检查此工作代码。此外，您还没有将Drop down的元素名称设置为select_style。这也是一个失误。

PHP代码

if(isset($_POST['submit-comment']) && $_POST['submit-comment']!='')  {
  $host= "localhost";
  $user="root";
  $pass="";
  $db="test";
  $con = mysqli_connect($host,$user,$pass,$db) or mysql_error();
    //====== Get Variable======= //
    $name  = mysqli_real_escape_string($con,$_POST['name']);
    $email = mysqli_real_escape_string($con,$_POST['email']);
    $rate  = mysqli_real_escape_string($con,$_POST['select_style']);
    $content = mysqli_real_escape_string($con,$_POST['content']);
  $insert="insert into reviews (name,email,rate,content) values ('$name','$email','$rate','$content')"; 
  if($name && $email && $content == true) {
    mysqli_query($con,$insert);
    $success = "<span class='success_testmonial'>Thank You! .. Your Raiting Has Been Submitted And We Will Post It As Soon We Verify It !</span>";
    echo $success;
  }
  else {
    $error = "<span class='error_testmonial'>Error : one or some fields has left empty .. Please fill all field and try again.</span>";
    echo $error;
  }
  mysqli_close($con);
}    

HTML

<form method="post" action="" id="form-contact" class="clearfix">
<div id="form-left">
  <label for="text-name">Name *</label><br />
  <input type="text" name="name" class="input" id="text-name" /><br />
  <label for="text-email">From *</label><br />
  <input type="text" name="email" class="input" id="text-email" /><br />
  <label for="text-phone">Rate us *</label><br />
  <div class="select-style">
  <select name="select_style">
  <option value="5.0">5.0</option>
  <option value="4.5">4.5</option>
  <option value="4.0">4.0</option>
  <option value="3.5">3.5</option>
  <option value="3.0">3.0</option>
  <option value="2.5">2.5</option>
  <option value="2.0">2.0</option>
  <option value="2.0">2.0</option>
  <option value="1.5">1.5</option>
  <option value="1.0">1.0</option>
  </select>
  </div>
</div>
<div id="form-right">
  <label for="text-comment">Review <span></span></label><br />
  <textarea name="content" cols="10" rows="20" class="input textarea" id="text-comment"></textarea><br />
  <input type="submit" name="submit-comment" class="button" value="Rate Us" />
</div>
<p id="text-contact">
<br><br><font color="#980303">Please Note *</font> Thate Your Reviews Will Not Published Untill We Check it and sure that the review don't contain Bad words or bad language, and be sure that we will publish all reviews and we accept criticism! 
</form>

尝试将get变量放入if-else语句中提交后检查POST中是否有数据：

if($_POST['submit-comment'])  {
        $name = $_POST['name'];
        $email=$_POST['email'];
        $rate=$_POST['select_style'];
        $content=$_POST['content'];
        $insert="insert into reviews (name,email,rate,content) values ('$name','$email','$rate','$content')";
        if ($con->query($sql) === TRUE) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $sql . "<br>" . $conn->error;
        }
        var_dump($_POST);
}
$con->close();    

检查错误：

$check = mysqli_query($con,$insert);
var_dump($check);

如果你找到了，请告诉我

注意：

• 将插入查询和传递变量（POST）放入if语句isset(POST["submit-comment"]中，以消除未定义变量的错误。

• 您应该使用mysqli_* prepared statement来防止SQL注入。

答案：

如果您坚持保留您的代码，那么在将其用于查询之前，可以使用mysqli_real_escape_string()函数来丰富变量的内容。

您的PHP文件应该如下所示：

<?php
$host= "localhost";
$user="root";
$pass="freedoom19";
$db="cookindoor";
$con = mysqli_connect($host,$user,$pass,$db) or mysql_error();
//====== IF SUBMIT-COMMENT ======= //
if(isset($_POST['submit-comment']))  {
  if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["content"])) {
    //====== GET VARIABLES ======= //
    $name = mysqli_real_escape_string($con,$_POST['name']);
    $email = mysqli_real_escape_string($con,$_POST['email']);
    $rate = mysqli_real_escape_string($con,$_POST['select_style']);
    $content = mysqli_real_escape_string($con,$_POST['content']);
    $insert="INSERT INTO reviews (name,email,rate,content) VALUES ('$name','$email','$rate','$content')";
    mysqli_query($con,$insert);
    $success = "<span class='success_testmonial'>Thank You! .. Your Raiting Has Been Submitted And We Will Post It As Soon We Verify It !</span>";
  }
  else {
    $error = "<span class='error_testmonial'>Error : one or some fields has left empty .. Please fill all field and try again.</span>";
  }
}
mysqli_close($con);
?>

建议：

但是，如果您在mysqli_* prepared statement中执行它，那么您的插入查询将如下所示。虽然这只是一个简单的例子，但仍然可以执行：

if($stmt = $con->prepare("INSERT INTO reviews (name, email, rate, content) VALUES (?,?,?,?)")){ /* CHECK THE QUERY */
  $stmt->bind_param('ssss', $_POST["name"], $_POST["email"], $_POST["rate"], $_POST["content"]); /* BIND VARIABLES TO YOUR QUERY */
  $stmt->execute(); /* EXECUTE YOUR QUERY */
  $stmt->close(); /* CLOSE YOUR QUERY */
}