foreach warning in PHP - php

Question

admin TABLE
+--------------+---------------+
| Username | Password |
+--------------+---------------+
| JOHN | 123 |
| EDWARD | 123 |
+--------------+---------------+
my code
$result = mysqli_query($connection, $query);
echo" <table >";
$row = mysqli_fetch_assoc($result);
echo "<tr>";
foreach($row as $key => $val){
echo"<th>$key</th> ";
echo "</tr>";
///////////////////////////////
$result = mysqli_query($connection, $query);
echo"<tr>";
while($row = mysqli_fetch_assoc($result)){
foreach($row as $key => $val){
echo "<td>$val</td>";
}
echo "</tr>";
}
echo "</table>";
}
So, the problem is if there is more than 1 row in the result it outputs exactly what i want for e.g if i say:
SELECT Username from admin;
+--------------+
| Username |
+--------------+
| JOHN |
| EDWARD |
+--------------+
But if there is only one row it does not show the COLUMN name and gives this warning
Select Username from admin where Username = 'JOHN';
Warning: Invalid argument supplied for foreach() on line 65
+--------------+
| JOHN |
+--------------+

Answer 1

Sneaky typo/logical coding error: you have a curly-bracket (brace) that's screwing everything up. Look at the end of the foreach used to generate the headings:
foreach($row as $key => $val){
echo"<th>$key</th> ";
This should be:
foreach($row as $key => $val)
echo"<th>$key</th> ";
And then remove the spurious brace at the very end.
This works (so long as there is at least 1 row to get the heaedrs):
$result = mysqli_query($connection, $query);
echo " <table >";
$row = mysqli_fetch_assoc($result);
echo "<tr>";
foreach($row as $key => $val)
echo"<th>$key</th> ";
echo "</tr>";
$result = mysqli_query($connection, $query);
echo"<tr>";
while($row = mysqli_fetch_assoc($result))
{
foreach($row as $key => $val)
{
echo "<td>$val</td>";
}
echo "</tr>";
}
echo "</table>";