我有一个 Webview 因为我正在加载带有查询字符串周长的网络喜欢
I have a Webview In that I am loading web with Query string perimeters like
String url = map.get("utype") + map.get("sid") + "/login/login.php?"+"&"+"strid" + "=" + map.get("strid")+"&"+"appmod" + "=" + map.get("appmod")+"&" +"ttype" + "=" + map.get("ttype")+"&" +"payp" + "=" + map.get("payp");我正在 webview 中加载该查询字符串
I am loading that Query string in webview
webView.loadUrl(url);所以现在使用查询字符串我想将参数发布到 x-www-form-urlencoded
So Now Insted of using Query string I want to POST Parameters to x-www-form-urlencoded
这些是我的参数
strid = map.get("strid"); appmod = map.get("appmod"); ttype = map.get("ttype"); payp = map.get("payp");我关注了这个但它不起作用
谁能帮我解决这个问题..
can any one help me on this..
我想在 Web 视图中将数据 POST 到 x-www-form-urlencoded
I want to POST data to x-www-form-urlencoded in Web-view
发布后我将加载我的网络服务器(网站).
After posting I it will load my Web-server(website).
更新
我已经为帖子提供了这些值
I have Given these values for post
String postData="strid="+URLEncoder.encode(map.get("strid"), "UTF-8")+ "&appmod="+URLEncoder.encode(map.get("appmod"), "UTF-8")+ "&ttype="+URLEncoder.encode(map.get("ttype"), "UTF-8")+ "&payp="+URLEncoder.encode(map.get("payp"), "UTF-8");现在添加详细信息后,我已将其提供给 URL
Now after Adding The Details I have given this to at URL
webView.postUrl(url,postData.getBytes());但是还是不行
谁能告诉我有什么问题
推荐答案需要调用webView.postUrl(url, encodingData);
值必须使用 URLEncoder.encode(postData, "UTF-8") 或 "BASE64" 编码,具体取决于后端.
Values must be encoded with URLEncoder.encode(postData, "UTF-8") or "BASE64" depends on backend.
如果它不起作用,请确保您的网址格式正确.如果您不确定,您也可以将其分享到此处而无需暴露您的价值观.
If its not working make sure you have url in correct format. You can also share it to here without exposing your values if you are not sure.
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