0 
    <html>


    <a href="\\\\192.168.1.102\\\\Data\\ACT\\OTHERACTS\\HTMLFILES\\2009\\ftn1acsdiv2.htm");'><span class=GramE>1a</span></a></sup>[(</span><i><span
    style='mso-bidi-font-family:AsterV'>ca</span></i><span 
style='mso-bidi-font-family:
    AsterV'>)<span style='mso-tab-count:1'>  
</span>“firm” shall have the meaning
    assigned to it in section 4 of the Indian Partnership Act, 1932 (9 of 1932),
    and includes,—&lt;o:p></o:p>
</span></p>
     </html>

in this 1a[(ca)  is click able . this is Html content i want to print
 <a href="\\\\192.168.1.102\\\\Data\\ACT\\OTHERACTS\\HTMLFILES\\2009\\ftn1csdiv2.htm");'> its file path in another activity on click

函数我有动态href我将如何在android的另一个活动中获得它的路径

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2 回答 2

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你好@Joan 我认为只需提取链接并通过意图传递它就足够了。或首先通过意图将其发送到另一个活动，然后提取。如果我没有错误地理解这个问题。

于 2013-01-14T05:58:57.383 回答
0

根据您的评论，我理解的是，您想打开 webview 在新活动中包含的每个链接（可能有另一个 webivew）。正确的？

    webView.setWebViewClient(new WebViewClient() {            
        public boolean shouldOverrideUrlLoading(WebView view, String url) {

            Bundle b = new Bundle();                
            Intent intent = new Intent(MyThisActivity.this, MySecondActivity.class);
            b.putString("newURL", url);                
            startActivity(intent);
            return true;
        }
    });
于 2013-01-14T07:07:35.033 回答