我可以使用函数(数据）jquery获取和填充不同的字段吗 - can i fetch and populate different fields using function(data) jquery?

can i fetch and populate different fields using function(data) jquery?

当某个人填写卡号时，我想填充两个字段

 <input type="text" id="name" name="name" class="form-control" Value="<?php echo $grow['name']; ?>">
 <input type="text" id="address" name="address" class="form-control" Value="<?php echo $grow['address']; ?>">

但是这个代码一个接一个地填充字段。有谁能建议从数据库中填充两个字段是更好的代码吗。谢谢

jquery

<script type="text/javascript">
$(document).ready(function()
{
$("#krishi").keyup(function()
{
var k=$(this).val();
var q="name";

$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q='+q,
cache: false,
success: function(data)
{
    if(data){
    $("#name").val(data);
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q=address',
cache: false,
success: function(data)
{
    if(data){
    $("#address").val(data);
    }
    } 
});
    }else
        $("#name").val("");
    $("#address").val("");
    } 
});
});

});
</script>

getresult.php

<?php
define('INCLUDE_CHECK',true);
include("mysql.php");

$k=$_POST['k'];
$q=$_POST['q'];
$sql=mysql_query("select * from inward where krishi='$k'");
$row=mysql_fetch_array($sql);
echo $row[$q]; 

?>

尝试从数据库中提取名称和地址，并用json将其转换为

$k=$_POST['k'];
$q=$_POST['q'];
$sql=mysql_query("select address,name from inward where krishi='$k'");
$row=mysql_fetch_array($sql);
$result = array(
               'name'=>$row['name'],
               'address'=>$row['address']);
echo json_encode($result);

之后通过jquery 解析它们

$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q=address',
cache: false,
success: function(data)
{
    if(data){
    var parsedData = jQuery.parseJSON(data);
    $("#name").val(parsedData.name);
    $("#address").val(parsedData.address);
    }
    } 
});

Javascript代码：

<script type="text/javascript">
$(document).ready(function()
{
    $("#krishi").keyup(function()
    {
    var k = $(this).val();
    var q = "name";
    $.ajax({
        type: 'POST',
        url: "getresult.php",
        data: 'k='+k,
        cache: false,
        success: function(data)
        {
            var jsonArr = $.parseJSON(data);
            if(typeof response =='object')
            {
                $("#name").val(jsonArr.name);
                $("#address").val(jsonArr.address); 
            }
            else
            {
                $("#name").val("");
                $("#address").val("");  
            }
        }
    });
    });
});
</script>

PHP代码：

<?php
define('INCLUDE_CHECK',true);
include("mysql.php");
$k   = $_POST['k'];
$sql = mysql_query("select * from inward where krishi='$k'");
$row = mysql_fetch_assoc($sql);
echo json_encode(array('name' => $row['name'], 'address' => $row['address']); 
?>