从数据库中的字段中选择ID

Question

我有一个新闻数据库，我在每个新闻中都使用多只猫，所以我在一个名为cats的数据库中归档了文件，然后在其中插入数据，例如1,5,8,2这些数据就是cats id。

我用这段代码从数据库中选择猫

$select_newscats = $mysqli->query("SELECT * FROM news_cats where show_home = '1' and  kind = 1 or kind = 2 order by ord_show asc");

while ($rows_newscats = $select_newscats->fetch_array(MYSQL_ASSOC)){

    $id_newscats          = $rows_newscats ['id'];
    $title_newscats       = $rows_newscats ['title'];
    $ord_show_newscats    = $rows_newscats ['ord_show'];
    $icon_newscats        = $rows_newscats ['icon'];
    $kind_newscats        = $rows_newscats ['kind'];
    $description_newscats = $rows_newscats ['description'];

}

我想从数据库中选择新闻，其中猫中的$id_newscats归档在新闻表中，因此我遍历了此代码以获取新闻数据

$select_newscats = $mysqli->query("SELECT * FROM news_cats where show_home = '1' and  kind = 1 or kind = 2 order by ord_show asc");

while ($rows_newscats = $select_newscats->fetch_array(MYSQL_ASSOC)){

    $id_newscats          = $rows_newscats ['id'];
    $title_newscats       = $rows_newscats ['title'];
    $ord_show_newscats    = $rows_newscats ['ord_show'];
    $icon_newscats        = $rows_newscats ['icon'];
    $kind_newscats        = $rows_newscats ['kind'];
    $description_newscats = $rows_newscats ['description'];


    $select_news = $mysqli->query("SELECT * FROM news where $id_newscats  IN (cats)");
    while ($rows_news = $select_news->fetch_array(MYSQL_ASSOC)){

        $id_new          = $rows_news ['id'];
        $title_news       = $rows_news ['title'];

        echo "<div>{$title_news}</div>";
    }
}

但我对此代码有问题，例如，如果我有此猫的ID为5,2,7它只会选择第一个， 5,2,7它仅给我ID为5新闻，而忽略2,7所以我该如何解决并选择所有新闻哪里这个id $id_newscats在cats字段中谢谢。

Answer 1

看起来同时存在语法（1）和逻辑（2）错误。

1. SELECT * FROM news where $id_newscats IN (cats)错误的新闻。
   
   这是正确的语法。
   $select_news = $mysqli->query("SELECT * FROM news where cats IN ($id_newscats)")

2. 在查询中使用括号。

   SELECT * FROM news_cats where show_home = '1' and (kind = 1 or kind = order by ord_show asc"

Answer 2

您的IN查询顺序错误。

$select_news = $mysqli->query("SELECT * FROM news where cats IN ($id_newscats)")

Answer 3

您不是在外部while循环中建立IN列表！ 您得到一个行，然后在内部循环中只有一个ID可供搜索，因此不需要IN语法。 反正错了

您的外部查询也看起来不对，我认为您需要在OR周围加上括号。

$select_newscats = $mysqli->query("SELECT * 
                                   FROM news_cats 
                                   WHERE show_home = '1' 
                                    and  ( kind = 1 or kind = 2 ) 
                                   ORDER BY ord_show asc");

while ($rows_newscats = $select_newscats->fetch_array(MYSQL_ASSOC)){

    $id_newscats          = $rows_newscats ['id'];
    $title_newscats       = $rows_newscats ['title'];
    $ord_show_newscats    = $rows_newscats ['ord_show'];
    $icon_newscats        = $rows_newscats ['icon'];
    $kind_newscats        = $rows_newscats ['kind'];
    $description_newscats = $rows_newscats ['description'];

    // amended query, you only have one $id_newscats
    //  each time round this outer loop
    $select_news = $mysqli->query("SELECT * 
                                   FROM news 
                                   WHERE id = $id_newscats");

    while ($rows_news = $select_news->fetch_array(MYSQL_ASSOC)){

        $id_new          = $rows_news ['id'];
        $title_news       = $rows_news ['title'];

        // no need for {} round scalar variables
        //echo "<div>{$title_news}</div>";
        echo "<div>$title_news</div>";

        // or you could just dothis and dont bother 
        // creating an unecessary scalar variable
        //echo "<div>{$rows_news['title']}</div>";
    }
}

Answer 4

我懂了

改变这个

$select_news = $mysqli->query("SELECT * FROM news where $id_newscats  IN (cats)");

进入这个

$select_news = $mysqli->query("SELECT * FROM news where FIND_IN_SET('$id_newscats', cats) ");