webview的geolocation问题

无聊人士LH 2015-10-15 10:41:06

代码:
WebView webView = (WebView) findViewById(R.id.webview);
WebSettings webSettings = webView.getSettings();
webSettings.setJavaScriptEnabled(true);
webSettings.setJavaScriptCanOpenWindowsAutomatically(true);
webSettings.setDomStorageEnabled(true);
webSettings.setAppCacheEnabled(true);
webSettings.setDatabaseEnabled(true);
webSettings.setGeolocationEnabled(true);

webView.setWebViewClient(new WebViewClient());
webView.setWebChromeClient(new WebChromeClient()
@Override
public void onGeolocationPermissionsShowPrompt(final String origin, final Callback callback) {
Toast.makeText(getApplicationContext(), "callback回调" ,Toast.LENGTH_LONG).show();
callback.invoke(origin, true, false);
}

});

webView.loadUrl("http://h.aidaijia.com/");

权限:

<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_GPS" />
<uses-permission android:name="android.permission.ACCESS_ASSISTED_GPS" />
<uses-permission android:name="android.permission.ACCESS_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />

html测试:
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
<p id="demo">点击这个按钮，获得您的坐标：</p>
<button onclick="getLocation()">试一下</button>
<script type="text/JavaScript">
var x=document.getElementById("demo");
function getLocation(){
if (navigator.geolocation){
navigator.geolocation.getCurrentPosition(showPosition,getPositionError);
} else{
x.innerHTML="Geolocation is not supported by this browser.";}
}

function showPosition(position){
x.innerHTML="Latitude: " + position.coords.latitude + "<br />Longitude: " + position.coords.longitude;
}

function getPositionError(error){
switch(error.code){
case error.TIMEOUT :
alert( " 连接超时，请重试 " );
break;
case error.PERMISSION_DENIED :
alert( " 您拒绝了使用位置共享服务，查询已取消 " );
break;
case error.POSITION_UNAVAILABLE :
alert( " 我们暂时无法为您所在的星球提供位置服务 " );
break;
}
}
</script>
</body>
</html>

问题:

1.webview打开后无法定位,如果是测试的html,执行到.getCurrentPosition()后无响应

2.微信/UC浏览器能定位,但是Chrome/2345浏览器/部分android内置浏览器打不开

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