我正在尝试将当前select语句插入到带有php变量的数据库表中。它是这样工作的,没有变量:
mysql_query("INSERT INTO table (column) SELECT column FROM table1");
有谁知道我该怎么修理它吗?
PHP: 5.3.10 and
try this
$sample="SELECT column FROM table1"; // your select query
$query="INSERT INTO `table` (column) $sample";
mysqli_query('database connection code',$query);
您可以使用mysql_insert_id()获取最后插入的ID值,并尝试单独插入查询
<?PHP
$lastId = mysql_insert_id();
$query = mysql_query("Your Select Query with $lastId");
?>