即時通訊新的PHP所以我確信這是一個簡單的。即時得到這個錯誤「未定義的變量」通知
Notice: Undefined variable: conn in C:\Dev\Webserver\Apache2.2\htdocs\EclipsePHP\thecock\php\db.php on line 23
此代碼
<?php
$host = "localhost"; $database = "dbname"; $username = "user"; $password = "pass";
$conn = new mysqli($host, $username, $password, $database);
if (! $conn) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}else{
echo("all ok!");
}
function getContent($id) {
$sql = "SELECT content FROM blocktext WHERE id=$id";
if ($rs = $conn->query($sql)) { # line 23
if ($row = $rs->fetch_assoc()) {
echo stripslashes($row['content']);
}
$rs->close();
}
}
?>
如何解決這個通知?
第23行是?.... – Pentium10 2010-01-27 10:31:45
可能的重複[爲什麼我在PHP中獲得「未定義變量」通知?](http://stackoverflow.com/questions/1050359/why-am-i-getting -an-undefined-variable-notice-in-php) – outis 2012-07-08 18:02:01