我可以使用函数（数据）jQuery获取并填充不同的字段？

Question

我想填充两个字段，当有人填满卡号。

<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填充字段。任何人都可以建议更好的代码来填充数据库中的两个字段。谢谢

jQuery的

<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]; 


?> 

Answer 1

尝试提取从数据库既名称和地址和JSON他们

$k=$_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); 

是解析后他们通过jQuery

$.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); 

    } 

    } 
}); 

Answer 2

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']); 
?>