php ajax 相關的下拉列表不從表中加載數據

Question

我正在嘗試使用 php 和 ajax 創建一個依賴下拉列表。 What I am expecting is when the 'Make' of car is selected the relevant car models should automatically load on the 'Model' dropdown. 我已經完成了汽車“制造”的預加載。 但“模型”下拉菜單仍然為空。 我使用了一個故事並在 sql 語句中使用（選擇 model where make= selected make）。 這是我的代碼

php

<form method="GET">
                            <div class="form-group">
                                <select class="form-control" name="make" id="make">
                                    <option value="" disabled selected>--Select Make--</option>
                                        <?php
                                            $stmt=$pdo->query("SELECT DISTINCT make FROM cars  WHERE cartype='general' ");
                                        while($row=$stmt->fetch(PDO::FETCH_ASSOC)){
                                        ?>
                                        <option value="<?= $row['make']; ?>"> <?= $row['make']; ?></option>
                                        
                                        <?php  } ?>
                                </select>
                            </div>
                            <div class="form-group">
                                <select class="form-control" name="model" id="model">
                                    <option value="" disabled selected>--Select Model--</option>
                                </select>
                            </div>
.......
....
.....

腳本

<script type="text/javascript">
        $(document).ready( function () {
           // alert("Hello");
        $(#make).change(function(){
            var make = $(this).val();
            $.ajax({
                url:"filter_action.php",
                method:"POST",
                data:{Make:make},
                success: function(data){
                    $("#model").html(data);
            });
        });
    });
    
</script>

filter_action.php

<?php 
    include('db_config2.php');
    $output='';
    $stmt=$pdo->query("SELECT DISTINCT model FROM cars  WHERE cartype='general' AND make= '".$_POST['Make']."'");
    $output .='<option value="" disabled selected>--Select Model--</option>';
        while($row=$stmt->fetch(PDO::FETCH_ASSOC)){
            $output .='<option value="'.$row["model"].'">'.$row["model"].'</option>'    ;
        }
        echo $output;

?>

Answer 1

There appeared to be a couple of mistakes in the Javascript that would have been obvious in the developer console and your PHP had left the mySQL server vulnerable to sql injection attacks.

<script>
    $(document).ready( function () {
    
        // The string should be within quotes here
        $('#make').change(function(e){
            var make = $(this).val();
            
            $.ajax({
                url:"filter_action.php",
                method:"POST",
                data:{'Make':make},
                success: function(data){
                    $("#model").html(data);
                };//this needed to be closed
            });
        });
    });
</script>

在 sql 中直接使用用戶提供的數據會使您的數據庫受到 sql 注入攻擊。 為了緩解這種情況，您需要采用"Prepared Statements" - 因為您正在使用 PDO 無論如何這應該是理所當然的事情。

<?php
    if( $_SERVER['REQUEST_METHOD']=='POST' && !empty( $_POST['Make'] ) ){
    
        # The placeholders and associated values to be used when executing the sql cmd
        $args=array(
            ':type'     =>  'general',  # this could also be dynamic!
            ':make'     =>  $_POST['Make']
        );
        # Prepare the sql with suitable placeholders
        $sql='select distinct `model` from `cars`  where `cartype`=:type and `make`=:make';
        $stmt=$pdo->prepare( $sql );
        
        # commit the query
        $stmt->execute( $args );
        
        # Fetch the results and populate output variable
        $data=array('<option disabled selected hidden>--Select Model--');
        while( $rs=$stmt->fetch(PDO::FETCH_OBJ) )$data[]=sprintf('<option value="%1$s">%1$s', $rs->model );
        
        # send it to ajax callback
        exit( implode( PHP_EOL,$data ) );
    }
?>

Answer 2

我已經嘗試過使用php pdo 。

首先我創建了 3 個文件。

1. db.php

2. htmlDropdown.php

3. modelAjax.php

在這里， db.php文件可以包含我的數據庫連接代碼。 和htmlDropdown.php文件包含我的汽車和模型下拉列表。 和modelAjax.php文件包含 ajax 以獲取所有模型。

db.php

<?php

$host_name = 'localhost';
$user_name = 'root';
$password = '';
$db_name = 'stackoverflow';

$conn = new PDO("mysql:host=$host_name; dbname=$db_name;", $user_name, $password);

?>

htmlDropdown.php

<?php include "db.php"; ?>
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Cars</title>
    <!-- jQuery cdn link -->
    <script src="https://code.jquery.com/jquery-3.5.1.js" integrity="sha256-QWo7LDvxbWT2tbbQ97B53yJnYU3WhH/C8ycbRAkjPDc=" crossorigin="anonymous"></script>
    <!-- Ajax cdn link -->
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery-ajaxy/1.6.1/scripts/jquery.ajaxy.min.js" integrity="sha512-bztGAvCE/3+a1Oh0gUro7BHukf6v7zpzrAb3ReWAVrt+bVNNphcl2tDTKCBr5zk7iEDmQ2Bv401fX3jeVXGIcA==" crossorigin="anonymous"></script>
</head>
<body>
    <?php
        $car_sql = 'SELECT car_name FROM cars';  //select all cars query
        $cars_statement = $conn->prepare($car_sql);
        $cars_statement->execute();
    ?>
    <select name="car" id="car">
        <option value="">Cars</option>
        <?php
            while ($cars = $cars_statement->fetch()) { // fetch all cars data
            ?>
                <option value="<?php echo $cars['car_name']; ?>"><?php echo $cars['car_name']; ?></option>
            <?php
            }
        ?>
    </select><br><br>

    <select name="model" id="model">
        <option value="">Model</option>
    </select>
</body>
</html>
<script>

    $(document).ready(function () {

        $('#car').on("change", function () {

            let car = $(this).val(); // car value

            $.post("http://local.stackoverflowanswer1/cars/modelAjax.php", { car_name : car }, function (data, status) { // ajax post send car name in modelAjax.php file
                
                let datas = JSON.parse(data); // convert string to json object

                let options = '';

                options = '<option>Model</option>';

                $.each(datas.model, function (key, value) {
                    options += "<option>"+value.modal_name+"</option>";
                });

                $('#model').html(options);

            });
        });
    });
</script>

模型Ajax.php

<?php

include "db.php";

if ($_POST['car_name']) 
{
    $car_id_sql = "SELECT id FROM cars WHERE car_name LIKE ?"; // get id from given car name
    $id_statement = $conn->prepare($car_id_sql);
    $id_statement->execute([$_POST['car_name']]);

    $id = $id_statement->fetch();

    $model_sql = "SELECT modal_name FROM models WHERE car_id = ?"; // get model name from given id
    $model_statement = $conn->prepare($model_sql);
    $model_statement->execute([$id['id']]);

    $models = $model_statement->fetchAll();

    echo json_encode(["model" => $models]); // i have a conver array to json object
}

?>