I did somthing like that in my Learning Chatbot Project. It basicly "tears up" each row to find the individual strings, using ¶ as a delimiter (or whatever they are called).
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Molybdenum wrote:
I did somthing like that in my Learning Chatbot Project. It basicly "tears up" each row to find the individual strings, using ¶ as a delimiter (or whatever they are called).
I looked at the project and I's impressive. I'm still figuring what it does but I'm guessing it just iterates over the list and if it's equal to ¶ it adds what it's already iterated over to a variable?
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I think it just takes the output list's item (the onpe it found), and puts the seperated data into another list, seperated. Ex.
Output: 123¶55¶I AM AWESOME
...
.Script
..-..-.
Outputlist:
123
55
I AM AWESOME
Then it can use the data.
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Molybdenum wrote:
I think it just takes the output list's item (the onpe it found), and puts the seperated data into another list, seperated. Ex.
Output: 123¶55¶I AM AWESOME
...
.Script
..-..-.
Outputlist:
123
55
I AM AWESOME
Then it can use the data.
Yea, that's what it does, but how is what I was talking about
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Oh. It basicly has a list (Wordlist or something) and sequentially reads that characters of the to-be-seperated line. If its a ¶, it adds a black line to wordlist, else, it puts the read character onto the last line. Sort of like a text editor, if the key just pressed was enter (analogous to the ¶), add a new line, else, put the keystroke (analogous to any character besides the ¶) on the line.
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Whoops, typos. (and I can't edit because I'm still a new scratcher)
*sequentially reads the characters
*it adds a blank line to wordlist
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Yes, it is. Try this script.
add (join (Variable 1) (join (,) (join (Variable 2) [and so on...]))) to [List v]You can keep doing this to add more variables.
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That won't work for a variable number of variables...
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For adding to a line it seems like something like this would work? If I'm understanding what you're talking about.
if <item (1 v) of [List v] = []> replace item (1 v) of [List v] with (Addition to row) else replace item (1 v) of [List v] with (join (item (1 v) of [List v]) (join (,) (Addition to row))) end
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