Prepared statement issue

[phppup 0]

After deciding to venture into the realm of prepared statements, I have this line in my script

Quote

if ($stmt = $conn->prepare('SELECT username FROM users WHERE username = ?')) {

Everything was working fine. 

I reviewed my code to adjust it to my old habits, and realized that I had hardcoded the TABLE NAME rather than using a variable.

I updated my code to

Quote

$table = "users";

if ($stmt = $conn->prepare('SELECT username FROM $table WHERE username = ?')) {

and results from my SELECT statement vanished.

Is the use of a variable for a table's name outdated? Even possible??

[Barand 1,650]

Your query string is in single quotes therefore $table is not interpreted as "users".

Why are you putting the table name in a variable? It's often a sympton of a poorly designed database if it is necessary.

[phppup 0]

As mentioned, it's a hold-over of an old habit (although my database will probably be restructured next. LOL)

Still, the double quotes were the only choice that didn't cause an error message.

How do I get the variable in there? Or is it even worthwhile?

[Barand 1,650]

https://www.php.net/manual/en/language.types.string.php