我有一个代码可以在 HTML 中打印播放器的状态，但我想在旁边添加一个按钮，当我点击它时，我希望更改该状态。

(数据库和 HTML 上有很多播放器，我遇到的问题是，当我单击按钮时，所有播放器的状态都发生了变化，所以我需要找到一种方法来找到 steamid 并更改通过 steamid 按钮查看该播放器的状态，我不知道该怎么做，这是我正在尝试做的管理面板:http://vanityrp.site.nfoservers.com/apply/admin.php )

管理.php:

<?php
$servername = "localhost";
$username = "------";
$password = "------";
$dbname = "-----";

$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
 die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT avatar, name, status FROM Apps";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
 // output data of each row
 while($row = $result->fetch_assoc()) {
     echo "
<form action='../apply/steamauth/status.php' method='post'>
<div class='advert'><img src='".$row["avatar"]. "'>\n</div><br>
<div name='username2' class='advert'>Username: ".$row["name"]. "\n<br></div>
<div class='advert'>Status: <font color='orange'>".$row["status"]."</font></div>\n<br>
<input type='submit' value='Deny' name='deny'/>
<input type='submit' value='Accept' name='accept'/>
<input type='submit' value='Remove' name='remove'/>
</form>
<hr> 

";

}
} else {
}
$conn->close();
?>

现在是 status.php:

<?php
$servername = "localhost";
$username = "-----";
$password = "----";
$dbname = "-----";

$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
 die("Connection failed: " . $conn->connect_error);
}

$denied = $_POST['deny'];
$accepted = $_POST['accept'];
$remove = $_POST['remove'];
$username = $_POST['username2'];

if($accepted){
$sql = "UPDATE Apps SET status = replace(status,'Pending','Accepted.');";
$sql = "UPDATE Apps SET status = replace(status,'Denied','Accepted.');";
}elseif($denied){
$sql = "UPDATE Apps SET status = replace(status,'Pending','Denied.');";
$sql = "UPDATE Apps SET status = replace(status,'Accepted','Denied.');";
}elseif($remove){
#$sql = "DELETE FROM Apps WHERE id=$username";
}
$result = $conn->query($sql);

if ($conn->query($sql) === TRUE) {
echo "App has been denied/accepted successfully.";

header( "refresh:3;url=http://vanityrp.site.nfoservers.com/apply/admin.php" );
} else {
echo "Error: " . $sql . "<br>" . $conn->error;

header("refresh:3;url=http://vanityrp.site.nfoservers.com/apply/admin.php");
}

?>

最佳答案

尝试在您的代码中进行以下更改

<form action='../apply/steamauth/status.php' method='post'>
<div class='advert'><img src='".$row["avatar"]. "'>\n</div><br>
<input type="hidden" name="userid" value='".$row["id"]."'/> <!--added a hidden variable "user your id column name in value"-->
<div name='username2' class='advert'>Username: ".$row["name"]. "\n<br></div>
<div class='advert'>Status: <font color='orange'>".$row["status"]."</font></div>\n<br>
<input type='submit' value='Deny' name='deny'/>
<input type='submit' value='Accept' name='accept'/>
<input type='submit' value='Remove' name='remove'/>
</form>

并在 status.php 中

$denied = $_POST['deny'];
$accepted = $_POST['accept'];
$remove = $_POST['remove'];
$username = $_POST['username2'];
$id = $_POST['userid']; // get the hidden variable here

if($accepted){
  $sql = "UPDATE Apps SET `status` = 'Accepted.' where id = $id;"; //check the condition
}elseif($denied){
  $sql = "UPDATE Apps SET `status` = 'Denied.' where id = $id;";
}elseif($remove){
  #$sql = "DELETE FROM Apps WHERE id=$username";
}

关于php - 在数据库中查找用户名并更改他的状态字符串，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34562804/