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Security Warning: This answer is not in line with security best practices. Escaping is inadequate to prevent SQL injection, use prepared statements instead. Use the strategy outlined below at your own risk.

You could do something basic like this:

$safe_variable = mysqli_real_escape_string($_POST["user-input"], $dbConnection);
mysqli_query($dbConnection, "INSERT INTO table (column) VALUES ('" . $safe_variable . "')");

This won't solve every problem, but it's a very good stepping stone. I left out obvious items such as checking the variable's existence, format (numbers, letters, etc.).

Security Warning: This answer is not in line with security best practices. Escaping is inadequate to prevent SQL injection, use prepared statements instead. Use the strategy outlined below at your own risk.

You could do something basic like this:

$safe_variable = mysqli_real_escape_string($dbConnection, $_POST["user-input"]);
mysqli_query($dbConnection, "INSERT INTO table (column) VALUES ('" . $safe_variable . "')");

This won't solve every problem, but it's a very good stepping stone. I left out obvious items such as checking the variable's existence, format (numbers, letters, etc.).

You could do something basic like this:

$safe_variable = mysqli_real_escape_string($_POST["user-input"], $dbConnection);
mysqli_query($dbConnection, "INSERT INTO table (column) VALUES ('" . $safe_variable . "')");

This won't solve every problem, but it's a very good stepping stone. I left out obvious items such as checking the variable's existence, format (numbers, letters, etc.).

Security Warning: This answer is not in line with security best practices. Escaping is inadequate to prevent SQL injection, use prepared statements instead. Use the strategy outlined below at your own risk.

You could do something basic like this:

$safe_variable = mysqli_real_escape_string($_POST["user-input"], $dbConnection);
mysqli_query($dbConnection, "INSERT INTO table (column) VALUES ('" . $safe_variable . "')");

This won't solve every problem, but it's a very good stepping stone. I left out obvious items such as checking the variable's existence, format (numbers, letters, etc.).

Deprecated Warning: This answer's sample code (like the question's sample code) uses PHP's MySQL extension, which was deprecated in PHP 5.5.0 and removed entirely in PHP 7.0.0.

Security Warning: This answer is not in line with security best practices. Escaping is inadequate to prevent SQL injection, use prepared statements instead. Use the strategy outlined below at your own risk. (Also, mysql_real_escape_string() was removed in PHP 7.)

You could do something basic like this:

$safe_variable = mysqli_real_escape_string($_POST["user-input"], $dbConnection);
mysqli_query("INSERT INTO table (column) VALUES ('" . $safe_variable . "')");

This won't solve every problem, but it's a very good stepping stone. I left out obvious items such as checking the variable's existence, format (numbers, letters, etc.).

You could do something basic like this:

$safe_variable = mysqli_real_escape_string($_POST["user-input"], $dbConnection);
mysqli_query($dbConnection, "INSERT INTO table (column) VALUES ('" . $safe_variable . "')");

This won't solve every problem, but it's a very good stepping stone. I left out obvious items such as checking the variable's existence, format (numbers, letters, etc.).