Create a Hyperlink using String resources

Question

I try to make a TextView as hyperlink. This one worked as expected :

content.text = "<a href=${args.article.url}>Content</a>".fromHtml()

But I get a lint warning regarding using string resources. But this one does now show TextView as hyperlink:

content.text = getString(R.string.content, args.article.url).fromHtml()

And this is String resource :

<string name="content"><a href="%s">Content</a></string>

Is there any solution to fix it using String resource?

@Suppress("DEPRECATION")
fun String.fromHtml() = if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
    Html.fromHtml(this, Html.FROM_HTML_MODE_COMPACT)
} else {
    Html.fromHtml(this)
}

Use getText() instead of getString()/fromHtml(): stackoverflow.com/a/19311520/115145 — CommonsWare
– CommonsWare, Commented May 10, 2019 at 21:46
@CommonsWare, when I use getText(R.string.content, args.article.url).fromHtml() I receive an error under args.article.url — Ali
– Ali, Commented May 10, 2019 at 21:51
Oh, right, getText() doesn't support placeholders. You could switch to the single-parameter getText() and use TextUtils.expandTemplate() or TextUtils.replace(). — CommonsWare
– CommonsWare, Commented May 10, 2019 at 21:55
Yes, that was right. Thanks. You can share your answer and I mark it as accepted. — Ali
– Ali, Commented May 10, 2019 at 22:15
I recommend that you answer your own question, showing what you wound up with! — CommonsWare
– CommonsWare, Commented May 10, 2019 at 22:16

Ali · Accepted Answer · 2019-05-10 22:20:04Z

1

This is how to set text in TextView :

content.text = TextUtils.expandTemplate(getText(R.string.content), args.article.url).toString().fromHtml()

And this is how to set String resources :

<string name="content"><![CDATA[ <a href="$s">Content</a>]]></string>

answered May 10, 2019 at 22:20

Ali

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