CODEWITHSUNDEEP
phplaravel-4eloquent
HelpPR
HelpPR

Reputation: 43

Laravel Like Query Issue

$a = table::select('name')->where('name', 'LIKE', '%'.$term.'%')->get();

$term holds the value entered by user. My Issue is, the like does not work and it returns all results from the table irrespective of the word "$term".

If I do a query like - 

$a = table::select('name')->where('name', 'LIKE', '%abc%')->get();

this works fine, but every time I pass Php variable, it does not work as expected. Can any one enlighten a bit on this.

EDIT --

I found the issue. My $term was not getting the value. Such a foolish issue. Thank You every one who replied.

Upvotes: 2

Views: 311

Answers (3)

PHP Worm...
PHP Worm...

Reputation: 4224

Even though there is nothing wrong in your sql query I suggest you to check the value of $term

You may also try this:

$a = table::select("name")->where("name", "LIKE", "% $term %")->get();

OR

$a = table::select("name")->where("name", "LIKE", "% {$term} %")->get();

Upvotes: 1

Ashwani
Ashwani

Reputation: 742

Since query looks correct. This may be it work.

$a = table::select('name')->where('name', 'LIKE', "%.$term.%")->get();

Upvotes: 1

Akshay Hegde
Akshay Hegde

Reputation: 16997

Not sure really whats wrong, it looks fine, you can also make use of sprintf function, like below

 $a = table::select('name')->where('name', 'LIKE', sprintf("%%%s%%",$term))->get();

 // OR

 $a = table::select('name')->where('name', 'LIKE', "%$term%")->get();

Upvotes: 1

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