android - 在代码中使用 findAllAsync 两次

Question

我想搜索一个网页来找到两个字符串。我使用这段代码:

webView.setWebViewClient(new WebViewClient() {
    @Override
    public void onPageFinished(WebView view, String url) {
        webView.findAllAsync("str1");
        webView.findAllAsync("str2");
    }
}); 

和这个 FindListener :

public void onFindResultReceived(int activeMatchOrdinal, int numberOfMatches, boolean isDoneCounting) {
    if(numberOfMatches > 0)     
      (...) //do something
}

但在这种情况下，onFindResultReceived 仅针对 str2 调用，并且如果 str1 的 numberOfMatches 大于0 什么都没做。是什么原因，应该如何解决？

Answer 1

来自 the docs

Finds all instances of find on the page and highlights them, asynchronously. Notifies any registered WebView.FindListener. Successive calls to this will cancel any pending searches.

换句话说，您的第二个调用取消了第一个。解决方案是等到每个搜索完成后再开始下一个。下面是一个可以为任意数量的字符串实现所需结果的示例

private int mCurrentSearchIndex = -1;
private String mSearchTerms = { "str1", "str2" } 

webView.setWebViewClient(new WebViewClient() {
    @Override
    public void onPageFinished(WebView view, String url) {
        nextSearch();
    }
}); 

private void nextSearch() {
    mCurrentSearchIndex ++; //the fisrt tiem this is called the index gets set to 0
    if (mCurrentSearchIndex < mSearchTerms.length) {
        //get the search string corresponding to this index and then search it
        webView.findAllAsync(mSearchTerms[mCurrentSearchIndex]);
    }
}

public void onFindResultReceived(int activeMatchOrdinal, int numberOfMatches, boolean isDoneCounting) {
    if (isDoneCounting) {
        if (numberOfMatches > 0) {   
            ...
        }
        //here we know the previous find finished, so its safe to start another
        nextSearch();
    }
}