PHP解析代码错误

-1

请帮我解决这个错误。请看看我的代码..PHP解析代码错误

的错误是：

"Parse error: syntax error, unexpected '=' in\\...delete_user.php on line 21"

线21具有这样的：

if ($_SERVER['REQUEST_METHOD'] = = 'POST') {

请帮助。我不知道如何检查语法，因为我是新来的PHP代码..

感谢，简

<?php # Script 1.7 delete_user.php 

$page_title = 'Delete a User'; 
//include ('includes/header.html');// 
echo '<h1>Delete a User</h1>'; 

// Check for a valid user ID value:// 
if ((isset($_GET['id'])) && (is_numeric($_GET['id']))) { $id = $_GET['id']; 
} elseif ((isset($_POST['id'])) && (is_numeric($_POST['id']))) { $id = $_POST['id']; 
    } else { echo '<p class="error">This page has been accessed in error.</p>'; 
     //include ('includes/footer.html');// 
    exit(); 

} 


require_once ('../mysqli_connect.php'); 



if ($_SERVER['REQUEST_METHOD'] = = 'POST') { 


if ($_POST['sure'] = = 'Yes') { $q = "DELETE FROM users10 WHERE user_id=$id LIMIT 1";   
$r = @mysqli_query($dbc, $q); 




if (mysqli_affected_rows($dbc) = = 1) { echo '<p>The user has been deleted.</p>'; 
} else { echo '<p class="error">The user could not be deleted due to a system error. 
</p>'; 
     echo '<p>' . mysqli_error($dbc) . '<br />Query: ' . $q . '</p>'; 
} 

} else { echo '<p>The user has NOT been deleted.</p>'; 

} 

} else { 


$q = "SELECT CONCAT(last_name, ',', first_name) FROM users10 WHERE user_id=$id"; 
$r = @mysqli_query($dbc, $q); if (mysqli_num_rows($r) = = 1) { $row =  
mysqli_fetch_array($r, MYSQLI_NUM); 


echo "<h3>Name: $row[0]</h3> Are you sure you want to delete this user?"; 



    echo '<form action="delete_user.php" method="post"> 
    <input type="radio" name="sure" value="Yes" /> Yes 
    <input type="radio" name="sure" value="No" checked="checked" /> No 
    <input type="submit" name="submit" value="Submit" /> 
    <input type="hidden" name="id" value="' . $id . '" /> 
    </form>'; 

    } else { echo '<p class="error">This page has been accessed in error.</p>'; } 

    } 
    mysqli_close($dbc); 


    ?>

来源

2013-10-07 jane

[参考？ - 这是什么错误PHP意味着]可能重复（http://stackoverflow.com/questions/12769982/reference-what-这个错误意味着在php中） – BlitZ

这个问题似乎是脱离主题，因为它是关于一个错字。 – j08691

我想你可能只是有一个额外的空间，在这里：

if ($_SERVER['REQUEST_METHOD'] = = 'POST') {

试试这个（改变 “= =” 到 “==”）：

if ($_SERVER['REQUEST_METHOD'] == 'POST') {

来源

2013-10-07 17:20:54 Will

嗨quasivivo，我试过你的建议和厕所！它的作品神奇。就这样？呵呵。对于我对代码的忽略感到抱歉...我真的很感谢你的帮助和时间来回答我的问题...非常感谢...更多的权力...欢呼.....简 – jane

尝试删除使您= =之间的空间就==

来源

2013-10-07 17:20:58 ameagher

嗨ameagher，非常感谢，我有尝试了你的建议，而且效果很好......所以这个小错误造成了不同，我花了数小时试图弄清楚......然后，就像这样？ heheh。再次，非常感谢和更多的权力....简 – jane

PHP解析代码错误

回答

相关问题