Question

我在这里愚蠢，因为我不知道是什么搞砸了我的节目。

我意外地删除了我的数据库，所有其他细节都消失了，但我马上从头开始创建了一个数据库。在我完成了一张桌子（工作正常）后，我遇到了这个问题

              //Selection of DB
     mysql_select_db("isladb", $con);


     //checking of fields
     $cusname =$_POST[cus_name];
     $cotnum =$_POST[cott_num];
     $eq_name = $_POST[eq_name];
     if($cusname == "" || $cotnum== ""|| $eq_name== ""){

        echo "<script>alert(\"Please complete the form before submiting.\");</script>";
        echo "<script>window.history.back();</script>"; 
     }
     elseif(is_numeric($cusname)){
        echo "<script>alert(\"Enter a valid name.\");</script>";
        echo "<script>window.history.back();</script>"; 
     }
     elseif($cotnum>20){
        echo "<script>alert(\"Exceeded number of cottages\");</script>";
        echo "<script>window.history.back();</script>"; 
     }


        else {

     //getdate
     date_default_timezone_set('Asia/Hong_Kong');
     $date=date('y.m.d H:i:s');
     //calculation of timeout
     $inc = 8 * 3600;
     $timeout = date('y.m.d H:i:s A', time()+ $inc);
     //calculation of total cost
     if ($eq_name == "Inner Tube") {
         $cost = 200;
     } elseif ($eq_name == "Floating Cottage") {
         $cost = 500;
     } 
     $hours = 8;
     $total_hours = $hours + $ext;
     $ext = 0;
     $total_cost = $cost + $ext;

     mysql_query("INSERT INTO equiptable (cus_ID, cus_name, time_in, total_hours, 
     total_cost, eq_name, time_out, cott_num) 
     VALUES (NULL, '$cusname', '$date', '$total_hours', '$total_cost', '$eq_name', 
     '$timeout', '$cotnum')");
     mysql_close($con);

     }

这是它之前的形式

     <form action="handler_eq.php" method="post" style="text-decoration:none; color:000;">
     Customer Name: <input type="text" name="cus_name" value=""/></br></br>
     Cottage Number:


              <?php
              $con=mysql_connect("localhost","root","");
             if(!$con)
             {die('Error'.mysql_error());}
             mysql_select_db("isladb",$con);
             $result = mysql_query("SELECT * FROM cottagetable");
             echo"<SELECT name='cott_num' >";
             echo"<option>Select </option>\n";

                 while ($row = mysql_fetch_array($result)){
                 $cotnum = $row["cott_num"];
                 $vacancy = $row["vacancy"];
                 if($vacancy=="Occupied")
                {     echo "              <option value=\"$cotnum\">$cotnum</option>";}
               else{};

                 }echo "          </select> ";
             ?>
        </SELECT>
     Equipment Name:
     </br>
     <input style="display:inline; width:0px;" name="eq_name" type="radio" value="Inner Tube" >Inner Tube<br/> 
     <input style="display:inline;width:0px;;" name="eq_name" type="radio" value="Floating Cottage">Floating Cottage</br></br>
     <input type="submit" value="Submit" onClick="return confirm('Submit form?')">
     </form>

我对此代码块感到非常困惑。我一整天都在努力工作，并决定在这里发表我的问题以征求意见。

问题：我分配给变量的值没有插入到我的数据库中 dbname：islaDB（已经检查了5次） tablename：equiptable（这也是）

[小屋租赁用品的学校项目]

希望你们能帮助我解决这个问题：O提前谢谢

为什么这个INSERT不会在MySQL中运行

0 个答案: