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?
查询从我的表格中取出出生日期并计算从今天开始的年龄。
如何在PHP中回应年龄?
答案 0 :(得分:3)
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'];
不要忘记替换table_name。
答案 1 :(得分:0)
试试这段代码
我看到你错过了表名的代码
$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";
$result = mysql_query($sql);
while($query_data = mysql_fetch_row($result))
$age= $query_data[0];
print $age;
答案 2 :(得分:0)
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!";
}