CODEWITHSUNDEEP
phppdo
Andrew G. Johnson
Andrew G. Johnson

Reputation: 26993

How do I create a PDO parameterized query with a LIKE statement?

Here's my attempt at it:

$query = $database->prepare('SELECT * FROM table WHERE column LIKE "?%"');

$query->execute(array('value'));

while ($results = $query->fetch()) 
{
    echo $results['column'];
}

Upvotes: 129

Views: 120631

Answers (9)

MrT2022
MrT2022

Reputation: 11

The only way I could get this to work was to put the %$search% into another variable.

    if(isset($_POST['submit-search'])){
         $search = $_POST['search'];
   }
    
        $query = 'SELECT * FROM posts WHERE post_title LIKE :search';
        $value ="%$search%";
        $stmt= $pdo->prepare($query);
        
        $stmt->execute(array(':search' => $value));

I don't know if this is the best way to do it, in the while loop I used:

while ($r = $stmt->fetch(PDO::FETCH_ASSOC)){

Upvotes: -2

zdarma
zdarma

Reputation: 37

I had a similar need but was using a variable grabbed from a form. I did it like this to get results from my PostgreSQL DB, using PHP:

<?php
     $player = $_POST['search'];  //variable from my search form
     $find = $sqlPDO->prepare("SELECT player FROM salaries WHERE player ILIKE ?;");
     $find->execute(['%'.$player.'%']);

     while ($row = $find->fetch()) {
         echo $row['player']."</br>";
     }
?>

The "ILIKE" makes the search non-case sensitive, so a search for cart or Cart or cARt will all return the same results.

Upvotes: 0

Kzqai
Kzqai

Reputation: 23102

For those using named parameters, here's how to use LIKE with % partial matching for MySQL databases: 

WHERE column_name LIKE CONCAT('%', :dangerousstring, '%')

where the named parameter is :dangerousstring.

In other words, use explicitly unescaped % signs in your own query that are separated and definitely not the user input.

Edit: Concatenation syntax for Oracle databases uses the concatenation operator: ||, so it'll simply become:

WHERE column_name LIKE '%' || :dangerousstring || '%'

However there are caveats as @bobince mentions here that:

The difficulty comes when you want to allow a literal % or _ character in the search string, without having it act as a wildcard.

So that's something else to watch out for when combining like and parameterization.

Upvotes: 113

gavin stanley
gavin stanley

Reputation: 1102

I got this from php delusions

$search = "%$search%";
$stmt  = $pdo->prepare("SELECT * FROM table WHERE name LIKE ?");
$stmt->execute([$search]);
$data = $stmt->fetchAll();

And it works for me, very simple. Like he says , you have to "prepare our complete literal first" before sending it to the query

Upvotes: 3

kjdion84
kjdion84

Reputation: 10064

This works:

search `table` where `column` like concat('%', :column, '%')

Upvotes: 5

Ozkar R
Ozkar R

Reputation: 19

PDO escapes "%" (May lead to sql injection): The use of the previous code will give the desire results when looking to match partial strings BUT if a visitor types the character "%" you will still get results even if you don't have anything stored in the data base (it may lead sql injections)

I've tried a lot of variation all with the same result PDO is escaping "%" leading unwanted/unexcited search results.

I though it was worth sharing if anyone has found a word around it please share it

Upvotes: 0

Vijaysinh Parmar
Vijaysinh Parmar

Reputation: 927

You can also try this one. I face similar problem but got result after research.

$query = $pdo_connection->prepare('SELECT * FROM table WHERE column LIKE :search');

$stmt= $pdo_connection->prepare($query);

$stmt->execute(array(':search' => '%'.$search_term.'%'));

$result = $stmt->fetchAll(PDO::FETCH_ASSOC);

print_r($result);

Upvotes: 17

Blazer
Blazer

Reputation: 14277

$query = $database->prepare('SELECT * FROM table WHERE column LIKE ?');
$query->bindValue(1, "%$value%", PDO::PARAM_STR);
$query->execute();

if (!$query->rowCount() == 0) 
{
    while ($results = $query->fetch()) 
    {
        echo $results['column'] . "<br />\n";
    }       
} 
else 
{
    echo 'Nothing found';
}

Upvotes: 21

Andrew G. Johnson
Andrew G. Johnson

Reputation: 26993

Figured it out right after I posted:

$query = $database->prepare('SELECT * FROM table WHERE column LIKE ?');
$query->execute(array('value%'));

while ($results = $query->fetch())
{
    echo $results['column'];
}

Upvotes: 143

Related Questions