在php變量中輸出mysql查詢結果

Question

我在PHP中有這個變量（ $sql ）：

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

如何將生成的數據作為PHP變量輸出？

該查詢從我的表中獲取出生日期並計算從今天開始的年齡。

如何在PHP中回應年齡？

Answer 1

試試這個：

$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'];

不要忘記替換table_name。

Answer 2

試試這段代碼

我看到你的代碼你錯過了表名

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

 $result = mysql_query($sql);
    while($query_data = mysql_fetch_row($result))
    $age= $query_data[0];
    print $age;

Answer 3

你可以嘗試如下：

$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!";
}

您可以參考http://www.w3schools.com