无法制作表格。

Question

我想要做以下事情。我输入值并点击提交。如果该值超出范围0和299，则将文本'blabla'保存到变量$ error中。如果它在0到299的范围内，则将该值更新到MySQL表（行 - SkinID）中; 这是我尝试过的并没有工作。 这是我的HTML表单：

<form action="skinid.php" method="POST"> 
          <!--<fieldset class="textarea-field"> 
           <textarea title="Message">Message</textarea> 
          </fieldset>--> 
<br> 
          <fieldset> 
           <span class="ico user-ico"></span> 
           <input name="skinid" onkeypress="return isNumberKey(event)" type="text" class="field" value="Enter new Skin ID. (-2000$)" title="Enter new Skin ID. (-2000$)" /> 
          </fieldset> 
          <center><input type="submit" class="submit btn blue-btn" value="Update" /></center> 
         </form> 
    <?php $_SESSION['skinid'] = $_POST['skinid']; ?> 

这是skinid.php文件：

<?php 
session_start(); 
$newskinid = $_SESSION['skinid']; 
if($newskinid < '0' || $newskinid > '299'){$error = 'The Skin ID must be in the range of 0 and 299.';} 
else { 
$con=mysqli_connect("localhost","USERNAME","PASSWORD","DB-NAME"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
mysqli_query($con,"UPDATE users SET SkinID=$newskinid 
WHERE Username='$_SESSION[user]'"); 
mysqli_close($con);} 
?> 

Answer 1

尝试skinid.php：

<?php 
session_start(); 
$_SESSION['skinid'] = $_POST['skinid']; 
$newskinid = $_SESSION['skinid']; 
if($newskinid < '0' || $newskinid > '299') { 
    echo 'The Skin ID must be in the range of 0 and 299.'; 
} 
else { 
    echo 'dhgj'; 
} 
?> 
<form method="post"> 

<br> 
<fieldset> 
    <span class="ico user-ico"></span> 
    <input name="skinid" onkeypress="return isNumberKey(event)" type="text" class="field" value="Enter new Skin ID. (-2000$)" title="Enter new Skin ID. (-2000$)" /> 
</fieldset> 
<center><input type="submit" class="submit btn blue-btn" value="Update" /></center> 
</form> 

Answer 2

除非第一个代码示例也从skinid.php，在HTML表单你$_POST['skinid']是空的，因此在skinid.php中你的$_SESSION['skinid']也将是空的。

相反，在skinid.php使用： $newskinid = $_POST['skinid'];