Mysqli为特定值获取assoc循环

Question

我有下表呼叫爱好：

name,  hobby1,  hobby2,  hobby3,   hobby4

Jenny, Skiing,  Bowling, Swimming, Dancing
Jane,  Skiing,  Jogging, Eating,   Dancing
Bonny, Skiing,  Bowling, Swimming, Dancing
Alice, Running, Bowling, Swimming, Drinking
Amber, Running, Bowling, Eating,   Dancing

我可以使用fetch_assoc函数来获取拥有hobby1的人的名字：

$sql = "SELECT name, hobby1 FROM hobbies";
$result = mysqli_query($con, $sql);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo "Name: " . $row['name'].", ".$row['hobby1']. "<br>";
    }
} else {
    echo "0 results";
}

答案：

Name: Jenny, Skiing
Name: Jane, Skiing
Name: Bonny, Skiing
Name: Alice, Running
Name: Amber, Running

如何修改代码只列出喜欢滑雪的人？

Answer 1

您可以将where子句设为

SELECT name, hobby1 FROM hobbies where hobby1 = 'Skiing'

Answer 2

如果您要求使用Skiing获取hobby1，那么这是您需要的查询

$sql = "SELECT name, hobby1 FROM hobbies WHERE hobby1 = 'Skiing'";