Php 尝试在一个SQL之后更新SQL
我从DOB字段按日期计算年龄,然后我想根据DOB将其推入具有正确年龄的年龄。因此,当我调试DOB时,计算年龄是可行的,但它无法更新年龄代码:Php 尝试在一个SQL之后更新SQL,php,sql,Php,Sql,我从DOB字段按日期计算年龄,然后我想根据DOB将其推入具有正确年龄的年龄。因此,当我调试DOB时,计算年龄是可行的,但它无法更新年龄代码: <?php $servername = "localhost"; $username = "usernameexmaple"; $password = "passworking"; $dbname = "dbnameworking"; // Create connection // Create connection $conn = new my
<?php
$servername = "localhost";
$username = "usernameexmaple";
$password = "passworking";
$dbname = "dbnameworking";
// Create connection
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id as ID, YEAR(CURRENT_TIMESTAMP) - YEAR(dob) - (RIGHT(CURRENT_TIMESTAMP, 5) < RIGHT(dob, 5)) as age
FROM regio_users";
$sql2 = ("UPDATE regio_users SET age = '$newage' WHERE id ='$newid' ");
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$newage = $row['age'];
$newid = $row['ID'];
$sql2 = ("UPDATE regio_users SET age = '$newage' WHERE id ='$newid' ");
$result2 = $conn->query($sql);
if ($result2){
echo "done"."<br>";
}
}
}
else {
echo "0 results";
}
$conn->close();
?>
正如saty所指出的,使用正确的变量名
检查自动提交是否打开。如果没有,请确保提交数据。检查PHP中的语法
这可以通过一行SQL完成,而不是使用PHP在所有行中循环以仅更新年龄:
UPDATE `regio_users` SET `age` = YEAR(CURRENT_TIMESTAMP) - YEAR(`dob`) - (RIGHT(CURRENT_TIMESTAMP, 5) < RIGHT(`dob`, 5));
UPDATE`regio\u users`SET`age`=YEAR(CURRENT\u TIMESTAMP)-YEAR(`dob`)-(RIGHT(CURRENT\u TIMESTAMP,5)
您已经使用了$result2=$conn->query($sql)代码>这是不正确的。您必须使用$result2=$conn->query($sql2)
as$sql2
是您形成的新查询。$result2=$conn->query($sql2)代码>使用$sql2
大量运行更新查询库,在我在此发布之前必须再次阅读我的sscript。很抱歉浪费时间,非常感谢你帮我把它赶出来。