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
> <javascript:>> 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 <javascript:>
>
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