SQL LIKE not searching

Question

I'm trying to find a value from my database using SQL LIKE:

SELECT id FROM titles WHERE skuid LIKE '%:cusa%'

Where :cusa is defined CUSA06536 as a dynamic value in my PHP code.

It should return:

But instead it returns nothing. No errors are outputted. Am I doing something wrong? Are dynamic values not supported?

PHP script:

public function findCusa($cusa) {
    $find = $this->query("SELECT id FROM titles WHERE skuid LIKE '%:cusa%'");
    $find->execute(array(":cusa" => $cusa));

    if($find->rowCount() > 0) {
        echo 'found!';
    }

    echo 'not found';
}

Then called as $ps->findCusa('CUSA06536'); which returns not found.

Include your PHP script. It sounds like you're not correctly binding the `:cusa` variable. *(`SELECT id FROM titles WHERE skuid LIKE '%CUSA06536%'` should work fine.)* — MatBailie, Jul 03 '18 at 15:42
Possible duplicate of [LIKE operator with $variable](https://stackoverflow.com/questions/1843640/like-operator-with-variable) — Evandro de Paula, Jul 03 '18 at 15:43
if your ":cusa" mean the parameter, than you need something like '%' + :cusa + '%'. — P. Waksman, Jul 03 '18 at 15:49
scaisEdge got it, also the Karol Dowbecki has nice answer. At least for me. — P. Waksman, Jul 03 '18 at 15:58

score 3 · Accepted Answer · answered Jul 03 '18 at 15:51

3

Don't add the % in the LIKE statement, instead do it in the PHP code. Most likely your DB framework is getting confused when you use a quoted bind parameter.

$find = $this->query("SELECT id FROM titles WHERE skuid LIKE :cusa");
$find->execute(array(":cusa" => '%' . $cusa . '%'));

answered Jul 03 '18 at 15:51

Karol Dowbecki

43,645
9
78
111

True, I added it in the query itself. This works, accepted! – J. Doe Jul 03 '18 at 16:00

score 1 · Answer 2 · answered Jul 03 '18 at 15:53

1

try using a proper string eg: using concat

    ("SELECT id FROM titles WHERE skuid LIKE concat('%', :cusa, '%')");

answered Jul 03 '18 at 15:53

ScaisEdge

131,976
10
91
107

SQL LIKE not searching

2 Answers2