当我使用AJAX在Symfony3中提交URL时,它会分解为数组
HTML 和 AJAX 代码:
<input type="text" name="url" id="url" class="form-control" placeholder="Search url" >
<script>
$('#search_submit').click(function(){
var feed_url = $("#url").val();
$.ajax({
type: "POST",
url: "/getdata",
data: "feed_url="+feed_url,
dataType: "json",
success: function(msg)
{
if(msg.status == 1){
alert("display")
}
else{
alert("No Product found");
}
},
beforeSend: function()
{
$("#loading-sp").show();
}
});
return false ;
});
</script>
控制器:
/**
* @Route("/getdata")
*/
public function getfeedAction(Request $request)
{
print_r($request->request->get('feed_url'));exit;
}
它只显示this而不是完整的网址。但是,如果我打印$_POST
,则会显示:
Array
(
[feed_url] => pf.tradetracker.net/?aid=1
[type] => xml
[encoding] => utf-8
[fid] => 251713
[categoryType] => 2
[additionalType] => 2
[limit] => 20
)
提前致谢。
答案 0 :(得分:0)
我已通过在javascript中添加encodeURIComponent来解决此问题
data: "feed_url="+encodeURIComponent(feed_url)