php - 如何从两个表中获取特定数据？

Question

我有下表
评论表:

id  user_id question_id comment 
1     2        3          hii   

用户表:

id    u_name   user_image 
2     naveen     a.jpg 

当选择 Question_id=3 时，如何打印如下值:

user_image     u_name    comment
a.jpg          naveen      hii <br>

sql查询:

$sql = "SELECT u.user_image, u.name, c.comment
FROM users u INNER JOIN comment c
    ON u.id = c.user_id
WHERE c.question_id = 2"; 

$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
    // output data of each row
while($rows = mysqli_fetch_assoc($result)) {
echo '<td class="small">'.$rows['name'].'</td>';
echo '<td class="small">'.$rows['user_image'].'</td>';
echo '<td class="small">'.$rows['comment'].'</td>';

}
} 
else {
    echo "0 results";
}

Answer 1

使用INNER JOIN:

SELECT u.user_image, u.u_name, c.comment
FROM users u INNER JOIN comments c
    ON u.id = c.user_id
WHERE c.question_id = 3

SQLFiddle