我的代码需要一些帮助,当我点击提交代码时,它总是说"用户成功踢了#34;。但是,查询未成功执行,用户仍在数据库中。
编辑:修复了mysql / msqli问题。现在查询答案"出了问题" "用户成功踢了#34;但我不会从mysql_error报告中得到任何错误。我该怎么办?
include 'connect.php';
if(empty($_POST['user_id']))
{
echo '<form method="post" action="">
User_id: <input type="text" name="user_id" />
<input type="submit" value="Kick user" />
</form>';
}
else
{
$sql = "DELETE FROM users
WHERE
user_id = '" .mysqli_real_escape_string($_POST['user_id']) . "'
";
$result = mysqli_query($con, $sql);
if($result)
{
//something went wrong, display the error
echo 'Something went wrong!.';
echo mysqli_error(); //debugging purposes, uncomment when needed
}
else
{
echo 'User successfully kicked!';
}
}
我的connect.php看起来像这样:
<?php
$con=mysqli_connect ("localhost","root","","dps");
if (mysqli_connect_errno()) {
echo "failed to connect mysql: ". mysqli_connect_error();
}
?>
答案 0 :(得分:2)
您要在$sql
部分设置if
var,并尝试在else
部分中访问它。这不起作用。
你也混合库是坏的(mysql与mysqli不一样,使用mysqli,因为 mysql已被弃用,将从php中移除)
include 'connect.php';
if(empty($_POST['user_id']))
{
echo '<form method="post" action="">
User_id: <input type="text" name="user_id" />
<input type="submit" value="Kick user" />
</form>';
}
else
{
$sql = "DELETE FROM users
WHERE
user_id = '" .mysqli_real_escape_string($con, $_POST['user_id']) . "'
";
$result = mysqli_query($con, $sql);
if(!$result)
{
//something went wrong, display the error
echo 'Something went wrong!: ';
echo mysqli_error($con) //debugging purposes, uncomment when needed
}
else
{
echo 'User succesfully kicked!';
mysqli_free_result($result);
}
}
您可能需要更改connect.php才能使用mysqli
编辑:现在应该修复一些错误
答案 1 :(得分:-1)
添加:
$result = mysqli_query($con, $sql);
之后:
$sql = "DELETE FROM
users
WHERE
user_id = '" . mysql_real_escape_string($_POST['user_id']) . "'
";