Php 在mysql查询中使用组_CONCAT时出错

我的代码运行良好，但当我添加

GROUP_CONCAT(CONCAT("<a href='index.php?id=", awards_user.userid, "'>", awards_user.username, "</a>")) AS userlist

GROUP_CONCAT（CONCAT（“”）作为用户列表

要获取用户名为的用户id，请显示错误

分析错误：语法错误，意外的T_字符串[在此行中]

SELECT awards.name as name, awards.link as link, 
       GROUP_CONCAT(DISTINCT awards_user.username) AS username,
       GROUP_CONCAT(DISTINCT awards_user.userid) AS userid,
       GROUP_CONCAT(CONCAT("<a href='index.php?id=", awards_user.userid, "'>", awards_user.username, "</a>")) AS userlist
   FROM awards LEFT JOIN awards_user ON (awards_user.awardid = awards.awardid)
   where awards.forumid = '".$_REQUEST['forumid']."'
   GROUP BY awards.awardid, awards.name, awards.link

选择awards.name作为名称，选择awards.link作为链接，
组\u CONCAT（不同的奖励\u user.username）作为用户名，
组\u CONCAT（不同的奖励\u user.userid）作为用户ID，
组_CONCAT（CONCAT（“”）作为用户列表
从奖品左侧加入奖品\u用户打开（awards\u user.awardid=awards.awardid）
其中awards.forumid='“$\u请求['forumid']””
按awards.awardid、awards.name、awards.link分组

您应该用反斜杠转义双引号

“

”，否则php会将其理解为字符串的结尾：

$sql = "SELECT awards.name as name
             , awards.link as link
             , GROUP_CONCAT(DISTINCT awards_user.username) AS username
             , GROUP_CONCAT(DISTINCT awards_user.userid) AS userid
             , GROUP_CONCAT(CONCAT(\"<a href='index.php?id=\", awards_user.userid, \"'>\"
             , awards_user.username, \"</a>\")) AS userlist
        FROM awards LEFT JOIN awards_user ON (awards_user.awardid = awards.awardid)
        WHERE awards.forumid = '" . $_REQUEST['forumid'] . "'
        GROUP BY awards.awardid
               , awards.name
               , awards.link";

$sql=“选择awards.name作为名称
，awards.link作为链接
，GROUP_CONCAT（独特的奖励_user.username）作为用户名
，GROUP_CONCAT（DISTINCT awards_user.userid）作为userid
，组\u CONCAT（CONCAT（\“\”）作为用户列表
从奖品左侧加入奖品\u用户打开（awards\u user.awardid=awards.awardid）
其中awards.forumid='“$\u请求['forumid']””
按奖项分组
，awards.name
，awards.link“；

引号转义是一个问题。为什么不这样使用呢

$formid = $_REQUEST['forumid'];

$query = <<<EOT
SELECT awards.name as name, awards.link as link, 
       GROUP_CONCAT(DISTINCT awards_user.username) AS username,
       GROUP_CONCAT(DISTINCT awards_user.userid) AS userid,
       GROUP_CONCAT(CONCAT("<a href='index.php?id=", awards_user.userid, "'>", awards_user.username, "</a>")) AS userlist
   FROM awards LEFT JOIN awards_user ON (awards_user.awardid = awards.awardid)
   where awards.forumid = '$formid'
   GROUP BY awards.awardid, awards.name, awards.link
EOT;

---

同一行中出现同样的错误，Thnx您是否有完整的PHP代码来设置此错误？将“$\u请求['forumid']”串联起来。“看起来很奇怪。您需要学习PHP字符串处理和转义的基础知识。看起来您需要转义双引号。

”“$\u请求['forumid']。“

这不是问题的原因，第行

GROUP_CONCAT（CONCAT（“”）上的错误作为userlist

thnx太多了，我发现可以解决问题，但我会尝试你的代码，我会注意你的观察

$formid = $_REQUEST['forumid'];

$id_link_1 = "<a href='index.php?id=";
$id_link_2 = "'>";
$id_link_3 = "</a>";

$query = <<<EOT
SELECT awards.name as name, awards.link as link, 
       GROUP_CONCAT(DISTINCT awards_user.username) AS username,
       GROUP_CONCAT(DISTINCT awards_user.userid) AS userid,
       GROUP_CONCAT(CONCAT($id_link_1, awards_user.userid, $id_link_2, awards_user.username, $id_link_3)) AS userlist
   FROM awards LEFT JOIN awards_user ON (awards_user.awardid = awards.awardid)
   where awards.forumid = '$formid'
   GROUP BY awards.awardid, awards.name, awards.link
EOT;