我不能發佈會議在PHP

Question

人喜及後可變我有下面的代碼創建一個簡單的PHP悄悄話我發送下面表格

<form enctype="multipart/form-data" method="POST" action="send.php"> 
<label>To</label> 
<select name="To" class="form-control"> 
<option value="user">user</option> 
<option value="user1">user1</option> 
</select> 
<input type="hidden" name="From" value="<?php $_SESSION['user'];?>"> 
<label>Subject</label> 
<input type="text" name="Subject" placeholder="Subject"> 
<label>Message</label> 
<textarea class="form-control" cols="40" name="Message" placeholder="Message" rows="10"></textarea> 
<input type="hidden" name="Date" value="<?php date("F j, Y, g:i a"); ?>"> 
<button type="submit" name="Send" class="btn btn-success">Send</button></form> 

和我的PHP代碼如下：

<?php 
session_start(); 
include 'db.php'; 
if(isset($_POST['Send'])){ 
    $To = $_POST['To']; 
    $From = $_POST['From']; 
    $Sub = $_POST['Subject']; 
    $Mess = $_POST['Message']; 
    $Date = $_POST['Date']; 

$sql = "INSERT INTO pm from, to, sub, mess, date VALUES '$From', '$To', '$Sub', '$Mess', '$Date'"; 
$result = mysqli_query($con, $sql); 

if($result == true){ 
$sent = "Message sent Successfully"; 
header("Location: Home.php?message=$sent"); 
} 
else{ 
$ntsent = "Message not sent Successfully"; 
header("Location: Home.php?message=$ntsent"); 
} 
mysqli_close($con); 
} 
?> 

它總是說消息沒有發送成功消息你能猜出什麼是錯的嗎？

Answer 1

我認爲你的INSERT查詢是不完整的。試試這個

$sql = "INSERT INTO TABLE_NAME (pm, from, to, sub, mess, date) VALUES ('$From', '$To', '$Sub', '$Mess', '$Date'");