find all occurrences except for the last of a pattern in javascript

When considering the patterns below:

"profile[foreclosure_defenses_attributes][0][some_text]"
"something[something_else_attributes][0][hello_attributes][0][other_stuff]"

It is possible to extract the last part by utilizing non-capturing groups:

var regex = /(?:\w+(\[\w+\]\[\d+\])+)(\[\w+\])/;
str = "profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]";
match = regex.exec(str);
["profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]", "[properties_attributes][0]", "[other_stuff]"]

Nevertheless, the goal is to obtain everything except for the final segment. In simpler terms, everything excluding [some_text] or [other_stuff].

The solution does not seem apparent using noncapturing groups. What alternative approach can be taken?

javascriptregex

Answer №1

How about this?

 A more concise version that captures items from the end if [] are present.

var regex = /(.*)(?:\[\w+\])$/;
var result = "something[something_else_attributes][0][hello_attributes][0][other_stuff11][other_stuff22][other_stuff33][other_stuff44]".match(regex)[1];
    result;

Alternatively, you could use replace, although it may be less efficient.

 var regex = /(.*)(?:\[\w+\])$/;

 var result = "something[something_else_attributes][0][hello_attributes][0][other_stuff11][other_stuff22][other_stuff33][other_stuff44]".replace(regex, function(_,$1){ return $1});
     result;

Answer №2

Assuming those strings truly belong to you:

let regex = /(.*)\[/;

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