Why is this code failing? mysql like problem

Question

I want to search in the mysql database. I want the search sent for search to be searched in 2 columns in the table. But I did not get results with this code.

$Ara = $db->prepare("SELECT * FROM urun_bilgileri WHERE (urun_adi LIKE ? OR urun_kodu LIKE ?)");
$Ara->execute(array('%'.$Gelen.'%'));
while($Liste = $Ara->fetchAll(PDO::FETCH_ASSOC)){;
     $toplam_icerik = $Ara->rowCount();
}

where is the problem?

You need to have one value in your array for each parameter in the query. You have two parameters and only one value. Try changing `$Ara->execute(array('%'.$Gelen.'%'));` to `$Ara->execute(array('%'.$Gelen.'%', '%'.$Gelen.'%'));` — Nick, Oct 25 '18 at 06:42
Try this `$Ara = $db->prepare("SELECT * FROM urun_bilgileri WHERE (urun_adi LIKE :first OR urun_kodu LIKE :second)"); $Ara->bindValue(':first', '%' . $Gelen . '%', PDO::PARAM_STR);$Ara->bindValue(':second', '%' . $Gelen . '%', PDO::PARAM_STR); $Ara->execute(); $data = $Ara->fetchAll();` — Gopal Sharma, Oct 25 '18 at 06:46

score 1 · Accepted Answer · answered Oct 25 '18 at 06:43

1

try with this,need to pass two values

$Ara = $db->prepare("SELECT * FROM urun_bilgileri WHERE (urun_adi LIKE ? OR urun_kodu LIKE ?)");
$Ara->execute(array('%'.$Gelen.'%','%'.$Gelen.'%'));
while($Liste = $Ara->fetchAll(PDO::FETCH_ASSOC)){;
     $toplam_icerik = $Ara->rowCount();
}

answered Oct 25 '18 at 06:43

Rp9

1,955
2
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@TurgutYıldırım For that, you need FULL TEXT SEARCH see https://stackoverflow.com/questions/224714/what-is-full-text-search-vs-like – Gopal Sharma Oct 25 '18 at 06:56
Unfortunately, I couldn't find the answer. – Turgut Yıldırım Oct 25 '18 at 07:12
does SELECT * FROM urun_bilgileri WHERE (urun_adi LIKE '%your value%' OR urun_kodu LIKE '%your value%') works in phpmyadmin? – Rp9 Oct 25 '18 at 07:17

Maksym Fedorov · Answer 2 · 2018-10-25T07:35:00.253

1

You use two placeholders in the query, but pass only one parameter. You must pass two parameters, for example:

$keyword = '%'.$Gelen.'%';
$Ara = $db->prepare("SELECT * FROM urun_bilgileri WHERE (urun_adi LIKE ? OR urun_kodu LIKE ?)");
$Ara->execute(array($keyword, $keyword));

or you can use named placeholders. For example:

$keyword = '%'.$Gelen.'%';
$Ara = $db->prepare("SELECT * FROM urun_bilgileri WHERE (urun_adi LIKE :urun_adi OR urun_kodu LIKE :urun_kodu)");
$Ara->execute(array(':urun_adi' => $keyword, ':urun_kodu =>$keyword));

edited Oct 25 '18 at 07:35

answered Oct 25 '18 at 06:45

Maksym Fedorov

6,383
2
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Named placeholders __must__ have different names, `:param`, `:param` will not work. – u_mulder Oct 25 '18 at 07:15
There are 1 search boxes. If I enter 'product name' here, look for its column. But if I enter 'product_code' here, search for the 'product_code' column. – Turgut Yıldırım Oct 25 '18 at 07:26

score 0 · Answer 3 · answered Oct 25 '18 at 06:55

0

Enable PHP error_reporting and PDO error_reporting than PHP tells you this within an error message.

error_reporting(-1);
ini_set('display_errors', true);

and http://php.net/manual/de/pdo.error-handling.php

answered Oct 25 '18 at 06:55

hausl

160
16

Why is this code failing? mysql like problem

3 Answers3