Question

此Ajax调用在更改事件期间触发，因为它会警告$name1变量。

但name1变量未传递给data.php文件 data.php可以独立正常工作。

我该如何纠正？如何正确传递name1变量？

<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<script>
$("#name").on('change',function (e){
  var name1 = this.value;
  // alert("name1 = " + name1);
  $.ajax ({
    data:{name1: name1},
    type: 'POST',
    url: 'data.php',
    success: function (response){
      console.log(response);

      $('.products-wrp').html('');
      $('.products-wrp').hide();
      $('.products-wrp').html(response);
      $('.products-wrp').show();
    }
    error: function (){
                alert('No data found!');
            }
});

}）;

data.php

<?php
$name1 = $_POST['name1'];
$results = $mysqli_conn->query("SELECT product_name, product_desc, product_code, product_image, product_price FROM products_list where product_name='$name1'");
$products_list =  '<ul id ="products_list" class="products-wrp">';
while($row = $results->fetch_assoc()){
  $products_list .= <<<EOT
  <li>
    <form class="form-item">
      <h4>{$row["product_name"]}</h4>
      <div>
        <img src="images/{$row["product_image"]}" height="62" width="62">
      </div>
      <div>Price :{$currency}{$row["product_price"]}<div>
    </form>
  </li>
  EOT;
}
$products_list .= '</ul></div>';
echo $products_list;
?>

Answer 1

$results = $mysqli_conn->query("SELECT product_name, product_desc, product_code, product_image, product_price FROM products_list where product_name='$name1'");

你有： product_name='$name1'。不应该是： product_name='{$name1}'

变量不发布到查询中

1 个答案: