Question

所以我创建了一个PHP验证脚本。在测试中，我填写并提交了表单，但到目前为止，$ error返回未定义的索引，并且没有数据设置到数据库中。任何人都可以看看，并就其无法按预期运作的原因提出第二意见吗？在我眼里，它看起来都很好。

否则我的脚本运行正常（插入数据库）这只是我的验证脚本的一些内容。

 <?php  

    if (isset($_POST['Submit'])) {  

        if ($_POST['name'] != "") {  
            $_POST['name'] = filter_var($_POST['name'], FILTER_SANITIZE_STRING);  
            if ($_POST['name'] == "") {  
                $errors .= 'Please enter a valid name.<br/><br/>';  
            }  
        } else {  
            $errors .= 'Please enter a name.<br/>';  
        }  

           if (isset($_POST['Submit'])) {  

        if ($_POST['address'] != "") {  
            $_POST['address'] = filter_var($_POST['address'], FILTER_SANITIZE_STRING);  
            if ($_POST['address'] == "") {  
                $errors .= 'Please enter a valid address<br/><br/>';  
            }  
        } else {  
            $errors .= 'Please enter a address.<br/>';  
        }  

           if (isset($_POST['postcode'])) {  

        if ($_POST['postcode'] != "") {  
            $_POST['postcode'] = filter_var($_POST['postcode'], FILTER_SANITIZE_STRING);  
            if ($_POST['postcode'] == "") {  
                $errors .= 'Please enter a valid name.<br/><br/>';  
            }  
        } else {  
            $errors .= 'Please enter a name.<br/>';  
        }  

        if (!$errors) {  



$name = $_POST['name'];
$address = $_POST['address'];
$postcode = $_POST['postcode'];
$photo = $_POST['photo'];
$db1 = new dbmember();
$db1->openDB();
$numofrows = $db1->insert_member('', $name, $address, $postcode, $photo);
echo "Success. Number of rows affected:
<strong>{$numofrows}<strong>";



$sql="SELECT * from member";
$result=$db1->getResult($sql);
echo "<table class='table table-hover'>";
echo "<tr><th>Member ID</th><th>Name</th><th>Address</th><th>Postcode</th><th>Photo</th></tr>";

  while($row = mysqli_fetch_assoc($result))
{
 echo "<tr>";
 echo "<td>{$row['mid']}</td><td>{$row['name']}</td>";
 echo "<td>{$row['address']}";
echo "<td>{$row['postcode']}";
 echo"<td><img height='80' width='120' src='{$row['photo'] }' /></td>";  
 echo "</tr>";
 }
 echo "</table>";


$db1->closeDB();
}


        }
        }
        }
            echo "Records updated!<br/><br/>";  
        } else {  
            echo '<div style="color: red">' . $errors . '<br/></div>';  
        }  

?>  

<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post" name="myform" class = "well" id="myform" onsubmit="return validateForm(  );">

Please fill in the fields to add a new member
<p></p>
<input type="text" class="span3" placeholder="Enter member name"name="name" id="name" /><br />
<input type="text" class="span3"placeholder="Enter an address"name="address" id="address" /><br />
<input type="text" class="span3"placeholder="Enter a postcode"name="postcode" id="postcode" /><br />
<input type="text"class="span3" placeholder="Enter a picture (optional)"name="photo" /><br />
<p>
<button class="btn btn-primary" type="submit" value="Save" >Submit  </button>
</p>

</form>

Answer 1

您的按钮没有name="Submit"属性。您的PHP代码找不到$_POST['Submit']，因为它不存在。

因此，if (isset($_POST['Submit'])) {条件将返回false，这意味着永远不会执行验证，并且$ error变量永远不会设置为值。

麻烦我的PHP验证脚本

1 个答案: