Question

PHP提交网页之后根据下拉菜单的值更新数据库

$success = $unseccess = "";
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
$date_selected = $_POST['selecting'];
$time_selected = $_POST["selecting1"];
echo $date_selected;
$query2 = "SELECT count FROM selectTutorial WHERE dates='$date_selected' AND time='$time_selected'";
$result2 = mysqli_query($con,$query2);
$result2 || die("Unavailable selected! \n".mysqli_error($con));
while($row2 = mysqli_fetch_array($result2))
{
if($row2['count'] < 2)
{
$success = "Successful!";
echo $success;
$query3 = "UPDATE selectTutorial SET count=count+1 WHERE dates='$date_selected' AND time='$time_selected'";
$result3 = mysqli_query($con,$query3);
$result3 || die("Database access failed: ".mysqli_error($con));
}
else
{
$unseccess = "Unavailable selected!";
echo $unseccess;
}
}
}
这是我的代码。不管怎么测试都不能在网页上显示是否成功。请各位高手看看到底哪里出错了？
selecting和selecting都是我的<select>的name。
表格结构是这样的：
dates VARCHAR(10)
time VARCHAR(20)
count INT
我想要通过用户选择的dates还有time更新count的值。

Answer 1

$result2 || die("Unavailable selected! \n".mysqli_error($con));这段如果是数据库连接代码的话，那放的位置应该置顶，规范下应该是$result2=mysql_connect("","","") or die(mysql_error())；mysql_select(“数据库",$result2)；
$query3 = "UPDATE selectTutorial SET count=count+1 WHERE dates='$date_selected' AND time='$time_selected'"; 这个selectTutorial应该是一个表吧，count=count+1；这个是有问题的，你count只是一个字段，没有具体的值，应该换成row2['count']+1;
因为你这代码也不全，没法实测

Answer 2

$result2 || die("Unavailable selected! \n".mysqli_error($con));
while($row2 = mysqli_fetch_array($result2))
//上面这两句代码啥意思？

.....

if($row2['count'] < 2)
{
$success = "Successful!";
echo $success;
......

以上代码俺看不懂啥意思，呵呵
是不是写反了啊？
是不是这样：
if(count($row2)<2){
echo"Successful!";

}

才疏学浅，请指教

Answer 3

有没有打印出你的SQL语句看看有没有哪里出错了
把打印出的sql域名在数据库里面执行下看 能不能成功
数据链接没问题，语法没问题，可能就是 SQL语句有误