Passing quoted parameter to prepared statement PDO - php

I checked other questions, PHP and PDO documentation and many other examples, but I feel stuck, probably since I'm just getting started with PHP and MySQL.
I'm trying to first check if a line exist with a SELECT statement and then if line exists DELETE it.
However, the query is not going through, I'm using MySQL 8 on Ubuntu 18, and the queries work only with the single quotes after the WHERE clause:
SELECT bookname FROM libri WHERE bookname LIKE "test.pdf";
DELETE FROM libri WHERE bookname LIKE "test.pdf";
However seems like i'm unable to reproduce it with the prepared statements:
$sql = "SELECT bookname FROM libri WHERE bookname LIKE :deleteTerm;";
$stmt = $pdo->prepare($sql);
$stmt->bindValue(':deleteTerm', $deleteKey);
$stmt->execute();
I set $deleteKey as $deleteKey = $_POST['delete']; and trying to attach the single quotes as $stmt->bindValue(':deleteTerm', "'".$deleteKey."'"); it's not working.
Also tried to set quotes directly into variable name: $deleteKey = "'".$_POST['delete']."'"; and using the prepared statement above, but still not working.
Am i missing something,or getting something wrong? Maybe I have to use another query?

"select" part of the transaction you have done is unnecessary. because you mean delete it if it exists. you are trying to delete it in the next section. if you actually run the second part directly, it will delete it if it already exists.I also recommend you to try this in the clawed area.
$sql = " DELETE FROM libri WHERE bookname LIKE 'test.pdf'";

Related

Stupid SQL error with no leads

Ok, I can't figure out why this doesn't work at all.
$get_data = mysql_query("SELECT * FROM chatbox ORDER BY ID WHERE message = 'bla'");
while($data = mysql_fetch_assoc($get_data)) {
*blablabla code*
}
This does NOT work. It gives me a
"Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource" error. (with line number ofcourse)
When I remove the "WHERE message = 'bla'" part, it works fine. Tryed with and without ' things around bla and around messages. Message field does excist in my DB, so does the entry bla. Tryed it even with the ID field in my DB, with number 1, does not work at all. What is wrong with this simple line of code. Tryed to look almost everywhere, can't figure it out...
You are using WHERE clause after ORDER BY, it can not be work for you.
Modified Query:
SELECT * FROM chatbox WHERE message = 'bla' ORDER BY ID
You need to follow this sequence when you crease MYSQL SELECT Statement:
SELECT .. COLUMNS .. FROM .. WHERE .. ORDER
SELECT Manual Reference
Side Note:
Please use mysqli_* or PDO instead of mysql_* because this extension deprecated and not available in PHP 7.
Your query is wrong. You are using order by before where condition.
Try this query : "SELECT * FROM chatbox WHERE message = 'bla' ORDER BY ID"
And also migrate from mysql_* to mysqli_*.

See how query looks after bind_param() [duplicate]

