andFilterWhere

Yii 1.1 General Discussion
1

Hello,

I have this code inside my search model:




$query->andFilterWhere(['like', 'name', $this->name])

It generates LIKE '%name% , which is fine in most ocasions.

But I would like my system to be able to replace spaces by %, so I could search for "Homer Simpson" and the system would include "Homer J Simpson" in the result.

How do I set the like operator? I’ve tried to replace spaces by % inside the searched name without success.

Thank you

2

Many views but no answer…

This question must be too hard, or too easy to deserve an answer!

Anyway, if someone has the same problem… I already figured it out.

Answer:




        $search_name="'%".str_replace(" ","%",$this->name)."%'";

        $query->andWhere('name like '.$search_name)

As far as I was able to understand, you can’t use andFilterWhere to do this.

But within andWhere you can build your WHERE clause just like in SQL.