CODEWITHSUNDEEP
sqlcakephp
Charmie
Charmie

Reputation: 2610

using "like" in cakephp to return search results

this is the scenario:

i typed an address in a search box: "Blk 6 lot 34 brgy teniente tiago"

it must return all the "Blk 6 lot 34 brgy teniente tiago" from the column "address" of a particular table:

say for example the addresses are these:

        1   = "Blk 6 lot 34 brgy teniente tiago gma cavite"
        2   = "Blk 6 lot 34 brgy teniente tiago trece martirez cavite"
        3   = "Blk 100 lot 34 brgy teniente tiago gma cavite"
        4   = "Blk 6 lot 34 brgy teniente tiago gma cavite"
        5   = "whatever address it may be"

the 1-4 records should be displayed... but the thing is if you did not type the exact address, as in the EXACT ADDRESS, it won't return anything.

i am coding in cakephp and it looks like this:

array('UserDetail.address LIKE'=>'%'.$filterparams['address'].'%')

Upvotes: 0

Views: 1339

Answers (2)

thejartender
thejartender

Reputation: 9379

Lets say you have two similar values

Assign the similar word to a varaiable e.g., keyword

and try:

$this->Model->find('all', array(
'conditions' => array('Model.column LIKE' => '%keyword%')
));

Upvotes: 1

Charmie
Charmie

Reputation: 2610

Found my mistake. I actually did not store the value of my query into the variable i was trying to challenge in my controller. so here's what i got now:

$address = array('UserDetail.address LIKE'=>'%'.$filterparams['address'].'%');

Upvotes: 0

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