This question already has answers here:
Getting raw SQL query string from PDO prepared statements
(16 answers)
Closed 6 years ago.
In PHP, when accessing MySQL database with PDO with parametrized query, how can you check the final query (after having replaced all tokens)?
Is there a way to check what gets really executed by the database?
So I think I'll finally answer my own question in order to have a full solution for the record. But have to thank Ben James and Kailash Badu which provided the clues for this.
Short Answer
As mentioned by Ben James: NO.
The full SQL query does not exist on the PHP side, because the query-with-tokens and the parameters are sent separately to the database.
Only on the database side the full query exists.
Even trying to create a function to replace tokens on the PHP side would not guarantee the replacement process is the same as the SQL one (tricky stuff like token-type, bindValue vs bindParam, ...)
Workaround
This is where I elaborate on Kailash Badu's answer.
By logging all SQL queries, we can see what is really run on the server.
With mySQL, this can be done by updating the my.cnf (or my.ini in my case with Wamp server), and adding a line like:
log=[REPLACE_BY_PATH]/[REPLACE_BY_FILE_NAME]
Just do not run this in production!!!
You might be able to use PDOStatement->debugDumpParams. See the PHP documentation .
Using prepared statements with parametrised values is not simply another way to dynamically create a string of SQL. You create a prepared statement at the database, and then send the parameter values alone.
So what is probably sent to the database will be a PREPARE ..., then SET ... and finally EXECUTE ....
You won't be able to get some SQL string like SELECT * FROM ..., even if it would produce equivalent results, because no such query was ever actually sent to the database.
I check Query Log to see the exact query that was executed as prepared statement.
I initially avoided turning on logging to monitor PDO because I thought that it would be a hassle but it is not hard at all. You don't need to reboot MySQL (after 5.1.9):
Execute this SQL in phpMyAdmin or any other environment where you may have high db privileges:
SET GLOBAL general_log = 'ON';
In a terminal, tail your log file. Mine was here:
>sudo tail -f /usr/local/mysql/data/myMacComputerName.log
You can search for your mysql files with this terminal command:
>ps auxww|grep [m]ysqld
I found that PDO escapes everything, so you can't write
$dynamicField = 'userName';
$sql = "SELECT * FROM `example` WHERE `:field` = :value";
$this->statement = $this->db->prepare($sql);
$this->statement->bindValue(':field', $dynamicField);
$this->statement->bindValue(':value', 'mick');
$this->statement->execute();
Because it creates:
SELECT * FROM `example` WHERE `'userName'` = 'mick' ;
Which did not create an error, just an empty result. Instead I needed to use
$sql = "SELECT * FROM `example` WHERE `$dynamicField` = :value";
to get
SELECT * FROM `example` WHERE `userName` = 'mick' ;
When you are done execute:
SET GLOBAL general_log = 'OFF';
or else your logs will get huge.
What I did to print that actual query is a bit complicated but it works :)
In method that assigns variables to my statement I have another variable that looks a bit like this:
$this->fullStmt = str_replace($column, '\'' . str_replace('\'', '\\\'', $param) . '\'', $this->fullStmt);
Where:
$column is my token
$param is the actual value being assigned to token
$this->fullStmt is my print only statement with replaced tokens
What it does is a simply replace tokens with values when the real PDO assignment happens.
I hope I did not confuse you and at least pointed you in right direction.
The easiest way it can be done is by reading mysql execution log file and you can do that in runtime.
There is a nice explanation here:
How to show the last queries executed on MySQL?
I don't believe you can, though I hope that someone will prove me wrong.
I know you can print the query and its toString method will show you the sql without the replacements. That can be handy if you're building complex query strings, but it doesn't give you the full query with values.
I think easiest way to see final query text when you use pdo is to make special error and look error message. I don't know how to do that, but when i make sql error in yii framework that use pdo i could see query text

MySQL Injection by LIKE operator [duplicate]

