通过HTML表单提交写入数据库的数据,PHP弹出报错!

2025-04-14 15:12:36
推荐回答(3个)
回答1:

$con=mysqli_connect("localhost","root","");
mysql_select_db("my_db",$con);
$mc = isset($_POST['mc'])?$_POST['mc']:'';
$sl = isset($_POST['sl'])?$_POST['sl']:'';
$jg = isset($_POST['jg'])?$_POST['jg']:'';
$fl = isset($_POST['fl'])?$_POST['fl']:'';
$sql="INSERT INTO jxc ('MC','SL','JG','FL')
VALUES('".$mc."','".$sl."','".$jg."','".$fl."')";
mysqli_query($sql,$con);
mysqli_close($con);

?>
试一下

回答2:

交驱动器偕鼓碧绿兑

回答3:

你这不给报错内容不知道哪的问题