Php mysql连接未正确执行

Php mysql连接未正确执行,php,mysql,join,left-join,Php,Mysql,Join,Left Join,这是我正在执行的mysql查询。问题是,表mymembers和forum_posts都有id字段,所以当我使用$id并保存“id”时,成员的id存储在其中。下面是我正在显示的代码 $sql = mysql_query(" SELECT * FROM forum_posts LEFT JOIN mymembers ON forum_posts.post_author_id = mymembers.id WHERE section_id='$sid' ORDER

这是我正在执行的mysql查询。问题是,表mymembers和forum_posts都有id字段,所以当我使用$id并保存“id”时,成员的id存储在其中。下面是我正在显示的代码

$sql = mysql_query("
    SELECT * 
    FROM forum_posts 
    LEFT JOIN mymembers ON forum_posts.post_author_id = mymembers.id 
    WHERE section_id='$sid' ORDER BY date_time DESC LIMIT 25
");
$dynamicList = "";
$numRows = mysql_num_rows($sql);

while($row = mysql_fetch_array($sql)){
    $reply_author_id = $row["post_author_id"];
    $author = $row["username"];
    $id = $row["id"];
    $anon = $row["anon"];
    $post_body=$row["post_body"];
    $post_author_id = $row["post_author_id"];
    $date_time = $row["date_time"];
    $thread_title = $row["thread_title"];

最好的做法是列出查询中的所有字段

不要使用SELECT*将其更改为:

$sql = mysql_query(
"SELECT forum_posts.post_author_id AS post_author_id, forum_posts.id AS id,    mymembers.username, forum_posts.anon,forum_posts.post_body,forum_posts.thread_title, forum_posts.date_time FROM forum_posts LEFT JOIN mymembers ON forum_posts.post_author_id=mymembers.id WHERE section_id='$sid' ORDER BY date_time DESC LIMIT 25");

最好的做法是列出查询中的所有字段

不要使用SELECT*将其更改为:

$sql = mysql_query(
"SELECT forum_posts.post_author_id AS post_author_id, forum_posts.id AS id,    mymembers.username, forum_posts.anon,forum_posts.post_body,forum_posts.thread_title, forum_posts.date_time FROM forum_posts LEFT JOIN mymembers ON forum_posts.post_author_id=mymembers.id WHERE section_id='$sid' ORDER BY date_time DESC LIMIT 25");

无需更改数据库中的列名,只需在查询中重命名即可:

SELECT mymembers.id as id, forum_posts.id as fpID, etc...

无需更改数据库中的列名,只需在查询中重命名即可:

SELECT mymembers.id as id, forum_posts.id as fpID, etc...

您可以对列进行别名:
选择mymembers.id作为mymember\u id,选择forum\u posts.id作为postid,…
您可以对列进行别名:
选择mymembers.id作为mymember\u id,选择forum\u posts.id作为postid,…