Crazy behaviour of PDO when executing with LIKE statement - php

I got code something like that:
$stmt = $pdo->prepare('SELECT * FROM table WHERE user LIKE :user');
and later I got
$stmt->bindValue(':user', '%'.$user.'%', PDO::PARAM_STR);
$stmt->execute();
And it do not work. I am pretty sure that this is correct way to do a LIKE statement with MySQL, but It not works when I enter some part of username, but when I enter full username it goes like a charm.
Any ideas why LIKE statement don't want to do a simple regex?

What I use in my own code for LIKE queries is:
$stmt = $pdo->prepare('SELECT * FROM table WHERE user LIKE CONCAT(\'%\', :user, \'%\')');
$stmt->bindValue(':user', $user, PDO::PARAM_STR);
$stmt->execute();
using the CONCAT force MySQL to generate the comparison string after the $user variable has been escaped by PDO.

Related

Do i need to filter my variable before selecting something from database with PDO?

I have a variable $Search and i want this query
$sel=$con->prepare("SELECT * FROM users WHERE Username LIKE '%{:src}%'");
$sel->bindValue(":src",$Search);
$sel->execute();
I wanted to know if it's safe for me to do this query without doing any filtration on user's input.
Prepared statements quote your data for you, but for your query to work you need to do it this way:
$Search = "%$Search%";
$sel = $con->prepare("SELECT * FROM users WHERE Username LIKE :src");
$sel->bindValue(":src", $Search, PDO::PARAM_STR);
Or directly:
$sel = $con->prepare("SELECT * FROM users WHERE Username LIKE :src");
$sel->bindValue(":src", "%$Search%", PDO::PARAM_STR);
If you use bindParam() you need to use the first option as it needs a variable as a reference.

PDO Prepared Statements and MSSQL databases not functioning correctly

I'm having a problem running prepared queries on a MSSQL database using PDO. I can connect to the database and run SELECT queries with no parameters, but now I'm trying to run a simple SELECT query with one parameter, :user. However, the code does not return any values, despite the fact that there definitely is a database row with that value in. Here's the code I'm using:
$db = new PDO('dblib:host='.$dbHost.';dbname='.$dbName.';charset=utf8mb4',$dbUser, $dbPass);
$stmt = $db->prepare('SELECT * FROM customer WHERE email_address = :user ');
$stmt->bindValue(":user", $_SESSION["username"], PDO::PARAM_STR);
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
var_dump($result);
I receive no output from the var_dump. I know that in the database there is a correct row, so I tried:
$stmt = $db->prepare("SELECT * FROM customer WHERE email_address = 'the#email.com'");
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
var_dump($result);
And yet still no value was returned. Am I doing something wrong with PDO? If I type this exact query into the query bar it runs.
you forgot to execute your query.
right after the paramter binding, put this code:
$stmt->execute();
Ok, I'm an idiot. Forgot to execute the query. Amended code for people in the same predicament:
$db = new PDO('dblib:host='.$dbHost.';dbname='.$dbName.';charset=utf8mb4',$dbUser, $dbPass);
$stmt = $db->prepare('SELECT * FROM customer WHERE email_address = :user ');
$stmt->bindValue(":user", $_SESSION["username"], PDO::PARAM_STR);
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
var_dump($result);

PHP mysqli bind_param types

Hi I have the following:
$query = "select * from test_admin_users where school_id=? and username=?";
$stmt = $mysqli->prepare($query);
$stmt->bind_param("is", $school_id,$username);
$username='jones';
$school_id=11;
$stmt->execute();
Which works as expected. Being new to mysqli_ I played around with the bind_param types and found that
$stmt->bind_param("ss", $school_id,$username);
$stmt->bind_param("ii", $school_id,$username);
Both also give the expected results. Why are incorrect types being accepted?
Also, is there any way to use an identifier such as username instead of ? in the query template?
Thanks.

