php值获取中的Jquery ajax
我试图在php中使用jQueryAjax从数据库中获取值。当我运行并从产品/服务中选择“开发”时,值获取但所有获取值都连接在一起,joinvalue显示所有字段。但我想在不同字段中显示不同的值。如何解决这个问题。我的代码如下。请检查并解决这个问题 index.php check.php 使用 返回JSON格式的Ajax响应php值获取中的Jquery ajax,php,jquery,ajax,Php,Jquery,Ajax,我试图在php中使用jQueryAjax从数据库中获取值。当我运行并从产品/服务中选择“开发”时,值获取但所有获取值都连接在一起,joinvalue显示所有字段。但我想在不同字段中显示不同的值。如何解决这个问题。我的代码如下。请检查并解决这个问题 index.php check.php 使用 返回JSON格式的Ajax响应 <script> $(document).ready(function () {}); function my_validate_func() {
<script>
$(document).ready(function () {});
function my_validate_func() {
var pro_serv = $('#pro_serv').val();
if ($('#type').val() != "") {
$.ajax({
type: "POST",
url: 'check.php',
data: {pro_serv: pro_serv},
dataType: "json", //Newly Added
success: function (response) {
if(response.length != 0){ //Newly Added (Check, if array is not empty, then print the returned values to respective fields)
$('#desc').val(response['desc']); //Modified
$('#unitprice').val(response['unitprice']); //Modified
$('#tax').val(response['tax']); //Modified
}
}
});
}
}
</script>
check.php
然后学习什么是json。在check.php中,编写print\r$return\u值;死亡在返回json_之前编码$return_值;并检查@DoniNo value的值:如果我运行directcheck.php。第6行中的show parser error表示$return_values=[];
<?php
include('database/db.php');
$type=$_POST['pro_serv'];
$sql="Select * from product_service_details where type='$type'";
$result=mysql_query($sql);
if($dtset=mysql_fetch_array($result)) {
$desc=$dtset['desc'];
$unitprice=$dtset['unitprice'];
$tax=$dtset['tax'];
echo $desc;
echo $unitprice;
echo $tax;
}?>
<script>
$(document).ready(function () {});
function my_validate_func() {
var pro_serv = $('#pro_serv').val();
if ($('#type').val() != "") {
$.ajax({
type: "POST",
url: 'check.php',
data: {pro_serv: pro_serv},
dataType: "json", //Newly Added
success: function (response) {
if(response.length != 0){ //Newly Added (Check, if array is not empty, then print the returned values to respective fields)
$('#desc').val(response['desc']); //Modified
$('#unitprice').val(response['unitprice']); //Modified
$('#tax').val(response['tax']); //Modified
}
}
});
}
}
</script>
<?php
include('database/db.php');
$type=$_POST['pro_serv'];
$sql="Select * from product_service_details where type='$type'";
$result=mysql_query($sql);
$return_values = []; //Newly Added
if($dtset = mysql_fetch_array($result)) {
/* Modified */
$return_values = [
'desc' => $dtset['desc'],
'unitprice' => $dtset['unitprice'],
'tax' => $dtset['tax']
];
}
return json_encode($return_values); // Return as JSON format
?>