Defining a variable inside isset function

Question

I'm trying to save lines by trying to define a variable in isset() . Currently the only way I could make this work is this:

$setpass_accinfo = $dbconn->query("SELECT id,name,password FROM users WHERE id=" . $dbconn->real_escape_string(_GET('id')));
$setpass_accinfo_row = isset($setpass_accinfo->num_rows) ? $setpass_accinfo->fetch_assoc() : false;

So, I'm trying to combine those two lines into one, if you know what I mean. Something like this for example:

$setpass_accinfo = isset(($setpass_accinfo = $dbconn->query("SELECT id,name,password FROM users WHERE id=" . $dbconn->real_escape_string(_GET('id'))))->num_rows) ? $setpass_accinfo->fetch_assoc() : false;

The issue is that it gives me an error about using $setpass_accinfo and I don't know how to fix it or if it's even possible. Any thoughts?

Answer 1

This can get a little messy at times but:

$setpass_accinfo_row = !empty($setpass_accinfo = $dbconn->query("SELECT id,name,password FROM users WHERE id=" . $dbconn->real_escape_string(_GET('id')))) ? $setpass_accinfo->fetch_assoc() : false;

Connot use isset() on the result of an expression
When you use single equal in an if statement, the $variable is assigned the value of the right side of the equal, then the variable is evaluated in the empty() , in this case. And precedence dictates the it is defined before the ternary operation is reached.

Answer 2

You can use this:

$setpass_accinfo_row = $dbconn->query("SELECT ...")->fetch_assoc() ?: false;

• $dbconn->query() returns mysqli result
• $dbconn->query()->fetch_assoc() returns:
  • An associative array that corresponds to the fetched row
  • NULL if no row was found
• The ?: is the shorter form of ternary operator which, in this case, returns the array or false

Note that it does not check if the query executed successfully.