请帮我解决此错误。请看我的代码..
错误是:
"Parse error: syntax error, unexpected '=' in\\...delete_user.php on line 21"
第21行有:
if ($_SERVER['REQUEST_METHOD'] = = 'POST') {
请帮助。我不知道如何检查语法,因为我是PHP代码的新手。
谢谢, 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);
?>
答案 0 :(得分:4)
我想你可能只有一个额外的空间:
if ($_SERVER['REQUEST_METHOD'] = = 'POST') {
试试这个(将“= =”更改为“==”):
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
答案 1 :(得分:3)
尝试删除= =
==