Php 使用表A中的id计算表B中的行

我一定很累了，因为我知道我以前做过这个，但我不记得我是怎么做的，也不记得找到了那个片段。我有两张表格“问题”和“答案”。“问题”中的主要“id”在“答案”中也是“id”。我试图根据表格中的“id”来计算“答案”

     <?php

  $query = "SELECT * FROM questions ORDER BY id DESC";

  $result = mysql_query("$query");
  while($row = mysql_fetch_array($result)){
$id = $row['id'];
    $date = $row['date'];
    $question = $row['question'];
$time = date("U");
$likes = $row['likes'];
    $query2 = "SELECT * FROM answers WHERE id = $id";
    $num_rows = mysql_num_rows($query2);
    print("<span style='font-size: .7em';><a href='qplusone.php?id=$id'>+1</a>
 - Likes:$likes</span> $question - <a href='answer.php?id=$id&pp=$time'>answer it</a>
 or <a href='answers.php?id=$id&pp=$time'>read $num_rows answers</a> 
<span style='font-size: .7em';>($date)</span><br />");
  }

  ?>

---

要获得每个问题的答案数，您可以这样做

select q.id,
       count(*) as answers
from answer a
left join questions q on q.id = a.id 
group by q.id

要获得每个问题的答案数，您可以这样做

select q.id,
       count(*) as answers
from answer a
left join questions q on q.id = a.id 
group by q.id

首先，您从未执行过
$query2
。但比返回所有行并计数更好的是，只需返回mysql计数的行数

$query2 = "SELECT count(*) FROM answers WHERE id = $id";
$result = mysql_query($q);
$row = mysql_fetch_array($result);

$count = $row['count(*)']; // $count holds the number of matching records.

首先，您从未执行过
$query2
。但比返回所有行并计数更好的是，只需返回mysql计数的行数

$query2 = "SELECT count(*) FROM answers WHERE id = $id";
$result = mysql_query($q);
$row = mysql_fetch_array($result);

$count = $row['count(*)']; // $count holds the number of matching records.

SELECT COUNT（answers.id）FROM questions LEFT JOIN answers ON questions.id=answers.id WHERE questions.id=$id
SELECT COUNT（answers.id）FROM questions LEFT JOIN questions.id=answers.id WHERE questions.id=$id
噢，好的悲伤。计数嘘。我知道我现在没钱了。谢谢你，伙计。哇！我正在解雇自己——就在我醒来之后。哦，天哪。计数嘘。我知道我现在没钱了。谢谢你，伙计。哇！我正在解雇自己——就在我醒来之后。