将选择选项插入数据库

Question

这是我连接到服务器并将每个选择标记的每个选项值插入数据库中的表中的 php 代码

<?php 

$servername = "localhost";
$username = "projectuser";
$password ="";
$dbname = "surveyproject";

//establish connection
$connection = mysqli_connect($servername , $username , $password       ,$dbname);

//check for connection errors
if ($connection->connect_error) {
die("Connection failed: " . $connection->connect_error);
}
echo "connected succesfully";

//inserting answers into db
if (isset($_POST['submit'])) {
# code...
for($i=1; $i<3; $i++){
   if(isset($_POST['select'])){
     $select = $_POST['select']; 
     $query = "INSERT INTO surveyanswers (answer) VALUES     ('$select')";
     mysqli_query($connection,$query);
}
}

echo "<script>alert('succesfully submitted answers , thank you')</script>";

mysqli_close($connection);
}
?>

这是我的 html 代码，带有三个选择标签，我希望在每个选择标签的选项中选择的值更新到数据库中的表中

<select class="form-control" name="select">
  <option value="Very Good">Very Good</option>
  <option value="Good">Good</option>
  <option value="Average">Average</option>
  <option value="Poor">Poor</option>
  <option value="Very Poor">Very Poor</option>
</select>
<select class="form-control" name="select">
  <option value="Very Good">Very Good</option>
  <option value="Good">Good</option>
  <option value="Average">Average</option>
  <option value="Poor">Poor</option>
  <option value="Very Poor">Very Poor</option>
</select>
<select class="form-control" name="select">
  <option value="Very Good">Very Good</option>
  <option value="Good">Good</option>
  <option value="Average">Average</option>
  <option value="Poor">Poor</option>
  <option value="Very Poor">Very Poor</option>
</select>

Answer 1

制作selects 数组。

<select class="form-control" name="select[]">
  <option value="Very Good">Very Good</option>
  <option value="Good">Good</option>
  <option value="Average">Average</option>
  <option value="Poor">Poor</option>
  <option value="Very Poor">Very Poor</option>
</select>
<select class="form-control" name="select[]">
 <option value="Very Good">Very Good</option>
  <option value="Good">Good</option>
  <option value="Average">Average</option>
  <option value="Poor">Poor</option>
  <option value="Very Poor">Very Poor</option>
</select>
<select class="form-control" name="select[]">
  <option value="Very Good">Very Good</option>
  <option value="Good">Good</option>
  <option value="Average">Average</option>
  <option value="Poor">Poor</option>
  <option value="Very Poor">Very Poor</option>
  </select>

然后你可以循环遍历数组。因为它们都具有相同的名称，所以只发送最后一个 select。

之后，通过使用带有参数化查询的准备好的语句来解决您的 SQL 注入问题。 http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

那么你的 PHP 可能是：

$query = "INSERT INTO surveyanswers (answer) VALUES (?)";
if ($stmt = mysqli_prepare($connection,$query)) {
     foreach($_POST['select'] as $selectvalue){
          mysqli_stmt_bind_param($stmt, "s", $selectvalue);
          mysqli_stmt_execute($stmt);
     }
} else {
     printf("Error: %s.\n", mysqli_stmt_error($stmt));
}