How to show sql select parent and child in php - php

Question

Maybe a simple question, but after hours of reading i still didn't find an answer.
I'm trying to get the values of subrow next to those of parent row, without constantly repeating parent row:
For example:
Table 1: ParentId = 1, ParentName = ParentName
Table 2:
ChildId = 1, ParentId = 1, ChildName = ChildName1
ChildId = 2, ParentId = 1, ChildName = ChildName2
The result with JOIN currently is:
ParentId, ParentName, ChildId, ChildName1
ParentId, ParentName, ChildId, ChildName2
But i would like to display the data like:
<div>Div with parent name</div>
<div>Div with ChildName1</div>
<div>Div with ChildName2</div>
<div>Div with parent name2</div>
<div>Div with ChildName1</div>
<div>Div with ChildName2</div>
etc...
Can anyone tell me how to get to this?
Something like
echo$Parent1 foreach child as $child echo$ChildName?

Answer 1

You can sort by parent ID, then check if it has changed. If it has changed output the name of the parent. Something like this:
$query = "query selecting data with join... ORDER BY ParentId";
$result = $mysqli->query($query);
$lastParent = 0;
while ( $row = $result->fetch_assoc($result) )
{
if ( $row['ParentId'] != $lastParent )
{
echo '<div>' . $row['ParentName'] . '</div>';
}
$lastParent = $row['ParentId'];
echo '<div>' . $row['ChildName'] . '</div>';
}
Using that the parent name will only output when the ID changes, and as we're ordering by ID all the children will be grouped by their Parent and output together.

Answer 2

Issue fixed with following code
$result = mysqli_query($link, "SELECT * FROM TestParent a
LEFT JOIN TestChild b ON a.ParentId = b.ParentId
ORDER BY a.ParentId ASC, b.ChildName ASC");
$lastParent = 0;
while($row = mysqli_fetch_assoc($result))
{
if ( $row['ParentId'] != $lastParent ) {
echo '<div>' . $row['ParentName'] . '</div>';
}
$lastParent = $row['ParentId'];
echo '<div>' . $row['ChildName'] . '</div>';
}