Intent.ACTION does not dial after first # sign

Question

I am trying to make a call with a long number that looks something like this.

tel:883994555,,,32343#,,,#

with code that looks like this.

        Intent intent = new Intent(Intent.ACTION_CALL);
        Uri uri = Uri.parse(number);
        intent.setData(uri);
        startActivity(intent);

What I see is that, the phone does not dial after the first '#' sigh. Any one knows how to make this to work.

Thanks.

score 9 · Accepted Answer · answered Jun 03 '11 at 19:03

9

How I got this to work was to use ';' (semicolon) for the hard wait instead of the 'w', and ',' (comma) for the pause, and then encode the phone number first, like this -

Uri.parse(String.format("tel:%s", Uri.encode(number)))

answered Jun 03 '11 at 19:03

ScottTx

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the # needs to be escaped yes. :) Thanks for the answer. I can not vote up yet due to my low reputation ;) – prakash Jun 04 '11 at 04:08
... my reputation permits upvoting though – Konstantin Pribluda Jun 19 '11 at 10:34

score 2 · Answer 2 · answered May 24 '11 at 19:55

2

Uri.parse(String) parses an RFC 2396-compliant, encoded URI.

RFC 2396 says:

The character "#" is excluded because it is used to delimit a URI from a fragment identifier in URI references.

answered May 24 '11 at 19:55

pawelzieba

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So do I have escape that explicitly ? – prakash May 24 '11 at 21:09
Also, # is a dtmf character, so it need not be escaped, but why is the phone dialer ignoring anything after the # sign? – prakash May 24 '11 at 21:41
This is syntax of uri: `[scheme:]scheme-specific-part[#fragment]` after "#" is [fragment](http://en.wikipedia.org/wiki/Fragment_identifier) – pawelzieba May 24 '11 at 21:54

Intent.ACTION does not dial after first # sign

2 Answers2

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