Mysqli检索数据并显示它们-最好的方法

Question

Which one of the options below is better

//control file:  
$smpt = $con->query("SELECT id,name FROM customers");
//fetch all resuts 
$results = array();
while ($row = $result->fetch_assoc()) {
  $results[] = $row;
}    
$smpt->close();
//Note: PHP version < 5.3.0 so I can't use just $results->fetch_all('MYSQLI_ASSOC');

---

//view file:
foreach($results as $key=>$value) {
//display results
}
unset($results);

---

//control file:  
$smpt = $con->query("SELECT id,name FROM customers");
//fetch all resluts 

---

//view file:
while($row = $result->fetch_assoc()) {
//display results
}
$smpt->close();

---

I am trying to completely separate the logic from presentation... Current I am using option 2 because with option 1 the script go through 2 loops. one to fetch data and the other to display them. But which one is better to use?

Thanks

Answer 1

It's always best that views only deal with presentation concerns.

So between the two options you've mentioned, I'd choose option 1 so you don't have to close $smpt inside the view ($smpt has nothing to do with the presentation layer).

Answer 2

Option 1 allows you to re-use the $data variable so you can display the results twice, but the cost of this is that you potentially have a large amount of data stored in a variable. You can clear this by using unset($data) once you are 100% sure you've finished with it.

Option 2 requires less loops (only one instead of two) and doesn't need a extra variable to store the data. The cost of this is that the while loop can only be used once.

In the grand scheme of things, the differences in speed will be so minimal the user won't notice them, however if there are 20+ instances of this in once script then it could have a noticeable affect.

I would recommend Option 2 providing that you'd only need to use the while loop once. Also, you could benchmark your scripts to see how they perform against one another.