What is the syntax to write a PHP MySQL query where a specific column is equal to a variable obtained from a function? I have been trying to retrieve data from my database based on a particular variable, but I am struggling with the correct syntax.
Here is an example of what I am trying to achieve:
```php
function get_data($variable) {
$query = "SELECT * FROM my_table WHERE column_name = '$variable'";
// Rest of the code to execute the query and retrieve data
}
```
In this code, `my_table` is the name of my database table, `column_name` is the column where I want to compare the variable value, and `$variable` is the value obtained from a function.
However, when I execute this code, the query does not return any results. I suspect that there might be an issue in correctly incorporating the variable value into the query.
Could someone please guide me on the correct way to write this query in PHP? Any help would be highly appreciated.

User 3: Greetings! I encountered a similar situation while working on PHP MySQL queries based on variable values. Your code appears to be almost correct, but there's one vital aspect you should address to ensure the query functions properly.
It's worth noting that when incorporating a variable into a query, it's crucial to consider SQL injection vulnerabilities and proper data sanitization. To safeguard your query against such risks, I recommend using prepared statements with parameter binding.
Here's an example of how you can modify your code to implement prepared statements and securely include the variable value:
By employing prepared statements and binding the variable as a parameter, you protect your code from potential SQL injection attacks, which can be detrimental to your application's security. Remember to adjust the connection details ("localhost", "username", "password", "database") according to your setup.
Give this approach a try and see if it resolves your issue while maintaining robust security practices. If you have any further questions, feel free to ask!