我真的很努力与这件pf php代码,我想要插入一个消息到表中使用从另一个表中获取的数据。下面是我的代码:PHP/MySQL插入不工作
$get_user_sql = "SELECT username FROM members WHERE id = '$member'";
$get_user_res = mysqli_query($con, $get_user_sql);
while($username = mysqli_fetch_assoc($get_user_res)){
$user = $username["username"];
};
$get_message_sql = "SELECT message FROM posts WHERE id = '1377077348-5922'";
$get_message_res = mysqli_query($con, $get_message_sql);
while($postsmessage = mysqli_fetch_assoc($get_message_res)){
$postmessage = $postsmessage["message"];
};
$type = "liked";
$title = "New Like";
$message = "<a href=\"profile.php?member=$user\">$user</a> just liked your post:<br /><i>$postmessage</i>";
$insert_note_sql = "INSERT INTO notes (id, sender, recipient, type, title, message, date) VALUES('$id', '$member', '$recipient', '$type', '$title', '$message', '$time')";
$insert_note_res = mysqli_query($con, $insert_note_sql);
我发现,如果我从$消息行删除$ PostMessage的一切工作正常,但是当我再次行添加回去未插入。任何人都可以看到这个原因吗?
mysqli_error()说什么?你是否回应了查询以确保它符合你的期望?你为此做了哪些调试? –
您需要通过'mysqli_real_escape_string()' –
转义您的所有数据查询正在通过JQuery传递,所以我怎样才能回显错误呢? – user2516546