![]() The important thing here is that the parameter values are combined with the compiled statement, not an SQL string. Then when you call execute, the prepared statement is combined with the parameter values you specify. By specifying parameters (either a ? or a named parameter like :name in the example above) you tell the database engine where you want to filter on. The SQL statement you pass to prepare is parsed and compiled by the database server. This makes sure the statement and the values aren't parsed by PHP before sending it to the MySQL server (giving a possible attacker no chance to inject malicious SQL).Īlthough you can set the charset in the options of the constructor, it's important to note that 'older' versions of PHP (before 5.3.6) silently ignored the charset parameter in the DSN. What is mandatory, however, is the first setAttribute() line, which tells PDO to disable emulated prepared statements and use real prepared statements. And it gives the developer the chance to catch any error(s) which are thrown as PDOExceptions. ![]() This way the script will not stop with a Fatal Error when something goes wrong. In the above example the error mode isn't strictly necessary, but it is advised to add it. $dbConnection->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION) $dbConnection->setAttribute(PDO::ATTR_EMULATE_PREPARES, false) An example of creating a connection using PDO is: $dbConnection = new PDO('mysql:dbname=dbtest host=127.0.0.1 charset=utf8', 'user', 'password') To fix this you have to disable the emulation of prepared statements. Note that when using PDO to access a MySQL database real prepared statements are not used by default. If you're connecting to a database other than MySQL, there is a driver-specific second option that you can refer to (for example, pg_prepare() and pg_execute() for PostgreSQL). $stmt->bind_param('s', $name) // 's' specifies the variable type => 'string' Using MySQLi (for MySQL): $stmt = $dbConnection->prepare('SELECT * FROM employees WHERE name = ?') Using PDO (for any supported database driver): $stmt = $pdo->prepare('SELECT * FROM employees WHERE name = :name') You basically have two options to achieve this: This way it is impossible for an attacker to inject malicious SQL. These are SQL statements that are sent to and parsed by the database server separately from any parameters. It is possible to create SQL statement with correctly formatted data parts, but if you don't fully understand the details, you should always use prepared statements and parameterized queries. The correct way to avoid SQL injection attacks, no matter which database you use, is to separate the data from SQL, so that data stays data and will never be interpreted as commands by the SQL parser. This will make your logs considerably smaller. Rather than converting it back to a PHP array and then back to JSON, it would be better to add the JSON_PRETTY_PRINT flag to wherever you’re getting $json_string from in the first place, if possible.Īlternatively, just log $json_string directly (no need to encode it: it’s already a string, you can pass it to error_log() as it is), and worry about prettifying it only when you need to read your logs. $json_pretty_string = json_encode($log_array, JSON_PRETTY_PRINT) Įrror_log('test'. ![]() $log_array = json_decode($json_string, true) So what you need to do is to (a) turn $json_string back into a PHP array, and then (b) reencode that as JSON, this time using the JSON_PRETTY_PRINT flag. JSON_PRETTY_PRINT is having no effect here, because you’re not producing JSON: you’re producing something that javascript too would see as a string. ![]() You pass that through json_encode() and all you’re going to get is another string (a string of doubly-encoded JSON). To be clear, $json_string is a string: (as far as PHP is concerned, it’s a string if you passed that same string to javascript, it would be interpreted as an object). And that’s what you’re doing, so that’s what you’re getting. If you give it an array or an object, it will produce JSON if you give it a string, it will produce a javascript string. Json_encode() will actually not necessarily produce JSON: it will produce something that can be read by javascript. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |