如何返回[查询]结果作为文件的位置？ [英] How to return [search] result as file location?

问题描述

我正在搜寻能够解释。它使用SQLite数据库。字典格式是如下图所示（2列由 - 隔开）

I'm working on Search able Dictionary. It uses SQLite database. Dictionary formats are like below (2 column separated by "-")

Apple - One kind of fruit
Cricket - One type of game
Titanic - One type of Movie

下面苹果是列 A 和一类的水果是列 B

我已经更新了我的数据库像下面

I've updated my database like below

Apple - res/raw/Apple.txt
Titanic- res/raw/Titanic.txt

我想开 Apple.txt 文件并显示文本时，用户搜索苹果。但样本code我使用的是TextView的制作而已，不打开文件，缓冲区和ViewText。

I want to open Apple.txt file and show text when user search Apple. But the sample code i'm using is made for TextView only, not to open file, buffer and ViewText.

下面是我的code这显示了我的应用程序列B

Here is my code which shows Column B on my application

TextView details = (TextView) findViewById(R.id.details);
details.setMovementMethod(new ScrollingMovementMethod());

请建议我怎么能打开Apple.txt从搜索查询替换上述code？

Please suggest how can i open Apple.txt from search query by replacing above code?

[更新后]

推荐答案

您可以根据使用拆分 - ：

String[] res = line.split("\\s+-\\s+");

现在你必须是这样的：

[Apple, One kind of fruit]

现在，因为你在A列中有苹果，你知道你必须展示 apple.txt 。

Now, because you have Apple in column A, you know that you have to show the location of apple.txt.

或者（更好的方法），可以存放在B列的路径@Egor在评论中指出：

Alternatively (a better approach), you can store the path in column B as @Egor stated in the comments:

Apple - the_path_to_open.txt

而现在，当你拆，你会得到的路径中的 1 数组（这是列B）的元素，你可以打开它，以显示其结果。

And now, when you split, you'll get the path in the 1 element of the array (Which is column B) and you can open it to show its' results.

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