这是我的php代码，用于连接到服务器并将每个select标记的每个选项值插入数据库的表中
<?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，有三个选择标记，我希望在每个选择标记的选项中选择的值更新到数据库的表中
<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>
制作select的数组。
<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>
然后可以循环遍历数组。照原样，只发送最后一个select，因为它们都有相同的名称。
之后，通过使用带参数化查询的已准备好的语句来修复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));
}