SQL UPDATE语句不影响任何行（PHP）

我目前正在做一个项目，想做一个简单的网页，我可以编辑组。我在XAMPP中一切正常，并尝试将其上载到服务器，但它不会影响数据库中的任何行。以下是语句：

UPDATE user_groups 
SET name = 'TEST', 
name_short = 'test', 
color = 'green', 
category = 'MMORPG' 
WHERE id = 2

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

以及：

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

这就是答案。创建新组效果很好（本地创建和编辑工作，我没有更改任何声明，因为我上传了这两个）

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

编辑：

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

$sql\u update\u info=“update user\u group SET name='$new\u title'，name\u short='$new\u short'，color='$new\u color'，category='$new\u cat'其中id=$group\u id”；
$query\u update\u info=mysqli\u query（$mysqli，$sql\u update\u info）；
printf（“受影响的行（更新）：%d\n”，mysqli_受影响的行（$mysqli））；
回显“
”.$sql_update_info.”；

---

是单击按钮时PHP部分的外观。

1st:尝试使用prepared语句来避免sql注入

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

2nd:

Execute（）

将返回true或false，因此您需要按照如下方式处理错误

$sql_update_info = "UPDATE user_groups SET name = '$new_title', name_short = '$new_short', color = '$new_color', category = '$new_cat' WHERE id = $group_id";
$query_update_info = mysqli_query($mysqli, $sql_update_info);
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($mysqli));

echo '<br><span style="color:white;">'.$sql_update_info.'</span>';

$stmt = $mysqli->prepare("UPDATE user_groups SET name = ?, name_short = ?, color = ?, category = ? WHERE id = ?");

$stmt->bind_param('ssssi', $new_title, $new_short, $new_color, $new_cat, $group_id);

//The argument may be one of four types:

//i - integer
//d - double
//s - string
//b - BLOB
//change it by respectively 

$r = $stmt->execute();
if(!$r){  
    echo $stmt->error; 
}else{
   $row_count= $stmt->affected_rows; 
}
$stmt->close();
$mysqli->close();

---

共享相关的PHP代码，因为查询看起来还可以。你有自动提交功能吗？在本地主机上，您可能使用了MySQLi，但如果您将其上载到具有不同引擎的服务器上，它可能无法像在本地计算机上那样工作。如果存在错误，请尝试查看错误日志。一旦尝试在数据库中执行查询，您可以

echo$sql\u update\u info

并将其粘贴到此处。。