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我一直在尝试更新我的数据库，但它确实更新了，并且显示成功，并出现这样的 id 错误

注意：未定义变量：第 4 行 C:\wamp\www\Potifolio\update_ac.php 中的 player_id

成功：下面是我的代码，请帮助：

我的.php

<?php
include_once("connect.php"); 
$query = "SELECT * FROM players";
$result = mysql_query($query,$db);
?>
<table><tr>
<td>Name</td>
<td>Surname</td>
<td>Positon</td>
<td>Email</td>
<td>Passowrd</td>
<td>Email</td>
<td>Action</td>
</tr>
<?php
while($row = mysql_fetch_assoc($result)){
?>
<tr>
<td><?php echo $row['name'];?></td>
<td><?php echo $row['surname'];?></td>
<td><?php echo $row['position'];?></td>
<td><?php echo $row['password'];?></td>
<td><?php echo $row['username'];?></td>
<td><?php echo $row['name'];?></td>
<td ><a href= "update1.php?player_id=<?php echo $row['player_id'];?>">Update</a></td>
</tr>
<?php
}
?>

更新1.php

<?php 
include_once("connect.php"); 
$player_id = $_GET['player_id'];
$query = "SELECT * FROM players where player_id ='$player_id'";
$result = mysql_query($query,$db);
$row = mysql_fetch_assoc($result);

?>

<form name = "form1"action="update_ac.php" method ="post">
<input type = "text" name="name" value="<?php echo $row['name'];?>">
<input type = "hidden" name = "player_id" ID = "player_id" value="<?php echo $row['player_id'];?>" >
<input type = "submit" name="submit" value="submit">
</form>

update_ac.php

<?php 
include_once('connect.php');
$name= $_POST['name'];
$sql="UPDATE players SET name='$name' WHERE player_id = '$player_id'";
$result = mysql_query($sql);
if($result){
echo "Success";
}else {
echo "Error";
}
?>
4

1 回答 1

1

将您的查询更改为此

$sql="UPDATE players SET name='$name' WHERE player_id = '$_POST['player_id']'";

或者像这样

$player_id = $_POST['player_id'];
$sql="UPDATE players SET name='$name' WHERE player_id = '$player_id'";

在您的情况下，您没有将 $_POST['player_id'] 的实际值分配给变量 $player_id 。

所以完整的代码应该是这样的

<?php 
include_once('connect.php');
if (isset($_POST['submit'])) {
    $name= $_POST['name'];
    $player_id = $_POST['player_id'];
    $sql="UPDATE players SET name='$name' WHERE player_id = '$player_id'";
    $result = mysql_query($sql);
    if($result){
        echo "Success";
    }else {
        echo "Error";
    }
}
?>
于 2013-04-28T08:41:05.897 回答