为什么我的UPDATE查询不起作用？

Question

我有以下更新查询，但由于某种原因，它无法正常工作。 我认为这与“ id ='”。$ id。“'有关，但我尝试了三种不同的方法，但似乎无法使其正常工作。由于某种原因，这让人很痛苦，在此先感谢。

$id = $_GET['id'];
$speaker = mysql_real_escape_string($_POST['speaker']);
$message = $_POST['message'];
$title = mysql_real_escape_string($_POST['title']);
$date = $_POST['date'];
$day = $_POST['day'];
$password = mysql_real_escape_string($_POST['password']);
$complete = $_POST['complete'];

$title = ucwords(strtolower($title));

if ($complete && ($password == "*****"))
{       

    $db = mysql_connect($hostname, $username, $password) or die(mysql_error());
    mysql_select_db($dbname,$db) or die(mysql_error());

    mysql_query("UPDATE sermons SET speaker = '$speaker', message = '$message', title = '$title', date = '$date', day = '$day' WHERE id = '$id'");

    $num_rows = mysql_num_rows(mysql_query("SELECT speaker, message, title, date, day FROM sermons WHERE speaker = '$speaker' AND message = '$message' AND title = '$title' AND date = '$date' AND day = '$day'", $db));

    if ($num_rows == 1)
        echo "<script type='text/javascript'> alert('Sermon Information Entered Successfully!'); </script>";

    else
        echo "<script type='text/javascript'> alert('Error! Please Try Again.'); </script>";
}

else if ($complete && ($password != "*****"))
{
    echo "<script type='text/javascript'> alert('Incorrect Password! Please Try Again.'); </script>";
}

Answer 1

由于id是Integer，因此这是正确的代码，因此请尝试：

mysql_query（“更新讲道SET说话者='$说话者'，消息='$消息'，标题='$标题'，日期='$日期'，天='$天'WHERE id =”。$ id）;

我期望它会起作用。

Answer 2

将其写为答案是因为很难在注释中提供代码

尝试改变

mysql_query("UPDATE sermons SET speaker = '$speaker', message = '$message', title = '$title', date = '$date', day = '$day' WHERE id = '$id'");

至

mysql_query("UPDATE sermons SET speaker = '$speaker', message = '$message', title = '$title', date = '$date', day = '$day' WHERE id = '$id'") or die(mysql_error());

然后将错误消息添加到您的问题中。