This question already has answers here:
How can I prevent SQL injection in PHP?
(27 answers)
Closed 9 years ago.
I've below code in one of my php files to fetch data from DB:
$products = $this->db->get_rows('SELECT * from products WHERE shop_id='.$_SESSION['shop_id'].'AND tags,title,text LIKE \'%'.$_POST['search'].'%\'');
Is it problematic? I mean LIKE operator can be injected?
Edited
please provide examples of injecting in this way
Any operator can be injected without binding.
$_POST['search'] = "1%'; DROP TABLE myTable LIKE '%";
Would make
.... AND tags,title,text LIKE '%1%'; DROP TABLE myTable LIKE '%%'
Read on how to bind parameters.
Of course this can be injected, you need to sanitize your input. Right now you are taking raw post data and inserting it into your SQL statement.
You should run your POST data through some sort of data sanitization, something like mysql_real_escape_string or the like
Or at least prepared statements. let server side code do the work for you.
Never, ever, use database queries like that, don't construct a string with variables and use it for database activities.
Construct a string that will later on be prepared and executed, by inserting the variables into the string, making them not act like "commands" but as "values".
You can do it like this:
$query = "SELECT * from products WHERE shop_id = :shopId;"; // An example, you can finish the rest on your own.
Now, you can prepare the statement (I recommend using PDO for this).
$statement = $db->prepare($query); // Prepare the query.
Now you can execute variables into the prepared query:
$statement->execute(array(
':shopId' => $_SESSION['shop_id']
));
If you're inserting or updating, then you would have wanted to do:
$success = $statement->execute(array(
':shopId' => $_SESSION['shop_id']
));
which stores a boolean in $success, or you can fetch the values from a result if you're SELECTing:
$statement->execute(array(
':shopId' => $_SESSION['shop_id']
));
$result = $statement->fetch(PDO::FETCH_ASSOC);
if($result )
{
// You can access $result['userId'] or other columns;
}
Note that you should actually make that be a function, and pass $shopId into the function, but not the session itself, and check if the session actually exists.
I recommend googling on how to use PDO, or take a look on one of my examples: How to write update query using some {$variable} with example
This is really bad. Pulling vars into an SQL statement without cleaning or checking them is a good way to get pwnd. There are several things that people can inject into code. Another injection method to watch out for, 1=1 always returns true.
$products = $this->db->get_rows('SELECT * from products WHERE shop_id='.$_SESSION['shop_id'].'AND tags,title,text LIKE \'%'.$_POST['search'].'%\'');
//This example expects no result from the table initially so we would blind attack the DB to pull the admin record.
$_POST['search'] = "-1\'; union all select * from users limit 1;";
Someone call pull up the top account in the database (like the admin).
$user_id = $this->db->get_rows('SELECT * from users WHERE email="'.$_POST['email'].'" and password="'.$_POST['password'].'"');
//This always returns true so now I'm the admin again
$_POST['password'] = "x\' or 1=1 limit 1";
You also want to be careful what you print on screen.
$user_id = $this->db->get_rows('SELECT * from users WHERE email="'.$_POST['email'].'" and password="'.$_POST['password'].'"');
A message that you echo that says "No user name exists for $_POST['email']" could be replaced with something else.
$_POST['email']=";
$fp = fopen('index.php', 'w');
fwrite($fp, \"header('Location: http://badwebsite.com;');\";
fclose($fp);";
index.php could now people to a different website entirely where an infected page exists or an infected page on the site.
If you're checking IDs do something like:
if(preg_match('!^[0-9]$!',$_POST['id'])){
$id = $_POST['id'];
} else {
//flush
}
or count for the number of possible records... if you're only expecting one and you get all of the records in the DB then it's an injection attempt.
if(is_numeric($_POST['id'])){
$id = $_POST['id'];
$count = mysql_result(mysql_query("select count(*) from users where id='$id''),0);
}

concatenate mysql select query with php variable?

