Research question on multi-column query using PHP
P粉242126786
P粉242126786 2023-07-26 17:32:50
0
1
449
<p>I have a question. I need to query all text containing a substring given by using POST method on PHP. For example, if I enter an "a" or a number like "1" (a character not an integer), do I need to find everything that contains that character or substring and check every column in the table if that's possible? I've tried something like this, but the code is a bit messy. </p> <pre class="brush:php;toolbar:false;"><?php if ($_SERVER["REQUEST_METHOD"] == "POST") { $usersearch = $_POST["usersearch"]; try { require_once "includes/dbh.inc.php"; $query = "SELECT * FROM tlattine WHERE tipologia LIKE :usersearch OR nome LIKE :usersearch OR caratteristiche LIKE :usersearch OR tabstyle LIKE :usersearch OR tabcolor LIKE :usersearch OR topstyle LIKE :usersearch OR topcolor LIKE :usersearch OR provenienza LIKE :usersearch OR produttore LIKE :usersearch OR sku LIKE :usersearch ORDER BY tipologia, provenienza, year, dimensione;"; $stmt = $pdo->prepare($query); $stmt->bindParam(":usersearch", $usersearch); $stmt->execute(); $results = $stmt->fetchAll(PDO::FETCH_ASSOC); $pdo = null; $stmt = null; } catch (PDOException $e) { die("Query failed: " . $e->getMessage()); } } else{ header("Location: ../index.php"); } ?></pre> <p>Already tried using '%:usersearch%' or combining % in any form but it seems I'm missing something, I'm absolutely sure but I can't find it :(</p>
P粉242126786
P粉242126786

reply all(1)
P粉793532469

Seeing your query, I noticed that you used the parameter: usersearch multiple times.

Mentioned in the PHP documentation:

I prefer to avoid enabling impersonation mode and instead change named placeholders to positional placeholders (use '?' instead of ':usersearch') , then use


$stmt->bindParam(1, $usersearch, PDO::PARAM_STR);
$stmt->bindParam(2, $usersearch, PDO::PARAM_STR);
$stmt->bindParam(3, $usersearch, PDO::PARAM_STR);
// and so on

One last thing, if you are querying so that the column contains the $usersearch variable, you should probably add % at the beginning and end of the string.

$usersearch = "%" . $_POST["usersearch"] . "%"
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!