "John doe Love",
"pic_comments": "Marvelous you are"
"pic_comment_date": "13-01-2017"
}
]
}
]
其中注释数组是包含图片详细信息的对象的成员之一。数组没有键，你的json实际上会在解析过程中抛出JSONException。
要解决这个问题，请在创建json时为该数组提供一个键，并在解析时通过'comments'键对图片进行注释，
所以我的代码生成的json就是这样：
[
{
"pic_id": "picture100",
"pic_owner": "Stanley Kin",
"pic_insert_date": "23-02-2017",
"comments":[
{
"pic_commentor_name": "Kim Love",
"pic_comments": "You look beautifull"
"pic_comment_date": "31-01-2017"
},
{
"pic_commentor_name": "John doe Love",
"pic_comments": "Marvelous you are"
"pic_comment_date": "13-01-2017"
}
]
}
]
答案 1 :(得分：0)
我认为这就是你的意思
<?php
$pic[0] = array("pic_id"=> "picture100","pic_owner"=> "Stanley Kin","pic_insert_date"=> "23-02-2017",);
$pic[0]['comments'][] = array("pic_commentor_name"=> "Kim Love","pic_comments"=> "You look beautifull","pic_comment_date"=> "31-01-2017");
$pic[0]['comments'][] = array("pic_commentor_name"=> "John doe Love","pic_comments"=> "Marvelous you are","pic_comment_date"=> "31-01-2017");
$pic[1] = array("pic_id"=> "picture101","pic_owner"=> "Linky Nam","pic_insert_date"=> "21-04-2017");
$pic[0]['comments'][] = array("pic_commentor_name"=> "Lira Love","pic_comments"=> "You look Gorgeous","pic_comment_date"=> "31-01-2017");
$pic[0]['comments'][] = array("pic_commentor_name"=> "Bram Javan","pic_comments"=> "love the pics more.....","pic_comment_date"=> "11-01-2017");
$json = json_encode($pic);
echo $json;
?>
结果：
[
{
"pic_id": "picture100",
"pic_owner": "Stanley Kin",
"pic_insert_date": "23-02-2017",
"comments": [
{
"pic_commentor_name": "Kim Love",
"pic_comments": "You look beautifull",
"pic_comment_date": "31-01-2017"
},
{
"pic_commentor_name": "John doe Love",
"pic_comments": "Marvelous you are",
"pic_comment_date": "31-01-2017"
}
]
},
{
"pic_id": "picture101",
"pic_owner": "Linky Nam",
"pic_insert_date": "21-04-2017",
"comments": [
{
"pic_commentor_name": "Lira Love",
"pic_comments": "You look Gorgeous",
"pic_comment_date": "31-01-2017"
},
{
"pic_commentor_name": "Bram Javan",
"pic_comments": "love the pics more.....",
"pic_comment_date": "11-01-2017"
}
]
}
]
对于SQL使用此：
$con=@mysql_connect('host', 'user', 'password');
$query = "SELECT * FROM `picture_table`";
$result=mysql_query($query,$con);
$pic = array();
$counter = 0;
while($row=mysql_fetch_array($result,MYSQL_ASSOC)){
$pic[$counter] = array("pic_id"=> $row["pic_id"],"pic_owner"=> $row["pic_owner"],"pic_insert_date"=> $row["pic_insert_date"]);
$query = "SELECT * FROM `picture_comment_table`";
$result=mysql_query($query,$con);
while($comments_row=mysql_fetch_array($result,MYSQL_ASSOC)){
$pic[$counter]['comments'][] = array("pic_commentor_name"=> $comments_row['pic_commentor_name'],"pic_comments"=> $comments_row['pic_comments'],"pic_comment_date"=> $comments_row['pic_comment_date']);
}
$counter++;
}
或者这个：
$con=@mysql_connect('host', 'user', 'password');
$query = "SELECT `picture_table`.*
, GROUP_CONCAT(DISTINCT CONCAT_WS( '^^^^',`picture_comment_table`.`pic_commentor_name` , `picture_comment_table`.`pic_comments` , `picture_comment_table`.`pic_comment_date`)SEPARATOR '||||' ) AS `comment`
FROM `picture_table`
INNER JOIN `picture_comment_table` USING (`pic_id`)
GROUP BY `pic_id`";
$result=mysql_query($query,$con);
$pic = array();
$counter = 0;
while($row=mysql_fetch_array($result,MYSQL_ASSOC)){
$pic[$counter] = array("pic_id"=> $row["pic_id"],"pic_owner"=> $row["pic_owner"],"pic_insert_date"=> $row["pic_insert_date"]);
$comment_list = explode("||||",$row['comment']);
foreach ($comment_list as $value){
$comment_array = explode("^^^^",$value);
$pic[$counter]['comments'][] = array("pic_commentor_name"=> $comment_array[0],"pic_comments"=> $comment_array[1],"pic_comment_date"=> $comment_array[2]);
}
$counter++;
}