Question

我有两个选择框，想要根据第一个选择框的更改填充第二个选择框。所以我的第一个选择框的代码

      <select id="category-box" name="category" onchange="showCrops()" >
            <option value ="0">Select category</option>
            <?php
                $query = $con->query("Select*from categories");
                while($row = $query->fetch_object())
                {
                    echo "<option value = '".$row->category_id."'>".$row->category."</option>";
                }
            ?>
        </select>

用于ajax调用的onchange函数 function showCrops（）{

    var name = $('#category-box').val();

    $.ajax({
        type: "POST",
        url: "getCropName.php",
        data: {category:name},
        dataType: 'json',
        success: function(data, textStatus, jqXHR)
         {
            var opts = $.parseJSON(data);
            $.each(opts, function(i,d) {
                $('#crop-box').append('<option value="' + d.crop_id + '">' + d.crop_name + '</option>');
            });

        },
        error: function (jqXHR, textStatus, errorThrown)
        {
            console.log(textStatus);
        }
    });

}

获取响应的PHP代码

header('Content-type: text/html; charset=utf-8');
include("connect.php");

$category = $_POST['category'];

$sql1 = $con->query("SELECT category_id from categories where category ='".$category."' ");
$row1= $sql1->fetch_array();

$sql2 = $con->query("SELECT * from crop_category where category_id ='".$row1['category_id']."' ");


while($row2 = $sql2->fetch_assoc()){
             echo json_encode($row2);
           }

json的回应是

          {"crop_id":"1","category_id":"1","crop_name":"rice"}   {"crop_id":"2","category_id":"1","crop_name":"wheat"}

但是我得到了解析错误＆＃39;在主要的PHP页面上。我的代码中的问题是什么？我对javascript的了解较少，所以可能需要修正以填充第二个选择框。

Answer 1

只需更改php代码，即可生成正确的JSON字符串..

$arr = array();
while($row2 = $sql2->fetch_assoc()){
    $arr[] = $row2;
}

echo json_encode($arr);

处理json响应时遇到麻烦

1 个答案: