Question

所以我试图让桌子显示骑车人骑着什么山（为MS150做好准备）。我的问题是，当一个人点击按钮说他们已经骑过山......它只在某些时候工作。当他们说他们已经完成了一个小山时，按钮会被禁用，并且他们不能再次推动它。但是，有些记录没有正确显示。这是我的代码：

HTML：

$get_hills = "SELECT * FROM hills ORDER BY hill_region ASC, hill_id ASC"; 
$run_hills = mysqli_query($con, $get_hills); 
while($row_hills = mysqli_fetch_array($run_hills)) {

$hill_id = $row_hills['hill_id'];

$get_user = "SELECT * FROM users WHERE user_id = '".$_SESSION['user_id']."'";
$run_user = mysqli_query($con, $get_user); 
$row_user = mysqli_fetch_array($run_user); 

$get_ridden = "SELECT * FROM hills_ridden WHERE hill_id = '".$hill_id."'";
$run_ridden = mysqli_query($con, $get_ridden);
$row_ridden = mysqli_fetch_array($run_ridden); 
echo "
<tr>
<td>".$row_hills['hill_name']."</td>
<td>".$row_hills['hill_region']."</td>
<td>".$row_hills['hill_rise']."</td>
<td>".$row_hills['hill_length']."</td>
<td>".$row_hills['hill_avg_grade']."</td>
<td>".$row_hills['hill_total_points']."</td>
<td><a href='".$row_hills['hill_link']."' target='_blank'><button class='btn btn-inverse btn-xs'>Link</button></a></td>
<td>";
if($row_user['user_id'] == $row_ridden['user_id']) {
echo "<button class='btn btn-default btn-xs' disabled>You Rode This</button>";
} else {
echo"
<form method='post' action=''>
<input type='hidden' name='hill_id' value='".$row_hills['hill_id']."'>
<input type='hidden' name='points_gained' value='".$row_hills['hill_total_points']."'>
<input type='hidden' name='user_id' value='".$row_user['user_id']."'>
<button type='submit' name='hill_sub' class='btn btn-inverse btn-xs'>I Rode This</button>
</form>";
} echo"</td>
</tr>";
}

php：

if(isset($_POST['hill_sub'])) {

$get_user = "SELECT * FROM users WHERE user_id = '".$_SESSION['user_id']."'";
$run_user = mysqli_query($con, $get_user); 
$row_user = mysqli_fetch_array($run_user);

$insert_points = "INSERT INTO points (user_id, hill_id, points_gained, points_date) VALUES ('".$row_user['user_id']."', '".$_POST['hill_id']."', '".$_POST['points_gained']."', now())"; 

$insert_activity = "INSERT INTO activity (user_id, hill_id, activity_date) VALUES ('".$row_user['user_id']."', '".$_POST['hill_id']."', now())"; 

$insert_hill = "INSERT INTO hills_ridden (user_id, hill_id) VALUES ('".$row_user['user_id']."', '".$_POST['hill_id']."')"; 

$run_points = mysqli_query($con, $insert_points); 
$run_activity = mysqli_query($con, $insert_activity); 
$run_hills = mysqli_query($con, $insert_hill); 

if($run_hills) {
    echo "<script>window.open('record-hill.php?user_id=".$row_user['user_id']."', '_self')</script>"; 
} else {
    echo $run_hills->error; 
}
}

这里是sql db文件：

-- Table structure for table `hills_ridden`
--

CREATE TABLE IF NOT EXISTS `hills_ridden` (
`ridden_id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` varchar(5) NOT NULL,
`hill_id` varchar(5) NOT NULL,
PRIMARY KEY (`ridden_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=33 ;

任何帮助将不胜感激...我一直在尝试不同的查询，使其工作，但过去2天一直没有成功：/请！

Answer 1

首先提出建议：由于您的user_id在会话中，因此无需将其设为输入字段。在后一种情况下，可以改变形式user_id值以为不同用户创建条目。而是在插入数据的代码中，从$ _SESSION而不是$ _POST

获取user_id

（编辑：我现在看到你从$ _SESSION中获取。在处理插入的代码中，不需要根据会话user_id从用户表中进行选择，因为你只使用该表中的user_id ）

然后，您可以尝试进行一次查询而不是三次，跳过整个用户表，因为user_id在会话中：

SELECT 
    hills.*, 
    CASE hills_ridden.hill_id WHEN NULL THEN 0 ELSE 1 AS hill_ridden
FROM hills 
LEFT JOIN hills_ridden 
    ON hills_ridden.hill_id = hills.hill_id 
    AND hills_ridden.user_id = '".$_SESSION['user_id']."' 
ORDER BY hill_region ASC, hill_id ASC;

您的按钮<td>可能会变成：

if($row_hill['hill_ridden'] == 1) {
echo "<button class='btn btn-default btn-xs' disabled>You Rode This</button>";
} else {
echo"
<form method='post' action=''>
<input type='hidden' name='hill_id' value='".$row_hills['hill_id']."'>
<input type='hidden' name='points_gained' value='".$row_hills['hill_total_points']."'>
<input type='hidden' name='user_id' value='".$row_user['user_id']."'>
<button type='submit' name='hill_sub' class='btn btn-inverse btn-xs'>I Rode This</button>
</form>";
} echo"</td>
</tr>";

此代码未经测试，只是为了提供一个想法。

Answer 2

@maarten，谢谢你的想法......我试着/修补它，它给了我一些我需要的东西......但意识到最好的计划是最简单的，所以我得到了这个： / p>

$get_ridden = "SELECT * FROM hills_ridden WHERE hill_id = '".$hill_id."' AND user_id = '".$row_user['user_id']."'";
$run_ridden = mysqli_query($con, $get_ridden);
$row_ridden = mysqli_fetch_array($run_ridden); 
$num_ridden = mysqli_num_rows($run_ridden);

if($num_ridden >= 1) {
echo "<button class='btn btn-default btn-xs' disabled>You Rode This</button>";
} else {

这给了我正在寻找的东西...需要稍稍离开它然后用新鲜的眼睛看待它！

php mysql没有正确提取数据

2 个答案: