從表B中使用ID從表A中計數行php mysql

我必須精疲力竭，因爲我知道我以前做過這件事，但我不記得我是如何做的，或者是否找到該代碼段。我有兩個表格「問題」和「答案」。「questions」中的主要「id」在「答案」中也是「id」。我試圖根據該表中的「id」來計算「答案」。從表B中使用ID從表A中計數行php mysql

 <?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 />"); 
    } 

    ?>

來源

2012-10-08 mrmunk

'SELECT COUNT（answers.id）FROM問題LEFT JOIN答案ON questions.id = answers.id WHERE questions.id = $ id' –

嗯，首先，你永遠不執行$query2的。但是，要比返回所有行並對它們進行計數更好，只需返回mysql計算的行數即可。

$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.

來源

2012-10-08 23:36:34 zz1433

哦，好傷心。計數。嘖。我知道我現在已經不在了。謝啦。哇。我正在解僱自己 - 在我醒來之後。 – mrmunk

要獲得每個問題的答案的數量，你可以做到這一點

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

來源

2012-10-08 23:34:04

從表B中使用ID從表A中計數行php mysql

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