Question

我试图获取json文件的值并将其插入到mysql数据库中。

我从json文件中获取所有对象值，但是当我尝试在数组中获取数组的值时，我只需插入每一行：＆＃34; array＆＃34;

我使用以下代码（为此目的修复）：

 <?php
    set_time_limit(0);

    function ApiCall($http_server, $params = array())
    {
        $query_string   =   '?';

        if(is_array($params) && count($params)){
            foreach($params as $k=>$v){
                $query_string .= $k.'='.$v.'&';
            }
            $query_string   =   rtrim($query_string,'&');
        }
        return  file_get_contents($http_server.$query_string);
    }


    function getVideos($page = 0, $category = false)
    {
        $http       =   'http://api.domain.com/';
        $call       =   'reference';


        $params     =   array(
            'output'    =>  'json',
            'data'      =>  $call,
            'page'      =>  $page
        );

        if($category){
            $params['category']     =   $category;
        }

        $response   =   ApiCall($http , $params);

        if ($response) {
            $json   =    json_decode($response);
            if(isset($json->code) && isset($json->message)){
                throw new Exception($json->message, $json->code);
            }
            return $json;
        }
        return false;
    }


    $page                       =   1;
    $pages                      =   2;

    while(true) {

        try{
            if($page > $pages)
                break;

            $videoResponse  =   getVideos($page);
            $videos         =   $videoResponse->videos;             

            $con=mysqli_connect("","","","");
            if (mysqli_connect_errno())
                {
                    echo "Failed to connect to MySQL: " . mysqli_connect_error();
                }

            foreach($videos as $video) {
                $video_obj  =   $video->video;

                $video_id = $video_obj->video_id;
                $video_title = $video_obj->title;

                $video_tags = array();
                $video_tags[0] = $video_obj->tags-tag[0];
                $video_tags[1] = $video_obj->tags-tag[1];
                $video_tags[2] = $video_obj->tags-tag[2];
                $video_tags[3] = $video_obj->tags-tag[3];
                $video_tags[4] = $video_obj->tags-tag[4];
                $video_tags[5] = $video_obj->tags-tag[5];
                $video_tags[6] = $video_obj->tags-tag[6];
                $video_tags[7] = $video_obj->tags-tag[7];
                $video_tags[8] = $video_obj->tags-tag[8];
                $video_tags[9] = $video_obj->tags-tag[9];
                $video_tags[10] = $video_obj->tags-tag[10];

                $video_thumbnail = $video_obj->thumb;
                $video_embed = $video_obj->embed_url;
                $video_duration = $video_obj->duration;
                $video_url = $video_obj->url;

                $friendlyUrl = preg_replace('/^-+|-+$/', '', strtolower(preg_replace('/[^a-zA-Z0-9]+/', '-', $video_title)));

                $result = mysqli_query($con,"INSERT INTO movies (movie_id, title, tags, stars, thumbnail, thumbnails, embed, length, location, friendlyUrl) 
                                             VALUES ('". $video_id ."','". $video_title ."', '". $video_tags ."', '". $video_stars ."', '". $video_thumbnail ."', '". $video_thumbnails ."', 
                                             '". $video_embed ."', '". $video_duration ."', '". $video_url ."', '". $friendlyUrl ."')") 
                or die(mysqli_error());

            }

            $pages          =   $videoResponse->count % 20 == 0 ? $videoResponse->count / 20 : floor($videoResponse->count) / 20;
            $page++;

            unset($videoResponse);
            unset($videos);

            sleep(2);

        } catch(Exception $e){
            // Something is wrong with the response. You should handle that exception.
        }   
    }
?>

将$ video_tags插入数据库时，我没有得到数组值。

Answer 1

$video_tags是一个数组：

 $video_tags = array();
 $video_tags[0] = $video_obj->tags-tag[0];
 //                               ^ is that just a typo?
 // $video_tags[0] = $video_obj->tags->tag[0]; ???
 // etc.

所以你不能在这里的查询中插入它：

$result = mysqli_query($con,"INSERT INTO movies (...) 
             VALUES (... , '". $video_tags ."',  ...)") 
                               ^^^^^^^^^^^ here

理想情况下，您可以使用不同的表来存储标记，但如果您确实必须将它们存储在一个字段中，则需要serialize它们或将它们存储为例如json字符串：

$result = mysqli_query($con,"INSERT INTO movies (...) 
             VALUES (... , '". serialize($video_tags) ."',  ...)")

但是，这样几乎不可能搜索/过滤标签，因此您应该将它们存储在不同的表格中，以便将单个标签与单个电影相关联。

Answer 2

不确定这是否是问题，但在这一行：

$video_tags[0] = $video_obj->tags-tag[0];

您似乎要么在属性名称中使用短划线，要么忘记了标记变量上的美元符号。

使用PHP进行JSON解码

2 个答案: