如何在数据库中插入游戏的当前ID [英] How to Insert Current ID of game in database
问题描述
我的问题是,如何将我的游戏ID的当前ID插入数据库。
这是我的桌子:
创建表如果不存在`game_review`(
`id`int(11)NOT NULL,
`user_id`int(11)NOT NULL,
`game_id`int(11)NOT NULL,
` rating` int(10)NOT NULL,
`title` varchar(100)NOT NULL,
`review`文本NOT NULL
)ENGINE = InnoDB DEFAULT CHARSET = latin1;
ALTER TABLE`game_review`
添加PRIMARY KEY(`id`);
ALTER TABLE`game_review`
MODIFY`id`int(11)NOT NULL AUTO_INCREMENT;
我想插入当前的user_id和标题和描述,我已经知道该怎么做了。 (我不担心错误检查或安全性。)我只是无法插入我点击的游戏的当前ID。
这是我的代码:
<?php
require_once $ _SERVER [' DOCUMENT_ROOT']。' / Gamesite / core / init.php';
include(' head.php');
include(' navigation.php');
?>
< < span class =code-leadattribute> div class = 容器 样式 = padding-removed 200px; >
<? php
// 获取游戏的当前ID
$ game_id = $ _GET [' id'];
$ errors = array();
// 从init.php获取用户ID
$ user_id = $ user_data [' id'];
if (isset($ _ POST [' submit'])){
// 清理表单字段。
$ title =((isset($ _ POST [' title'])&& $ _POST [' title']!= ' ')?sanitize($ _ POST [< span class =code-string>' title']):' '跨度>);
$ review =((isset($ _ POST [' 审核])&& $ _POST [' review'] != ' ')?sanitize($ _ POST [' review']):' '< /跨度>);
$ rating =((isset($ _ POST [' 评级])&& $ _POST [' rating'] != ' ')?sanitize($ _ POST [' rating']):' '< /跨度>);
if (!empty($ errors)){
display_errors($ errors);
} else {
$ SQL_INSERT = 插入到game_review(`user_id`,`game_id`,`rating`,`title`,`review`)VALUES('$ user_id ','$ game_id','$ rating','$ title','$ review');
$ QUERY = $ db-> query($ SQL_INSERT);
// header('Location:Review.php');
}
}
?>
当我点击表单中的提交时,会显示此错误消息。
注意:未定义的索引:第14行的C:\ xampp \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
_SERVER [' DOCUMENT_ROOT']。' / Gamesite / core / init.php';
include(' head.php');
include(' navigation.php');
?>
< < span class =code-leadattribute> div class = 容器 样式 = padding-removed 200px; >
<? php
// 获取游戏的当前ID
< blockquote> game_id =
_GET [' id'跨度>];
My question is, how do I insert the current ID of my Game ID into the database.
Here is my table:
CREATE TABLE IF NOT EXISTS `game_review` (
`id` int(11) NOT NULL,
`user_id` int(11) NOT NULL,
`game_id` int(11) NOT NULL,
`rating` int(10) NOT NULL,
`title` varchar(100) NOT NULL,
`review` text NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
ALTER TABLE `game_review`
ADD PRIMARY KEY (`id`);
ALTER TABLE `game_review`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT;
I want to insert the current user_id and title and description, which i already know how to do. (I'm not worrying about errors checking or security yet.) I just cannot insert the current ID of the game that I clicked on.
Here is my code:
<?php
require_once $_SERVER['DOCUMENT_ROOT'].'/Gamesite/core/init.php';
include('head.php');
include('navigation.php');
?>
<div class="container" style="padding-removed 200px;">
<?php
// Get the current ID of the game
$game_id = $_GET['id'];
$errors = array();
// Get the user ID from init.php
$user_id = $user_data['id'];
if (isset($_POST['submit'])) {
// Sanitize form fields.
$title = ((isset($_POST['title']) && $_POST['title'] != '') ? sanitize($_POST['title']) : '');
$review = ((isset($_POST['review']) && $_POST['review'] != '') ? sanitize($_POST['review']) : '');
$rating = ((isset($_POST['rating']) && $_POST['rating'] != '') ? sanitize($_POST['rating']) : '');
if (!empty($errors)) {
display_errors($errors);
} else {
$SQL_INSERT = "insert into game_review (`user_id`, `game_id`, `rating`, `title`, `review`) VALUES ('$user_id', '$game_id', '$rating', '$title', '$review')";
$QUERY = $db->query($SQL_INSERT);
//header('Location: Review.php');
}
}
?>
When I click submit in the form, this error message show up.
Notice: Undefined index: id in C:\xampp\htdocs\Gamesite\game_includes\Review.php on line 14
_SERVER['DOCUMENT_ROOT'].'/Gamesite/core/init.php'; include('head.php'); include('navigation.php'); ?> <div class="container" style="padding-removed 200px;"> <?php // Get the current ID of the game
game_id =
_GET['id'];
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