Question

我有一个带表格的moviesdb

电影表;
数据表
人员表
字符表
模式表

电影表

===============================================
CODE : TITLE :        DESCRIPTION       YEAR
=====================================================
1      IRON MAN 3       xxxxx          2013
2      STAR WARS        xxxxx          2013

人员表

===============================================
CODE : NAME:                 BIRTH_DATE
=====================================================
1      SHANE BLACK         1965:04:04 00:00:00
2      ROBERT DOWNEY       1965:04:04 00:00:00
3      Gwyneth Paltrow     1965:04:04 00:00:00
4      Don Cheadle         1965:04:04 00:00:00

字符表

===============================================
PERSON  : NAME:             MOVIE
=====================================================
2      TONY STARK          1
3      PEPPER POTTS        1
4      Col. James Rhodes   1

模式表

===============================================
CODE : NAME : 
=====================================================
1      DIRECTOR 
2      WRITER
3      ACTOR
4      CHARACTER

数据表

===============================================
CODE : MOVIE :     PERSON   MODE
=====================================================
1      1             1       1
2      1             2       3 
3      1             3       3 
4      1             4       3

最终回答如下：

===============================================
CODE : MOVIE TITLE :  DIRECTOR  :     ACTORS           CHARACTERS
=====================================================
1      IRON MAN 3     SHANE BLACK   Robert Downey Jr.   Tony Stark
                                    Gwyneth Paltrow     Pepper Potts
                                    Don Cheadle         Col. James

MySQL命令是：

SELECT 
  movies.title,
  movies.year,
  persons.name,
  modes.name,
  persons.birth_date,
  characters.name
FROM
  data
  INNER JOIN movies ON (data.movie = movies.code)
  INNER JOIN persons ON (data.person = persons.code)
  INNER JOIN modes ON (data.mode = modes.code)
  INNER JOIN characters ON (persons.code = characters.person)
  AND (characters.movie = data.movie)
WHERE
  MOVIES.code = '1'

SQL命令工作正常;

我遇到了PHP CODE的问题

$movie_query = mysql_query("
SELECT 
  movies.title,
  movies.year,
  persons.name,
  modes.name,
  persons.birth_date,
  characters.name
FROM
  data
  INNER JOIN movies ON (data.movie = movies.code)
  INNER JOIN persons ON (data.person = persons.code)
  INNER JOIN modes ON (data.mode = modes.code)
  INNER JOIN characters ON (persons.code = characters.person)
  AND (characters.movie = data.movie)
WHERE
  MOVIES.code = '1';");

WHILE ($rows = mysql_fetch_array($movie_query ))
{ 
    $movie_code = $rows['movies'.'code'];
    $movies_title = $rows['movies'.'title'];
    $movies_year = $rows['movies'.'year'];
    $movies_date_add = $rows['movies'.'date_add'];
    $movies_tagline = $rows['movies'.'tagline'];

    echo "$movie_code<br>$movies_title<br>$movies_year<br>$movies_date_add<br>$movies_tagline<br>";
}

我的PHP代码有什么问题..？（似乎SQL命令与MySQL语法不同）

Answer 1

1你需要使用PDO而不是使用mysql，请参阅上面的christian-giupponi评论，了解其中的内容

2你有一个“;”在你的sql语句中应删除

2您的查询结果不包括表名

3你无法访问表格中的数据，除非你选择它，所以$ row ['code']不起作用，除非你在sql中要求movies.code

4当您选择具有相同名称的字段时，您需要重新分配它们，否则您将会出现错误

$movie_query = mysql_query("
SELECT 
    movies.code,
    movies.title,
    movies.year,
    persons.name as pname,
    modes.name as mname,
    persons.birth_date,
    characters.name as cname
FROM
    data
    INNER JOIN movies ON (data.movie = movies.code)
    INNER JOIN persons ON (data.person = persons.code)
    INNER JOIN modes ON (data.mode = modes.code)
    INNER JOIN characters ON (persons.code = characters.person)
    AND (characters.movie = data.movie)
WHERE 
    MOVIES.code = '1'
");

WHILE ($rows = mysql_fetch_array($movie_query ))
{ 
    $movie_code = $rows['code'];
    $movies_title = $rows['title'];
    $movies_year = $rows['year'];
    $movies_date_add = $rows['date_add'];
    $movies_tagline = $rows['tagline'];

    echo "$movie_code<br>$movies_title<br>$movies_year<br>$movies_date_add<br>$movies_tagline<br>";
}

祝你好运，请记住，如果你现在不使用PDO，你将会受到影响

从多个mysql表中获取数据并以PHP（数组）显示

1 个答案: