Mysql Connection Error Reporting Issue

Folks,

I see tutorials teach something like this:

[code]
mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT);
if(!$conn = mysqli_connect(“localhost”,”root”,””,”buzzoff”))
{
die(‘Something went wrong. Please try again later!’);
}
$stmt = mysqli_stmt_init($conn);
$sql_count = “SELECT COUNT(id) FROM domains WHERE domain = ? OR domain_email = ?”;

if(!mysqli_stmt_prepare($stmt,$sql_count))
{
$_SESSION[‘error_report’] = __LINE__ .”n”;
$_SESSION[‘error_report’] = ‘Mysqli Error: ‘ .mysqli_error($conn) .”n”;
$_SESSION[‘error_report’] = ‘Mysqli Error Number: ‘ .mysqli_errno($conn) .”n”;
[/code]

Note the:
1st Code

[code]
mysqli_error() //empty inside brackets.
mysqli_errno() //empty inside brackets.
[/code]

But should not it be something like this instead ?
2nd Code

[code]
mysqli_error($conn) //NOT empty inside brackets.
mysqli_errno($conn) //NOT empty inside brackets.
[/code]

Here is the context …
3rd Code

[code]
mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT);
if(!$conn = mysqli_connect(“localhost”,”root”,””,”buzzoff”))
{
die(‘Something went wrong. Please try again later!’);
}
$stmt = mysqli_stmt_init($conn);
$sql_count = “SELECT COUNT(id) FROM domains WHERE domain = ? OR domain_email = ?”;

if(!mysqli_stmt_prepare($stmt,$sql_count))
{
$_SESSION[‘error_report’] = __LINE__ .”n”;
$_SESSION[‘error_report’] = ‘Mysqli Error: ‘ .mysqli_error($conn) .”n”;
$_SESSION[‘error_report’] = ‘Mysqli Error Number: ‘ .mysqli_errno($conn) .”n”;
[/code]

Which code should I stick to ?
The 3rd one showed error the other night but tonight no errors.
Strange.
Hence, enquiring.

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