PDO queryString with binded data - php

If I have a prepared statement like SELECT * FROM users WHERE userid = :userid, i can read this SQL statement via PDOStatement::$queryString. For logging i want to have the string, which is executed, e.g. ... WHERE userid = 42. How do i get this string?

PDOStatement->debugDumpParams is what you want. You may need to use output buffering though as the results are echoed out.

Related

PHP PDO mysql prepared statment and join

I have a question.
I have the following query:
$query = "select * from module,bloc where module.id_bloc = ?";
I tried to bind the value so I did:
$stmt = $this->db->prepare($query);
$stmt->bindValue(1, "bloc.id_bloc");
But, when I test I don't get any result on my browser.
It's weird because when I replace directly inside like the following code:
$query = "select * from module,bloc where module.id_bloc = bloc.id_bloc";
I get the the right result on my browser.
Could someone explain to me why it doesn't work when I am doing a bindValue?
It will not work because, when bound, a string will be quoted. (Or, for all intents and purposes, work as if it were quoted, however PDO may handle it behind the scenes.) Then, your query is interpreted as:
select * from module,bloc where module.id_bloc = 'bloc.id_bloc'
That is: It will be interpreted as a literal string, rather than a reference to a table column, and will obviously not give you the expected result. There is no need for binding it to begin with.
If, for some reason, you need to run a query with a variable table/column name from an unsafe source, you will have to manually format/sanitize it; see here for an example of how to do it.

Strange Error in mysql_query() function in PHP (And clause in Query)

When I add AND operator in mysql_query() function, it stops working and anything after that stops working!
For Example:
When i wrote this:
$query1 = mysql_query("SELECT * FROM chat1 where friendname = '$_POST[fname]' ");
$row= mysql_fetch_array($query1) or die(mysql_error());
echo "$row[message]";
The above query runs successfully !
But when i do this :
$query1 = mysql_query("SELECT * FROM chat1 where friendname = '$_POST[fname]' AND username = '$_POST[uname]' ");
$row= mysql_fetch_array($query1) or die(mysql_error());
echo "$row[message]";
I get Null output!
I think the "AND" operator is not working!!!
please help me with this!!
Have a look at my complete code and Database Snapshot!
Click here
If it is returning NULL then probably the record doesn't exists. Try to output this query on the screen and post the raw query here.
Maybe your search needs a LIKE instead of a =
Likely, the row(s) you are looking for do not exist.
The AND is a boolean operator that requires that both expressions have to evaluate to true. In the context of your query, that means for a row to be returned, both of the conditions have to be true on that single row.
I suspect that you may want an OR those two conditions. Did you want to return only rows that meet both criteria, or did you want any rows that have fname with a certain value, along with any rows that have uname of a specific value? If the first query is returning rows, then replacing AND with OR should return you some rows.
For debugging this type of problem, generate the SQL text into a variable, and then echo or var_dump the SQL text, before you send it to the database.
e.g.
$sql = "SELECT * FROM chat1 where friendname = '"
. mysql_real_escape_string($_POST['fname'])
."' ";
echo "SQL=" . $sql ; # for debugging
Take the text of SQL statement that's emitted to another client, to test the SQL statement, to figure out if the SQL statement is actually returning the resultset you expect it to return.
(In your code, reference the $sql in the function that prepares/executes the SQL statement.)
Follow this pattern for all dynamically generated SQL text: generate the SQL text into a variable. For debugging, echo or var_dump or otherwise emit or log the contents of the variable. Take the SQL text to another client and test it.
Dumping code that isn't working on to StackOverflow is not the most efficient way to debug your program. Narrow down where the problem is.
How to debug small programs http://ericlippert.com/2014/03/05/how-to-debug-small-programs/
NOTES
You probably want to verify that $_POST['fname']) contains a value.
It's valid (SQL-wise) for a SELECT statement to return zero rows, if there are no rows that satisfy the predicates.
Potentially unsafe values must be properly escaped if you include them in the text of a SQL statement. (A better pattern is to use prepared statements with bind placeholders, available in the (supported) mysqli and PDO interfaces.
Also, use single quotes around fname.... e.g.
$_POST['fname']
^ ^

how to select exact json value in mysql query condition

I am trying to select exact json value in mysql query condition, currently i am using LIKE and it work but like returns false positive.
my code is::
$id='1';
json data example:: ["1","12","38"]
$sql="SELECT * FROM `posts` WHERE `json_column` LIKE '%".$id."%' ";
with this query, it returns both 1 and 12. how can i get just 1?
Note the extra quotes:
$sql="SELECT * FROM posts WHERE json_column LIKE '%\"$id\"%'";
However, this is wide open to SQL inject if $id comes from user input. Be careful.
The more secure method would be to use parameterised queries with PHPs prepared statements as follows:
$stmt = $dbh->prepare("SELECT * FROM posts WHERE json_column LIKE '%\"?\"%'");
$stmt->bindParam(1, $id);

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);
}

Should I also sanitize data coming from $_GET if i use those without changing the state of the DB?

Should i also sanitize data coming from $_GET if those data are not going to change the state of database.
example of changing the state of database is INSERT, DELETE
example of not changing the state is SELECT.
the problem i think about not sanitizing is this
inside code
$word = $_GET['word'];
$query = "SELECT * FROM tbl_example WHERE name LIKE '%$word%'";
mysqli_query($query);
then the data you search is
testword%';DROP TABLE tbl_example; --
so final value of $query is
SELECT * FROM tbl_example WHERE name LIKE '%testword%';DROP TABLE tbl_example; -- %'
BUT i guess this wont work cause mysqli_query and mysql_query() can only execute single SQL statement.
Also if you sanitize $_GET then inside the database there is ' then you search ' will your search match the one in the DB?
Can you also show me examples of sql injection?
Let's say your table has 1 million rows and you dont sanitize $_GET as in example above then someone can exploit your search so that it returns all 1 million rows, and then your server will try to display all of those and it will go down. Thats just one example.
Example
$_GET["word"]="test%' or name LIKE '";
$word = $_GET['word'];
$query = "SELECT * FROM tbl_example WHERE name LIKE '%$word%'";
Query
SELECT * FROM tbl_example WHERE name LIKE '%test%' or name LIKE '%'
what you need to use is prepared queries instead of sanitize data coming from $_GET
you are using mysqli like mysql and even not escaping GET ... you need to use prepared queries
you need to use like
/* create a prepared statement */
if ($stmt = $mysqli->prepare("SELECT District FROM City WHERE Name=?")) {
/* bind parameters for markers */
$stmt->bind_param("s", $city);
/* execute query */
$stmt->execute();
/* bind result variables */
$stmt->bind_result($district);
/* fetch value */
$stmt->fetch();
printf("%s is in district %s\n", $city, $district);
/* close statement */
$stmt->close();
}
from php manual
good read
Sanitizing user's data in GET by PHP
It sounds like you're trying to find excuses to not sanitize data. It's sort of like saying "I don't need to put on my seat belt if I'm only driving a half mile, do I?" Why so hesitant from consistently using best practices?
Moreover, don't bother santizing your data. Instead, use parametrized queries like the other answers tell you to.
Others have given you examples that even a SELECT is susceptible to malicious input, but you seem to be wanting to even get around those reasons. Don't fight it. Get into good habits and do it right.

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