输出 mysql 查询结果在 php 变量中

Question

我在PHP中有这个变量（$sql）：

$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";

如何将生成的数据输出为 PHP 变量？

该查询是从我的表中获取出生日期，并从今天的日期计算年龄。

如何呼应PHP中的年龄？

Answer 1

试试这个：

$con = mysqli_connect("localhost","user","pass", "database_name"); //your connection
$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age FROM table_name WHERE id=".$row['id'];    
$query = mysqli_query($con, $sql);
$result = mysqli_fetch_assoc($query);
echo $result['age'];

不要忘记更换table_name。

Answer 2

试试这段代码

我看到你的代码你错过了表名

$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";
 $result = mysql_query($sql);
    while($query_data = mysql_fetch_row($result))
    $age= $query_data[0];
    print $age;

Answer 3

您可以尝试如下：

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
    echo "age: " . $row["age"]. "<br>";}
} 
else {
echo "No result!";
}

您可以从 http://www.w3schools.com 引用