Question

我收到错误：

Fatal error: Uncaught Error: Call to undefined function mysqli_connect() in C:\Apache\htdocs\movie\actor_form\demo.php:9 Stack trace: #0 {main} thrown in C:\Apache\htdocs\movie\actor_form\demo.php on line 9

当我尝试通过php表单将数据输入我的数据库时。这是我的两个脚本：

demo.php文件：

<?php

define('DB_Name', 'movies');
define('DB_USER', 'root');
define('DB_PASSWORD', 'root');
define('DB_HOST', 'localhost');

$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

if (!$link)  {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysqlI_select_db(DB_NAME, $link);

if (!$db_selected)  {
    die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}

$actfn = $_POST['Actor First Name'];
$actln = $_Post['Actor Last Name'];
$actag = $_POST['Actor Age'];
$actgn = $_POST['Actor Gender'];
$actht = $_POST['Actor Height'];

$sql = "INSERT INTO actor_demographics (Actor_First_Name, Actor_Last_Name, Actor_Age, Actor_Gender, Actor_Height)  VALUES ('$actfn', '$actln', '$actag', '$actgn', '$actht')";

if (!mysql_query($sql))  {
    die('ERROR: ' . mysql_error());
}

mysql_close();
?>

demo-form.php文件：

<form action="demo.php" method="post" />
<p>Actor First Name: <input type="text" name="Actor First Name" /><p>
<p>Actor Last Name: <input type="text" name="Actor Last Name" /><p>
<p>Actor Age: <input type="text" name="Actor Age" /><p>
<p>Actor Gender: <input type="text" name="Actor Gender" /><p>
<p>Actor Height: <input type="text" name="Actor Height" /><p>

<input type="submit" value="Submit" />
</form>

错误引用的第9行是：

$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

在此问题上还有其他帖子表明我需要删除;在命令

;extension=php_mysqli.dll

在我的php.ini文件中找到

我已尝试删除此分号并将其重新添加，但仍然收到错误。

非常感谢任何帮助。

谢谢

Php.ini编辑文件：

; Directory in which the loadable extensions (modules) reside.
; http://php.net/extension-dir
; extension_dir = "./"
; On windows:
extension_dir ="C:/php/ext"

;extension=php_exif.dll      ; Must be after mbstring as it depends on it
extension=php_mysqli.dll
;extension=php_oci8_12c.dll  ; Use with Oracle Database 12c Instant Client

Answer 1

试试这个

如果您正在使用本地服务器，则无需输入密码，只需将其设为“空白”即可。

始终使用SQL注入来提交数据库中的数据例如

$actfn=$link ->real_escape_string(trim($_POST['Actor_First_Name']));

如果您使用mysqli_query() expects at least 2 parameters, 1

，则会收到错误mysqli_query($sql)

你必须这样使用。

if (!mysqli_query($link,$sql))
  {
  echo("Error description: " . mysqli_error($link));
  }

mysqli_close($link);

在此处查看完整代码

  define('DB_NAME', 'movies');
    define('DB_USER', 'root');
    define('DB_PASSWORD', '');
    define('DB_HOST', 'localhost');

$link = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
if (!$link) {
    die("Database connection failed: " . mysqli_error());
}

// 2. Select a database to use 
$db_select = mysqli_select_db($link, DB_NAME);
if (!$db_select) {
    die("Database selection failed: " . mysqli_error());
}

  $actfn=$_POST['Actor_First_Name'];
  $actln=$_POST['Actor_Last_Name'];
  $actag=$_POST['Actor_Age'];
  $actgn=$_POST['Actor_Gender'];
  $actht=$_POST['Actor_Height'];

$sql = "INSERT INTO actor_demographics (Actor_First_Name, Actor_Last_Name, Actor_Age, Actor_Gender, Actor_Height)  VALUES ('$actfn', '$actln', '$actag', '$actgn', '$actht')";

if (!mysqli_query($link,$sql))
  {
  echo("Error description: " . mysqli_error($link));
  }

mysqli_close($link);
?>

演示from.php 我刚刚在name标记

中添加了下划线

<form action="demo.php" method="post" />
<p>Actor First Name: <input type="text" name="Actor_First_Name" /><p>
<p>Actor Last Name: <input type="text" name="Actor_Last_Name" /><p>
<p>Actor Age: <input type="text" name="Actor_Age" /><p>
<p>Actor Gender: <input type="text" name="Actor_Gender" /><p>
<p>Actor Height: <input type="text" name="Actor_Height" /><p>

<input type="submit" value="Submit" />
</form>

Php未捕获的异常错误

1 个答案: