试图通过PHP编辑SQL，但不断收到错误

Question

，据我可以看到我的代码是健全不过，我不断收到一个错误 这是错误

注意：未定义的变量：人 \ SQL \ modify.php上线

这里是我的代码..

<?php 
include 'includes/connection.php'; 

if (!isset($_POST['submit'])){ 
    $q = "SELECT * FROM people WHERE ID = $_GET[id]"; 
    $result = mysql_query($q); 
    $person = mysql_fetch_array($result);  
    } 
?> 
<h1>You are modifying A User</h1> 
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
Name<input type="text" name="inputName" value="<?php echo $person['Name']; ?>" /><br /> 
Description<input type="text" name="inputDesc" value="<?php echo $person['Description']; ?>" /> 
<br /> 
<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" /> 
<input type="submit" name="submit" value="Modify"/> 
</form> 
<?php 
if(isset($_POST['sumbmit'])) { 

$u = "UPDATE people SET `Name` = '$_POST[inputName]', `Description` = '$_POST[inputDesc]' WHERE ID = $_POST[id]"; 
    mysql_query($u) or die(mysql_error()); 

echo "User has been modify"; 
header("Location: index.php"); 

} 
?> 

任何想法或我只是盲？

Answer 1

<?php 
include 'includes/connection.php'; 

// set $person veriable 
if (!isset($_POST['submit'])){ 
    $q = "SELECT * FROM people WHERE ID = $_GET[id]"; 
    $result = mysql_query($q); 
    $person = mysql_fetch_array($result);  
} 
// if form submit you use update and redirect 
else { 
    $u = "UPDATE people SET `Name` = '$_POST[inputName]', `Description` =  '$_POST[inputDesc]' WHERE ID = $_POST[id]"; 
    mysql_query($u) or die(mysql_error()); 

    //echo "User has been modify"; // this not need, bcz execute header('location') redirect you current page 
    header("Location: index.php"); 
    exit(); //use it after header location 
} 
?> 
<h1>You are modifying A User</h1> 
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
Name<input type="text" name="inputName" value="<?php echo $person['Name']; ?>" /><br /> 
Description<input type="text" name="inputDesc" value="<?php echo $person['Description']; ?>" /> 
<br /> 
<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" /> 
<input type="submit" name="submit" value="Modify"/> 
</form> 

Answer 2

你只需要检查你是否真的有输出。

只是一个简单的例子：

if ($person): 
?> 
<h1>You are modifying A User</h1> 
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
Name<input type="text" name="inputName" value="<?php echo $person['Name']; ?>" /><br /> 
Description<input type="text" name="inputDesc" value="<?php echo $person['Description']; ?>" /> 
<br /> 
<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" /> 
<input type="submit" name="submit" value="Modify"/> 
</form> 
<?php 
endif; 

记住，总是验证您的输入和输出，而且如果查询您尝试运行做产生一个结果集。

Answer 3

这里的问题：

<?php 
include 'includes/connection.php'; 

if (!isset($_POST['submit'])){ 
    $q = "SELECT * FROM people WHERE ID = $_GET[id]"; 
    $result = mysql_query($q); 
    $person = mysql_fetch_array($result);  
    } 
?> 

此时$人只设置如果表单尚未提交;但后来：

Name<input type="text" name="inputName" value="<?php echo $person['Name']; ?>" /><br /> 
Description<input type="text" name="inputDesc" value="<?php echo $person['Description']; ?>" /> 

无论如何，您正在使用它。如果表格已经提交，那么你会得到你所看到的警告。你需要做的是一样的东西：

if (isset($_POST['submit'])){ 
    $name = $_POST['name']; 
    $description = $_POST['description'] 
} else { 
    $q = "SELECT * FROM people WHERE ID = $_GET[id]"; 
    $result = mysql_query($q); 
    $person = mysql_fetch_array($result);  

    $name = $person['name']; 
    $description= $person['description']; 
    } 

然后：

Name<input type="text" name="inputName" value="<?php echo $name ?>" /><br /> 
Description<input type="text" name="inputDesc" value="<?php echo $description; ?>" /> 

变量现在设置两种方式。

一些其他的东西 - 你没有做任何错误检查，看看你的查询是否工作;如果查询失败，您的代码将继续执行。

其次，mysql_函数已被弃用，并将在某个点停止工作;您应该考虑转而使用mysqli_*或PDO。