Building HTML table by appending to PHP string - php

Question

First let me start by saying this is a homework problem so the way I have to build this table is because of the requirements for the assignment. The requirements call for three php functions to build an html table. I have a table, td, and tr function. I think I am very close to figuring out hwo to get this to work. Just running into one small problem and was wondering if someone could help me out or explain why I am running into this problem. Here is my code:
<?php
//need a constant to keep track of the number of rows
define('NUM_ROWS', '3');
//function to start building table
function table()
{
$myTable = '<table border="1">' . tr($myTable) . '</table>';
return $myTable;
}
//function to build table rows
function tr( $myTable )
{
$myTable .= '<tr>' . td($myTable) . '</tr>';
return $myTable;
}
//function to build table data
function td($myTable)
{
$rowData = array("Planes", "Trains", "Automobiles");
for($i = 0, $length = count($rowData); $i < $length; $i++)
{
$myTable .= '<td>' . $rowData[$i] . '</td>';
}
return $myTable;
}
echo table();
?>
The problem I am having is in the table function. If I write the function the way it is shown above it will display the table but it also gives me an undefined variable error on $myTable when I call the tr function. If I change that function to look like the function below it doesn't display my table at all and just shows a blank page.
function table()
{
$myTable = '<table border="1">';
tr($myTable);
$myTable .= '</table>';
return $myTable;
}
Any ideas on what I am doing wrong or can do differently?

Answer 1

You are not actually using the passed variable nicely anyways so just don't try to pass the variable (which you were passing before it existed and hence was complaining about being undefined); just do this:
<?php
//need a constant to keep track of the number of rows
define('NUM_ROWS', '3');
//function to start building table
function table()
{
$myTable = '<table border="1">' . tr() . '</table>';
return $myTable;
}
//function to build table rows
function tr()
{
$tr = '<tr>' . td() . '</tr>';
return $tr;
}
//function to build table data
function td()
{
$td = '';
$rowData = array("Planes", "Trains", "Automobiles");
for($i = 0, $length = count($rowData); $i < $length; $i++)
{
$td .= '<td>' . $rowData[$i] . '</td>';
}
return $td;
}
echo table();
?>

Answer 2

you need to pass
you aren't passing any parameter to function table
function table($myTable)
{
$myTable = '<table border="1">' . tr($myTable) . '</table>';
return $myTable;
}

Answer 3

Try this.
//need a constant to keep track of the number of rows
define('NUM_ROWS', '3');
//function to start building table
function table($val)
{
$myTable="";
$myTable = '<table border="1">' . tr($val) . '</table>';
return $myTable;
}
//function to build table rows
function tr( $val )
{
$myTable="";
$myTable .= '<tr>' . td($val) . '</tr>';
return $myTable;
}
//function to build table data
function td($val)
{
$myTable="";
for($i = 0, $length = NUM_ROWS; $i < $length; $i++)
{
$myTable .= '<td>' . $val[$i] . '</td>';
}
return $myTable;
}
$rowData = array("Planes", "Trains", "Automobiles");
echo table($rowData);