Question

这是我的数据库：

id | liker | ques_id
1  | 15    | 2342
2  | 22    | 2342
3  | 22    | 2311
4  | 15    | 2389

我需要得到的是所有喜欢ques_id的人。所以结果应该是这样的：

Question 2342 has been liked by 15 and 22.
Question 2311 has been liked by 22 and so on

我当前的代码为每个liker和ques_id生成单独的行：

$sqlq=mysql_query("SELECT * FROM likes");
while($rowq=mysql_fetch_array($sqlq)){
  $qid=$rowq['ques_id'];

  $sql=mysql_query("SELECT * FROM likes where ques_id='$qid'");
  $num=mysql_num_rows($sql);
  $cont='';

  while($row=mysql_fetch_array($sql)){
    $liker=$row['liker'];
    $cont .="$qid being liked by $liker<br>";
  }
  echo $cont;
}

Answer 1

我没有测试过这个，但它应该让你开始：

$sqlq=mysql_query("SELECT DISTINCT(ques_id) FROM likes");
$cont='';
while($rowq=mysql_fetch_array($sqlq)){
  $qid=$rowq['ques_id'];

  $sql=mysql_query("SELECT * FROM likes where ques_id='$qid'");
  $num=mysql_num_rows($sql);
  $row=mysql_fetch_array($sql)
  $cont .= "$qid being liked by $liker ";

  while($row=mysql_fetch_array($sql)){
    $liker=$row['liker'];
    $cont .= " and $liker";
  }
  $cont .= ".<br>";
}
echo $cont;

Answer 2

你不需要对DB进行第二次查询，它只能迭代第一个结果。例如，您可以收集所有与关键字数组中特定ques_id相匹配的关键字，其中关键字为ques_id，如下所示：

$mathces = array();
$sqlq=mysql_query("SELECT * FROM likes");
while($rowq=mysql_fetch_array($sqlq)){
  $qid=$rowq['ques_id'];
  $liker=$rowq['liker'];

  $matches[ $qid ][] = $liker;
}

然后，您可以预先$mathces数组并构建字符串。

foreach ($matches as $qid => $likers) {
  $cont .= "$qid being liked by " . implode(' and ', $likers );
  if (count($likers) == 1)
    $cont .= ' and so on';
  echo "$cont<br>";
}

我没有测试我的代码。它需要额外的验证（例如，在第二个循环中为$linkers）

Answer 3

SELECT GROUP_CONCAT(liker), ques_id FROM likes GROUP BY ques_id

这将为每个ques_id将每个liker组合在一起。然后，您只需处理返回的行。

您应该避免使用mysql extension，因为它已被弃用;请改用PDO或mysqli。

$sql = 'SELECT GROUP_CONCAT(liker) as likes, ques_id FROM likes GROUP BY ques_id';
foreach ($conn->query($sql) as $row) {
    printf('Question %s is liked by %s', $row['ques_id'], $row['likes']);
}

如何根据ques_id检索liker id？

3 个答案: