Break String into Substrings if true, once true stop checking

[Xavier Gomez]

Hi,

here is a possibitity, not sure it is the most efficient one though.
It adds a delimiter tag after each of these patterns, then splits the string:

var textIn = “go [up] or [down] then [right] but not [left]”;
var D = “#”; // or something else
var strings = textIn.replace(/(\[(up|down|right|left)\])\s*(?!$)/g, “$1″+D).split(D); // an array of substrings of textIn

Xavier

[Bryce Poole]

Hey Xavier,

Thank you for the response, I think this might work to solve my problem! This seems to work well as long as the first [direction] is present, but sometimes the string may not contain any directions at all, in which case I’d like to keep it whole. Would it be possible to add some type of null catch for when there isn’t a second string present? Ideally I’d like to set the string to a very long max value as a default, something like 9999.

For example:
“go [up] first” or “go [up] then [down]” both work but “go wherever” returns an error when calling the second index in the string.

var textIn = thisComp.layer("Add Text Here").text.sourceText;
var D = "#"; // or something else
var strings = textIn.replace(/(\[(up|down|right|left)\])\s*(?!$)/g, "$1"+D).split(D); // an array of substrings of textIn
strings[1];// returns error if string does not contain [up], [down], [left], or [right]

[Bryce Poole]

I think I may have just solved it by adding a simple if statement to the declaration of the strings array.
Though there might still be a better way to do this? Thank you again for all the help!

var textIn = thisComp.layer("Add Text Here").text.sourceText;
var D = "#"; // or something else
var strings = textIn.replace(/(\[(up|down|right|left)\])\s*(?!$)/g, "$1"+D).split(D); // an array of substrings of textIn

if (textIn.match(/\[(up|down|left|right)\]/) ) {
strings[1];
}else{
strings[0];
}


[Xavier Gomez]

With this expression, the entries of the array ‘strings’ are all of the formm “something+[up/down/left/right]”, except maybe the very last one. So for that last entry you have to check again:

N = strings.length;
all entries: strings[n] (n=0,…., N-2) have one of your direction patterns at the end;
the last entry (which is strings[0] if N===1) can be an exception:

if (strings[N-1].match(/\[(up|down|left|right)\]/)){
doThis; // direction spotted
}
else{
doThat; // no direction spotted
};

But… there is maybe another solution that leads more straigthforwardly to the same thing.

//===================

Edit: actually if you remove the (?!$) in the regexp (which tells not to match if the [up/down] pattern if right at the end of the string, i added this so you dont end up with an empty entry) then the last entry of string NEVER has a direction and all other have one:

strings = textIn.replace(/([(up|down|right|left)])s*/g, “$1″+D).split(D);
N = strings.length;

//>>> now all entries strings[n] (n=0,… N-2) have a direction, the last one not (and can be an empty string). Maybe better that way.

[Bryce Poole]

Thank you Xavier! This solved it!

Though, I’m not totally sure I understand what all the RegExpr parts are doing, if you have time would you mind breaking it down for me, just so I can fully wrap my brain around this?

s*/g, “$1″+D).split(D); //specifically this bit

In either case this totally solved it, thank you very much!