Mysqli Prepare Statement - Returning False, but Why?

Question

I have a function that generates a prepared INSERT statement based on an associative array of column names and values to be inserted into that column and a table name (a sim

Answer 1

I'm copying the solution into this answer so this can be given an upvote, otherwise the question will appear in the "unanswered questions" forever. I'm marking this answer CW so I won't get any points.

@Andrew E. says:

I just turned on mysqli_report(MYSQLI_REPORT_ALL) to get a better understanding of what was going on - turns out that one of my field names was incorrect - you'd think that prepare() would throw an exception, but it fails silently.