【发布时间】:2018-04-13 04:13:23
【问题描述】:
我无法根据选择的选项插入值。因此,如果从下拉列表中选择 Type1,否则选择类型 2。如果选择类型 2,则会出现错误/甚至在选择类型 2 时尝试保持空值,但只会出现错误.when type 1 is selected data is inserted.
include_once('db.php');
if(isset($_POST['submit'])) {
$id = mysqli_real_escape_string($db ,$_POST['id']);
$Type = mysqli_real_escape_string($db ,$_POST['type']);
$Date = mysqli_real_escape_string($db ,$_POST['date']);
$Treat = mysqli_real_escape_string($db ,$_POST['treat']);
if($Type=="1")
{
$sql= "INSERT INTO set1(id,type,date,Treat)
VALUES('$id','$Type','$Date','$Treat')";
$query = mysqli_query($db, $sql) or die(mysqli_error($db));
if(!$query)
echo "You have to fill all values";
else
echo "successfully inserted";
}
else
{
$sql= "INSERT INTO set1(id, type, date)
VALUES('$id', '$Type', '$Date')";
$query = mysqli_query($db, $sql);
if(!$query)
echo mysqli_error();
else
//echo "successfully inserted";
echo"<script> alert('Data inserted Successfully'); window.close(); window.opener.location.reload(true);</script>";
}
}
?>
【问题讨论】:
-
请同时说明您的错误。
-
未定义索引:在 C:\wamp\www 中处理
-
@AlivetoDie 是的,请任何 cmets
-
您的数据库字段是否允许
NULL? -
您可以在从
mysqli_real_escape_string($db ,(isset($_POST['treat'])?$_POST['treat']:''))等数组访问之前检查密钥