Flutter String.contains(substring) fails when there is a space

Question

Using String.contains(substring) to generate a list out of another fails when there is a space in my input.

Here is my code :

List<String> finalSuggestions = suggestions.where((i) => i.contains(new RegExp(pattern,caseSensitive:false, unicode: true))).toList();

As suggestions is a list of strings, and pattern is the user input. everything works fine unless the users adds a space in the string he is typing . finalSuggesions becomes null. else .. as long as the pattern is only a single word it is all fine.

How about you erase all space before run 'suggestions.where'? — KuKu
– KuKu, Commented Sep 4, 2020 at 18:07
If none of your suggestions contains a space, then it makes sense that a space in the pattern would result in no matched suggestions. What exactly is your question here? — Abion47
– Abion47, Commented Sep 4, 2020 at 19:22
all of my suggestions have a space between 2 words .. but when the user finishes the 1st word then press space .. the result shows noting !! — Ahmed Wagdi
– Ahmed Wagdi, Commented Sep 4, 2020 at 21:03

farouk osama · Accepted Answer · 2020-09-06 06:20:57Z

1

you don't need to use RegExp, Because pattern a string and suggestions a list of String.

List<String> finalSuggestions = suggestions.where((i) => i.contains(pattern)).toList();

It shouldn't be affected by space in this case

edited Sep 6, 2020 at 6:20

answered Sep 4, 2020 at 18:17

farouk osama

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2 Comments

Abion47 Over a year ago

A space isn't a special character, and even if it was, special characters don't need to be escaped when they come from user input.

Abion47 Over a year ago

Now your code won't compile because pattern.toString is a method and needs to be called with (). Regardless, though, pattern is already a string, so calling toString on it is redundant.

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