PHP尝试连接MySQL数据库!但代码不起作用。
这是代码,当我将代码放在PHP文件中时,显示实际上显示了我编写的代码。
我被困住了,请帮忙!
<?php // sqltest.php
require_once '../../htdocs/login.php';
$conn = new mysqli($hn, $un, $pw, $db);
if ($conn->connect_error) die($conn->connect_error);
if (isset($_POST['delete']) && isset($_POST['isbn']))
{
$isbn = get_post($conn, 'isbn');
$query = "DELETE FROM classics WHERE isbn='$isbn'";
$result = $conn->query($query);
if (!$result) echo "DELETE failed: $query<br>" .
$conn->error . "<br><br>";
}
if (isset($_POST['author']) &&
isset($_POST['title']) &&
isset($_POST['category']) &&
isset($_POST['year']) &&
isset($_POST['isbn']))
{
$author = get_post($conn, 'author');
$title = get_post($conn, 'title');
$category = get_post($conn, 'category');
$year = get_post($conn, 'year');
$isbn = get_post($conn, 'isbn');
$query = "INSERT INTO classics VALUES" .
"('$author', '$title', '$category', '$year', '$isbn')";
$result = $conn->query($query);
if (!$result) echo "INSERT failed: $query<br>" .
$conn->error . "<br><br>";
}
echo <<<_END
<form action="sqltest.php" method="post"><pre>
Author <input type="text" name="author">
Title <input type="text" name="title">
Category <input type="text" name="category">
Year <input type="text" name="year">
ISBN <input type="text" name="isbn">
<input type="submit" value="ADD RECORD">
</pre></form>
_END;
$query = "SELECT * FROM classics";
$result = $conn->query($query);
if (!$result) die ("Database access failed: " . $conn->error);
$rows = $result->num_rows;
for ($j = 0 ; $j < $rows ; ++$j)
{
$result->data_seek($j);
$row = $result->fetch_array(MYSQLI_NUM);
echo <<<_END
<pre>
Author $row[0]
Title $row[1]
Category $row[2]
Year $row[3]
ISBN $row[4]
</pre>
<form action="sqltest.php" method="post">
<input type="hidden" name="delete" value="yes">
<input type="hidden" name="isbn" value="$row[4]">
<input type="submit" value="DELETE RECORD"></form>
_END;
}
$result->close();
$conn->close();
function get_post($conn, $var)
{
return $conn->real_escape_string($_POST[$var]);
}
?>
我正在读这本书,名为Learning PHP,MySQL,&amp; JavaScript的 第4版作者:Robin Nixon,但是当我写出确切的代码时,它并没有像书中那样显示,但我在照片中得到了它。我正在尝试使用xampp连接MySQL和PHP。我还创建了php.login文件。我写了我的用户名和密码,并使用此代码将其保存在同一目录中! 请帮忙