Php 只有当登录用户出现在动态列表中时，才会显示链接

对于下面的函数，我希望该链接仅在登录用户当前出现在editorlist.php上时出现。即。如果函数中的loginid对应于当前出现在editorlist.php中的任何用户名

出现在editorlist.php上是动态的

我该怎么做

提前感谢,

约翰

在editorlist.php上：

$sqlStr = "SELECT 
    l.loginid, 
    l.username, 
    l.created,
    DATEDIFF(NOW(), l.created) AS days,
    COALESCE(s.total, 0) AS countSubmissions, 
    COALESCE(c.total, 0) AS countComments,
    COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore,
    DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2
FROM login l    
LEFT JOIN (
    SELECT loginid, COUNT(1) AS total 
    FROM submission 
    GROUP BY loginid
) s ON l.loginid = s.loginid
LEFT JOIN (
    SELECT loginid, COUNT(1) AS total 
    FROM comment 
    GROUP BY loginid
) c ON l.loginid = c.loginid
GROUP BY l.loginid
ORDER BY totalScore2 DESC 
LIMIT 10";

  $result = mysql_query($sqlStr);

$arr = array(); 
echo "<table class=\"samplesrec1edit\">";
while ($row = mysql_fetch_array($result)) { 
    echo '<tr>';
    echo '<td class="sitename1edit1"><a href="http://www...com/.../members/index.php?profile='.$row["username"].'">'.stripslashes($row["username"]).'</a></td>';
    echo '<td class="sitename1edit2">'.($row["countSubmissions"]).'</td>';
    echo '<td class="sitename1edit2">'.($row["countComments"]).'</td>';
    echo '<td class="sitename1edit2">'.($row["days"]).'</td>';
    echo '<td class="sitename1edit2">'.($row["totalScore2"]).'</td>';
    echo '</tr>';
    }
echo "</table>";

---

您可以尝试从editorlist.php运行SQL查询以确定用户是否在列表中，而不是尝试从editorlist.php页面本身提取数据。

我会使您的编辑器SQL更易于管理，例如：

function getEditors($editor = false) {
    $sqlStr = "SELECT  l.loginid,  l.username,   l.created,
    DATEDIFF(NOW(), l.created) AS days,   COALESCE(s.total, 0) AS countSubmissions, 
    COALESCE(c.total, 0) AS countComments,  COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore,   DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2
FROM login l LEFT JOIN (
    SELECT loginid, COUNT(1) AS total    FROM submission 
    GROUP BY loginid) s ON l.loginid = s.loginid LEFT JOIN (  SELECT loginid, COUNT(1) AS total 
    FROM comment    GROUP BY loginid ) c ON l.loginid = c.loginid ";

  if( $editor !== false ) { //if we specified an editor, find it
     $sqlStr .= " WHERE `l.loginid` = '" . $editor . "'";
  }

   $sqlStr .= "GROUP BY l.loginid
   ORDER BY totalScore2 DESC 
   LIMIT 10";

$result = mysql_query($sqlStr);

  if( $editor !== false) { // if we specified an editor, return that editor or false
     if( $row = mysql_fetch_assoc($result)) {
        return $row;
     }
     return false;
  }else { // otherwise, return the array of editors 
    $editors = array();
     while( $row = mysql_fetch_assoc($result)) {
       $editors[] = $row;
     }
    return $editors;
  }
}

对于editorlist.php，您可以这样做

$editors = getEditors();
foreach( $editors as $editors ) {
    // echo your table row like you were doing
}

$editor = getEditors(12); 
if( $editor ) {
  // echo your vote html stuff
} 

在show_用户框中，您可以

$editors = getEditors();
foreach( $editors as $editors ) {
    // echo your table row like you were doing
}

$editor = getEditors(12); 
if( $editor ) {
  // echo your vote html stuff
} 

---

我不是说要含蓄地这么做。。但我认为它会让您正确地了解从何处跳转。

如果一个用户在整个网站中被列为编辑，我建议设置一个$\u SESSION[]变量，在您处理其登录时记录该用户是否为编辑

如果编辑器在控制方面受到限制，那么可以选择将会话变量设置为可以编辑的项目数组

只是思考——检查一次并记住，而不是每次都检查。仅有一个的时间