Question

这是我的PHP代码连接到服务器并将每个select标签的每个选项值插入数据库中的表

<?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);
}
?>

这是我的带有三个选择标签的html代码，我希望将每个select标签的选项中的值选择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select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>

然后你可以遍历数组。由于只发送了最后一个select，因为它们都具有相同的名称。

之后，通过使用带参数化查询的预准备语句来修复SQL注入问题。 http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

然后你的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));
}

将选择选项插入数据库

1 个答案: