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＆＃34; id =＆＃39;＆＃34;。$ id。＆＃34;＆＃39;有关。但我尝试了三种不同的方式，似乎无法让它发挥作用。我以前写过更新查询没有任何问题，但由于某种原因，这个问题很痛苦。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Integer，所以这是正确的代码尝试：

mysql_query（＆＃34; UPDATE布道SET演讲者=＆＃39; $ speaker＆＃39;，message =＆＃39; $ message＆＃39;，title =＆＃39; $ title＆＃39;，date = ＆＃39; $ date＆＃39;，day =＆＃39; $ day＆＃39; WHERE id =＆＃34;。$ id）;

我预计它会有效。

Answer 2

将此作为答案写为答案，因为在评论中提供代码很难阅读

尝试更改

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

要

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。

为什么我的UPDATE查询不起作用？

2 个答案: