如果条目不存在，我将尝试将条目从一个表复制到另一个表。下面的代码没有任何条目。我似乎无法确定。我在这里错过了什么？
<?php
$conn = mysqli_connect("localhost", "user", "Pass123", "DB");
$copy="SELECT * FROM TIME";
$doit=$conn->query($copy);
while($row = $doit->fetch_assoc()){
$check=mysqli_query($conn, "SELECT * FROM TIME2 WHERE O_LOG='{$row['LOG']}'");
if(!$check){
$sql = mysqli_query($conn, "INSERT INTO TIME2 (O_LOG, T_IN, T_OUT, EMP, ITEM, NOTE) VALUES ('".$row['LOG']."', '".$row['T_IN']."', '".$row['T_OUT']."', '".$row['EMP']."', '".$row['ITEM']."', '".$row['NOTE']."')");
}
}
?>
if (!$check)
应该是
if ($check->num_rows == 0)
您只是在测试查询是否出错，而不是它是否没有返回行。
但相反，您可以通过一个查询完成所有操作：
INSERT IGNORE INTO TIME2 (O_LOG, T_IN, T_OUT, EMP, ITEM, NOTE)
SELECT LOG, T_IN, T_OUT, EMP, ITEM, NOTE
FROM TIME
IGNORE修饰符告诉它静默地跳过任何可能导致重复密钥错误的插入(我假设LOG是唯一密钥(。
如果LOG不是唯一的密钥，您仍然可以使用一个查询：
INSERT INTO TIME2 (O_LOG, T_IN, T_OUT, EMP, ITEM, NOTE)
SELECT t.LOG, t.T_IN, t.T_OUT, t.EMP, t.ITEM, t.NOTE
FROM TIME AS t
LEFT JOIN TIME2 AS t2 ON t.LOG = t2.O_LOG
WHERE t2.O_LOG IS NULL
您可以尝试下面的SQL语句，它将自动比较并插入任何丢失的条目到克隆表：
INSERT INTO `TIME2`
SELECT * FROM `TIME` o
WHERE (
SELECT COUNT(*) FROM `TIME2` c WHERE c.`O_LOG` = o.`LOG`
) = 0
INSERT INTO TIME2 (select * from TIME where LOG not in(select O_LOG from TIME2 ))
或
INSERT INTO TIME2 (O_LOG, T_IN, T_OUT, EMP, ITEM, NOTE) (select LOG, T_IN, T_OUT, EMP, ITEM, NOTE from TIME where LOG not in(select O_LOG from TIME2 ))