PHP将下拉值保存到mysql

Question

首先让我说，我是新手，我的代码对于那些知道自己在做什么的人来说可能是可笑的......

我正在尝试从我的 mysql 数据库中检索数据并在匹配的下拉列表中显示信息。我让这部分使用一些丑陋的基本代码。

$url = $_SERVER["REQUEST_URI"];
$url_path = parse_url($url, PHP_URL_PATH);
$devid = pathinfo($url_path, PATHINFO_BASENAME);

$con = mysqli_connect("localhost","xxxxxx","xxxxxx","xxxxxxxx");
  //Run a query
  $result = mysqli_query($con,"SELECT type FROM `xxxxx`.`xxxxxx` WHERE device_id='$devid'");


$pulldown1 = '<select name="extension_type">';

while($row = mysqli_fetch_array($result))
{
if($row['type'] == "Unlimited")
{
$pulldown1 .= "<option selected value=\"Unlimited\">Unlimited Extension</option>
<option value=\"Metered\">Metered Extension</option>
<option value=\"Virtual\">Virtual Extension</option>
\n";
}
if($row['type'] == "Metered"){
$pulldown1 .= "<option selected value=\"Metered\">Metered Extension</option>
<option value=\"Unlimited\">Unlimited Extension</option>
<option value=\"Virtual\">Virtual Extension</option>
\n";
}
if($row['type'] == "Virtual"){
$pulldown1 .= "<option selected value=\"Virtual\">Virtual Extension</option>
<option value=\"Unlimited\">Unlimited Extension</option>
<option value=\"Metered\">Metered Extension</option>
\n";
}

}
$pulldown1 .= '</select>';

echo $pulldown1;

mysqli_close($con);

现在我希望能够更改值并让它保存相应的值并更新数据库中的该列。这就是我卡住的地方。有人可以指导我正确的方向吗？

Answer 1

You could use jQuery to send an AJAX post request when the select box has it's value changed.反过来，这可以更新您的 mysql 数据库条目。如果这是您想要的事情，请告诉我，我将通过代码示例进一步扩展我的答案。

JS代码示例

$("select").change(function () {
    // Send ajax request
    $.ajax
    ({
       type: "POST",
       url: "/change-value.php",
       dataType: 'json',
       data: { location: $(this).val() },
       cache: false,
       success: function(data)
       {
       }
    });

});

// PHP change-value.php 示例

<?php

// Get the changed value
$value = $_POST['value'];

// Code to update you mysql database entry would go here...

Answer 2

根据 php.net 上的 PHP 文档修改：

<?php
if ( !empty($_POST['extension_type']) )
{
   // Your MySQL connection code should be at the top of the script.
   $query = "UPDATE `xxxxx` SET `extension_type` = ? WHERE `[your_id_field_here]` = ?";
   /* Prepared statement, stage 1: prepare */
   if ( !($stmt = $mysqli->prepare($query) ) {
       echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
   }
   /* Prepared statement, stage 2: bind and execute */
   if ( !$stmt->bind_param("s", $_POST['extension_type']) || !$stmt->bind_param("i", intval($id)) ) {
       echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
   }
   if ( !$stmt->execute() ) {
       echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
   }
}
?>

正如@Simon Carlson 所说，您还需要考虑 SQL 注入。如果您有一组扩展类型列表，您可以通过检查 $_POST 值与已知的有效值来改进上面的代码。