mysqli prepared statement without bind_param

I have this code for selecting fname from the latest record on the user table.
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$sdt=$mysqli->('SELECT fname FROM user ORDER BY id DESC LIMIT 1');
$sdt->bind_result($code);
$sdt->fetch();
echo $code ;
I used prepared statement with bind_param earlier, but for now in the above code for first time I want to use prepared statement without binding parameters and I do not know how to select from table without using bind_param(). How to do that?
If, like in your case, there is nothing to bind, then just use query()
$res = $mysqli->query('SELECT fname FROM user ORDER BY id DESC LIMIT 1');
$fname = $res->fetch_row()[0] ?? false;
But if even a single variable is going to be used in the query, then you must substitute it with a placeholder and therefore prepare your query.
However, in 2022 and beyond, (starting PHP 8.1) you can indeed skip bind_param even for a prepared query, sending variables directly to execute(), in the form of array:
$query = "SELECT * FROM `customers` WHERE `Customer_ID`=?";
$stmt = $db->prepare($query);
$stmt->execute([$_POST['ID']]);
$result = $stmt->get_result();
$row = $result->fetch_assoc();
The answer ticked is open to SQL injection. What is the point of using a prepared statement and not correctly preparing the data. You should never just put a string in the query line. The point of a prepared statement is that it is prepared. Here is one example
$query = "SELECT `Customer_ID`,`CompanyName` FROM `customers` WHERE `Customer_ID`=?";
$stmt = $db->prepare($query);
$stmt->bind_param('i',$_POST['ID']);
$stmt->execute();
$stmt->bind_result($id,$CompanyName);
In Raffi's code you should do this
$bla = $_POST['something'];
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$stmt = $mysqli->prepare("SELECT `fname` FROM `user` WHERE `bla` = ? ORDER BY `id` DESC LIMIT 1");
$stmt->bind_param('s',$_POST['something']);
$stmt->execute();
$stmt->bind_result($code);
$stmt->fetch();
echo $code;
Please be aware I don't know if your post data is a string or an integer. If it was an integer you would put
$stmt->bind_param('i',$_POST['something']);
instead. I know you were saying without bind param, but trust me that is really really bad if you are taking in input from a page, and not preparing it correctly first.

mySQLi Prepared Statement Select with Escape Characters

I am trying to select from a mySQL table using prepared statements. The select critera is user form input, so I am binding this variable and using prepared statements. Below is the code:
$sql_query = "SELECT first_name_id from first_names WHERE first_name = ?";
$stmt = $_SESSION['mysqli']->prepare($sql_query);
$stmt->bind_param('s', $_SESSION['first_name']);
$stmt->execute();
$stmt->store_result();
if ($stmt->num_rows == '1') {
$stmt->bind_result($_SESSION['first_name_id']);
$stmt->fetch();
} else {
$stmt->close();
$sql_query = "INSERT INTO first_names (first_name) VALUES (?)";
$stmt = $_SESSION['mysqli']->prepare($sql_query);
$stmt->bind_param('s', $_SESSION['first_name']);
$stmt->execute();
$_SESSION['first_name_id'] = $_SESSION['mysqli']->insert_id;
}
$stmt->close();
Obviously my code is just determining whether or not the first_name already exists in the first_names table. If it does, it returns the corresponding ID (first_name_id). Otherwise, the code inserts the new first_name into the first_names table and gets the insert_id.
The problem is when a user enters a name with an escape character ('Henry's). Not really likely with first names but certainly employers. When this occurs, the code does not execute (no select or insert activity in the log files). So it seems like mySQL is ignoring the code due to an escape character in the variable.
How can I fix this issue? Is my code above efficient and correct for the task?
Issue #2. The code then continues with another insert or update, as shown in the code below:
if (empty($_SESSION['personal_id'])) {
$sql_query = "INSERT INTO personal_info (first_name_id, start_timestamp) VALUES (?, NOW())";
} else {
$sql_query = "UPDATE personal_info SET first_name_id = ? WHERE personal_info = '$_SESSION[personal_id]'";
}
$stmt = $_SESSION['mysqli']->prepare($sql_query);
$stmt->bind_param('i', $_SESSION['first_name_id']);
$stmt->execute();
if (empty($_SESSION['personal_id'])) {
$_SESSION['personal_id'] = $_SESSION['mysqli']->insert_id;
}
$stmt->close();
The issue with the code above is that I cannot get it to work at all. I am not sure if there is some conflict with the first part of the script, but I have tried everything to get it to work. There are no PHP errors and there are no inserts or updates showing in the mySQL log files from this code. It appears that the bind_param line in the code may be where the script is dying...
Any help would be very much appreciated.
you should validate/escape user input before sending it to the db.
checkout this mysql-real-escape-string()

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