Question

我得到了一些人在这里得到的同样的错误。但我的问题仍未解决。在我的所有模块的大学项目中我得到了同样的错误。

当我在PHP中运行以下查询

时

$username = $_SESSION['username'];
$query = "SELECT id FROM user WHERE username = '$username'";
$qpost = mysqli_query($connection,$query);
$post_user = mysqli_fetch_assoc($qpost);
$forum_size = strlen($forum);
$dt = time();
$datetime = strftime("%Y-%m-%d %H:%M:%S", $dt);
$enteruser = $post_user['id'];
$Query = "INSERT INTO posts('id','user_id','post','date_added') VALUES(NULL,'$enteruser','$forum','$datetime')";`

我收到以下错误，

＃1064 - 您的SQL语法出错;查看与您的MySQL服务器版本相对应的手册，以便在＆＃39;＆＃39; id＆＃39;，＆＃39; user_id＆＃39;，＆＃39; post＆＃39;，＆##附近使用正确的语法39; date_added＆＃39;）VALUES（NULL，＆＃39; 3＆＃39;，＆＃39; erye yeyeyeyetyery ery ery＆＃39;在第1行

当我在mysql shell中运行它时，我得到了同样的错误。当我在phpmyadmin中运行该查询时，如下所示，

INSERT INTO posts('id','user_id','post','date_added') VALUES (NULL,3,'erye yeyeyeyetyery ery ery ','2015-07-24 18:53:48');

我得到了同样的错误。你能帮我解决一下这个问题。我以各种方式检查了查询，我可以提出错误。我不能继续解决这个问题！谢谢你的帮助！

Answer 1

您的表格标题上不需要撇号。

        $username = $_SESSION['username'];
        $query = "SELECT id FROM user WHERE username = '$username'";
        $qpost = mysqli_query($connection,$query);
        $post_user = mysqli_fetch_assoc($qpost);
        $forum_size = strlen($forum);
        $dt = time();
        $datetime = strftime("%Y-%m-%d %H:%M:%S", $dt);
        $enteruser = $post_user['id'];
        $Query = "INSERT INTO `posts` (id, user_id, post, date_added) VALUES (NULL,'$enteruser','$forum','$datetime')";

Answer 2

鉴于id为primary-key和auto-increment，您可以将其排除。您还可以对now()

使用本机Mysql函数date_added

$Query = "INSERT INTO `posts` (user_id, post, date_added) VALUES ('$enteruser','$forum',now())";

Answer 3

试试这个......

$username = $_SESSION['username'];
$query = "SELECT id FROM user WHERE username = '$username'";
$qpost = mysqli_query($connection,$query);
$post_user = mysqli_fetch_assoc($qpost);
$forum_size = strlen($forum);
$dt = time();
$datetime = strftime("%Y-%m-%d %H:%M:%S", $dt);
$enteruser = $post_user['id'];
$Query = "INSERT INTO posts ('user_id','post','date_added') VALUES('$enteruser','$forum','$datetime')";`

＃1064（42000）INSERT INTO中的MySQL错误查询PHP MySQL

3 个答案: