Jump to content

Prepared statement issue


phppup
 Share

Recommended Posts

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Is the use of a variable for a table's name outdated?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

Link to comment
Share on other sites

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?

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.