I am trying to concatenate a MySQL SELECT query with PHP variable but got an error.
My PHP statement which gives an error is:
$result=mysql_query("SELECT user_id,username,add FROM users WHERE username =".$user."AND password=".$add);
and error as:
( ! ) Notice: Undefined variable: info in C:\wamp\www\pollBook\poll\login.php on line 18
Call Stack
I don't understand where I missed the code.
When I write query without WHERE clause it works fine.
The reason why your code isn't working
You are attempting to use a variable, $info, that has not been defined. When you attempt to use an undefined variable, you're effectively concatenating nothing into a string, however because PHP is loosely typed, it declares the variable the second you reference it. That is why you're seeing a notice and not a fatal error. You should go through your code, and ensure that $info gets a value assigned to it, and that it is not overwritten at some point by another function. However, more importantly, read below.
Stop what you are doing
This is vulnerable to a type of attack called an SQL Injection. I'm not going to tell you how to concatenate SQL strings. It's terrible practice.
You should NOT be using mysql functions in PHP. They are deprecated. Instead use the PHP PDO Object, with prepared statements. Here's a rather good tutorial.
Example
After you've read this tutorial, you'll be able to make a PDO Object, so I'll leave that bit for you.
The next stage is to add your query, using the prepare method:
$PDO->prepare("SELECT * FROM tbl WHERE `id` = :id");
// Loads up the SQL statement. Notice the :id bit.
$actualID = "this is an ID";
$PDO->bindParam(':id', $actualID);
// Bind the value to the parameter in the SQL String.
$PDO->execute();
// This will run the SQL Query for you.
You are missing space before "AND " and you should use single quotes as suggested in other answers.
$result=mysql_query("SELECT user_id,username,add FROM users WHERE *username =".$user."AND* password=".$add);
Updated:
echo $sql = "SELECT user_id,username,add FROM users WHERE username ='".$user."' AND password='".$add."'";
$result=mysql_query($sql);
although there is no $info variable used in the query but you need to correct the query:
$result=mysql_query("SELECT user_id,username,add FROM users WHERE username ='" . $user . "' AND password='" . $add . "'");
First from the error its looks like one of your variables is not defined. .. check it. Second surround your parameters with ' for safer syntax.
This is because the variables you are using might not have defined above
So first initialize your variables or if its coming from somewhere else(POST or GET) then check with isset method
So complete code would be
$user = 123; // or $user = isset($user)?$user:123;
$add = 123456; // or $add = isset($add)?$add:123456;
And then run your query
$result=mysql_query("SELECT user_id,username,add FROM users WHERE username =".$user."AND password=".$add);

Sql query problem

I have the below sql query that will update the the values from a form to the database
$sql=
"update leads set
category='$Category',
type='$stype',
contactName='$ContactName',
email='$Email',
phone='$Phone',
altphone='$PhoneAlt', mobile='$Mobile',
fax='$Fax',
address='$Address',
city='$City',
country='$Country',
DateEdited='$today',
printed='$Printed',
remarks='$Remarks'
where id='$id'";
$result=mysql_query($sql) or die(mysql_error());
echo '<h1>Successfully Updated!!.</h1>';
when i submit I dont get any errors and the success message is displayed but the database isnt updated . When i echo the $sql, all the values are set properly. and when i ech the $result i get the value 1.
can someone please tell me what am i doing wrong here??
Have you tried running the echo of $sql directly using some DB tool? It may provide a more informative error. Alternatively, if that works you may have an issue where the transaction isn't being committed. Often a connection is set to automatically commit transactions, but that may not be the case here. Try adding a commit.
And have you ever heard of SQL injection attacks?
If you have a query that is not giving the expected result or receiving an error, and the problem isn't obvious, you should generally take a look at the final query just before it's run. Try using this right before running the query:
echo $sql;
exit;
Viewing the actual query often makes it obvious what the problem is, especially when the query includes variables. If the problem still isn't obvious, you can paste the query as is into a query browser to get feedback directly from the database engine.
Interestingly, using parametrized queries, you won't get to see the parameter values, as the parameters get replaced by MySQL, not PHP, however, you'll still get to see the entire prepared query.
Also, you can see the number of affected rows from your UPDATE statement with the mysql_affected_rows() function. You could put this immediately after the query is run:
echo ("Updated records:", mysql_affected_rows());
Spaces are often forgotten when concatenating queries.
$sql = "SELECT * FROM ducks";
$sql .= "WHERE duck = 'goose'";
When echoing the above query, we see:
SELECT * FROM ducksWHERE duck <> 'goose'
I'm guessing that the WHERE clause in your UPDATE statement isn't matching an "id = '$id'".
Also, is the id column really a string? You've put single quotes around the value. MySQL will cast the string to an integer if needed, but if it's an integer, save the database some work and remove the single quotes.
try to echo $sql and run it directly in any database console, may be there is no record with id = $id
SQL Injection can be the answer. Not an intentional attack (at this moment), but if your parameters have some unexpected information like quotes or other reserved characters you can have strange results. So, try to run this SQL directly in your database administration utility.
Try doing this
"""update leads set
category="$Category",
type="$stype", etc...; """
See if that works

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