mysql - 为什么我的更新查询不起作用？

Question

我有以下更新查询，但由于某种原因它不起作用。我认为这与“id = '”.$id.”有关，但我尝试了三种不同的方法，但似乎无法让它工作。我之前编写过更新查询，没有任何问题，但对于出于某种原因，这很痛苦。提前致谢。

$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

因为 id 是整数，所以这是正确的代码，请尝试:

mysql_query("更新布道集演讲者 = '$speaker', 消息 = '$message', 标题 = '$title', 日期 = '$date', day = '$day' WHERE id =".$id )；

我预计它会起作用。