Hi Danielo
Yes, it's the unencoded line breaks that look like the problem. JSON
requires line breaks in strings to be encoded to `\n`, while you've got
literal line breaks in there.
Best wishes
Jeremy
On Tue, Nov 4, 2014 at 12:02 PM, Danielo Rodríguez <rdani...@gmail.com>
wrote:
> Hello Jeremy,
>
> I'm just using the maketid widget from Matabelle,which is based on
> Stephans widget.
> Let me know if what I posted below is what you asked
>
> 1.
> 2. make:
>
> "{"text":"$(code)$","bag":"default","revision":"0","type":"text/vnd.tiddlywiki","title":"_ipx/templates/insertCode/codeTemplate","modified":"20141031075025632","created":"20141031074930842","comments":"$(comments)$"}"
> 3. makeClone: undefined
> 4. modificationField: undefined
> 5. newtags: undefined
> 6. t: undefined
> 7. tempTest: "{"text":"Now↵Try
>
> ↵LB","bag":"default","revision":"0","type":"text/vnd.tiddlywiki","title":"_ipx/templates/insertCode/codeTemplate","modified":"20141031075025632","created":"20141031074930842","comments":"no"}"
> 8.
>
>
> El martes, 4 de noviembre de 2014 12:58:53 UTC+1, Jeremy Ruston escribió:
>>
>> Hi Danielo
>>
>> I don't think that that code is a great approach. There's too much scope
>> for variable substitution to break the JSON formatting. In particular, JSON
>> needs string constants to be encoded if they include double quotes, and the
>> variable substitution mechanism isn't capable of that. It would be much
>> better to apply variable substitution to each field in turn.
>>
>> Can you log the complete original JSON string as well as the value that
>> comes out of substituteVariableReferences?
>>
>> Best wishes
>>
>> Jeremy
>>
>>
>>
>> On Tue, Nov 4, 2014 at 11:54 AM, Danielo Rodríguez <rdan...@gmail.com>
>> wrote:
>>
>>> Hello,
>>>
>>> I traced down the code and the problem is on the JSON.parse part.
>>> After the substituteVariableReferences execution, JSON.parse will
>>> receive the following string:
>>>
>>> "{"text":"Now↵Try ↵LB","bag":"default","revision":"0","type":"text/
>>> vnd.tiddlywiki","title"
>>>
>>> Which does not seems to be a valid JSON string.
>>> The code that fails is like this
>>>
>>> // Make the clone of the template
>>> var make = this.wiki.getTiddlerAsJson(this.maketidTemplate);
>>> debugger;
>>> var tempTest =this.substituteVariableReferences(make);
>>> var makeClone = JSON.parse(tempTest);
>>>
>>> The problem is that substituteVariableReferences returns that string
>>> with those unexpected characters. Maybe the substituteVariableReferences
>>> should return \n instead?
>>>
>>> El martes, 4 de noviembre de 2014 12:38:07 UTC+1, Jeremy Ruston escribió:
>>>>
>>>> Hi Danielo
>>>>
>>>> There's no problem that I can see with the snippet you posted above.
>>>> Can you put up a minimal test case?
>>>>
>>>> Best wishes
>>>>
>>>> Jeremy
>>>>
>>>>
>>>> On Tue, Nov 4, 2014 at 11:27 AM, Danielo Rodríguez <rdan...@gmail.com>
>>>> wrote:
>>>>
>>>>> The error is much more concise in firefox:
>>>>>
>>>>> SyntaxError: JSON.parse: bad control character in string literal at
>>>>> line 1 column 13 of the JSON data
>>>>>
>>>>> Why?
>>>>>
>>>>> --
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>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Jeremy Ruston
>>>> mailto:jeremy...@gmail.com
>>>>
>>>
>>
>>
>> --
>> Jeremy Ruston
>> mailto:jeremy...@gmail.com
>>
>
--
Jeremy Ruston
mailto:jeremy.rus...@gmail.com
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