Calling another function within a function - positioning the output - php

Question

I have the following function:
public function createRatingButtons(){
for($count=1;$count<=5;$count++){
echo "<button onclick=''><b>$count</b></button> ";
}
}
The above function simply creates 5 separate buttons, labelled 1, 2, 3, 4, and 5!
I am calling the above function below:
public function displayGrades($student,$unit){
$studentID = $this->getStudentID($student);
$unitID = $this->getUnitID($unit);
$obj = new dbConfig();
$obj->dbConnect();
$sql="SELECT * FROM grading where studentID = '$studentID' && unitID = '$unitID'";
$result = mysqli_query($obj->conn,$sql) or die(mysqli_error($sql));
$row = mysqli_fetch_assoc($result);
echo "
<table>
<tr>
<td>Independence</td>
</tr>
<tr>
<td>Current Rating: " . $row['independence'] . "</td>
</tr>
<tr>
<td>" . $this->createRatingButtons() . "</td>
</tr>
<tr>
<td>Participation</td>
</tr>
.
.
.
PROBLEM: It displays the buttons, but they do not appear in that column where I have called the function. The buttons appear above the table. I want them to appear in the same column where I am calling the function.
How do I fix this problem? Please can anyone help. Thank you in advance

Answer 1

Your are echoing inside an echo, and this is not possible.
Maybe you should try to build a string of buttons, by doing this :
public function createRatingButtons(){
$result = "";
for($count=1;$count<=5;$count++){
$result.="<button onclick=''><b>$count</b></button> ";
}
return $result;
}