Question

$story_query = "SELECT table_name, id FROM planning WHERE parent = '$novelnum'";
$story_result = db_query($story_query);
while($story_row = db_fetch_array($story_result)) {
    $taleTable_Name = $story_row['table_name'];
    $postid[] = $story_row['id'];

    $q2 = "Select * from $taleTable_Name where approved='Y' order by id";
    $bset2 = db_query($q2);
    while($rset2 = db_fetch_array($bset2)) {
        $i[] = $rset2['id'];
        $t[] = $rset2['thread'];
        $s[] = $rset2['subject'];
        $a[] = $rset2['author'];
        $d[] = $rset2['datestamp'];

    }
}

if(isset($d)) {
    $fc = count($d);
    if($fc > 20) {
        $xs = $fc - 20;
    }
    else {
        $xs = 0;
    }

    for($c=$xs;$c<$fc;$c++) {
        if($s[$c] != "") {
            $newpost .= $d[$c];
            $newpost .= " <a href='../forums/read.php?f=";
            $newpost .= end($postid);
            $newpost .= "&i=";
            $newpost .= $i[$c];
            $newpost .= "&t=";
            $newpost .= $t[$c];
            $newpost .= "'>" ;
            $newpost .= $s[$c];
            $newpost .= "</a> by "; 
            $newpost .= $a[$c];
            $newpost .= $taleTable_Name;
            $newpost .= "<br>\n";
        }
    }
}
else {
    $newpost = "There are no posts for this scroll yet.";
}

上面的代码正确地向我展示了使用$ taleTable_Name找到的所有条目，但是当我打印$ newpost时，它只显示$ postid的最后一个变量。我想要id（$ postid匹配table_name（$ taleTable_Name），以便创建的url实际上进入正确的论坛。

Answer 1

首先，我不知道你的变量名是什么意思。我试过猜测，但你应该考虑重命名它们。为了保持维护程序员的理智。

查看你的代码，我知道它在做什么，但我不知道它应该做什么，或者为什么。下面是尝试重命名您的变量名称，但我不知道它是否正确，或接近右侧：

$story_query = "SELECT table_name, id FROM planning WHERE parent = '$novelnum'";
$story_result = db_query($story_query);
while($story_row = db_fetch_array($story_result)) {
    $taleTable_Name = $story_row['table_name'];
    $postid[] = $story_row['id'];

    $query2 = "Select * from $taleTable_Name where approved='Y' order by id";
    $bset2 = db_query($query2);
    while($rowset2 = db_fetch_array($bset2)) {
        $id[] = $rowset2['id'];
        $thread[] = $rowset2['thread'];
        $subject[] = $rowset2['subject'];
        $author[] = $rowset2['author'];
        $datetime[] = $rowset2['datestamp'];

    }
}

if(isset($datetime)) {
    $forumcount = count($datetime);
    if($forumcount > 20) {
        $xs = $forumcount - 20;
    }
    else {
        $xs = 0;
    }

    for($postnum=$xs;$postnum<$forumcount;$postnum++) {
        if($subject[$postnum] != "") {
            $newpost .= $datetime[$postnum];
            $newpost .= " <a href='../forums/read.php?f=";
            $newpost .= end($postid);
            $newpost .= "&id=";
            $newpost .= $id[$postnum];
            $newpost .= "&thread=";
            $newpost .= $thread[$postnum];
            $newpost .= "'>" ;
            $newpost .= $subject[$postnum];
            $newpost .= "</a> by "; 
            $newpost .= $author[$postnum];
            $newpost .= $taleTable_Name;
            $newpost .= "<br>\n";
        }
    }
}
else {
    $newpost = "There are no posts for this scroll yet.";
}

以下变量是什么意思？：$xs，$c，&i，&t和$fc？这些应该是有意义的名字。我已经重命名了你的其他变量，但对于这些我猜对了。我可能会离开。有意义的变量名称可以帮助某人阅读没有注释的代码。事实上，如果你有一个有意义的变量名，I may not need comments - 虽然'20'的magic number在做什么？这个号码有什么意义呢？

Answer 2

我注意到你的代码有一件事是没有打印出$ newpost的地方。在for循环结束之前，需要使用$ newpost完成某些操作。在这种情况下，您只是每次丢弃结果（或更正确地，覆盖它）。您需要使用$ newpost（打印它，将其存储在另一个值，将其发送回浏览器等）。

Answer 3

怎么样？

unset($post); // ensure start afresh

$story_query = "SELECT table_name, id FROM planning WHERE parent = '$novelnum'";
$story_result = db_query($story_query);
while($story_row = db_fetch_array($story_result)) {
    $taleTable_Name = $story_row['table_name'];
    $postid         = $story_row['id'];

    $q2 = "Select * from $taleTable_Name where approved='Y' order by id";
    $bset2 = db_query($q2);
    while($rset2 = db_fetch_array($bset2)) {
        $post[$postid] = $rset2;
    }
}

if(isset($post)) {
    /***
     * Page control here    
     ****/

    foreach($post as $id => $msg) {
        if($row['subject'] != "") {
            $newpost .= $msg['datestamp'];
            $newpost .= " <a href='../forums/read.php?f=";
            $newpost .= $id;
            $newpost .= "&i=".$msg['id'];
            $newpost .= "&t=".$msg['thread'];
            $newpost .= "'>" ;
            $newpost .= $msg['subject'];
            $newpost .= "</a> by "; 
            $newpost .= $msg['author'];
            $newpost .= $taleTable_Name;
            $newpost .= "<br>\n";
        }
    }
}
else {
    $newpost = "There are no posts for this scroll yet.";
}

PHP：在MySQL的数据数组中保存两个匹配项

3 个答案: