PHP To Show 2 Tables On One Page - php

Question

I am using this syntax to display a PHP Table on my page. I now need to add in a second table directly above this one, but all the syntax I try throws a 500 error. How can I with 1 connection to MSSQL run 2 Select statements and populate 2 individual html tables?
$option = array();
$option['driver'] = 'mssql';
$option['host'] = 'IP Address';
$option['user'] = 'UserName';
$option['password'] = 'Password';
$option['database'] = 'DB';
$option['prefix'] = '';
$db = JDatabase::getInstance($option);
$query = $db->getQuery(true);
$query = "SELECT name, hiredate, bday, payrate, hourlypay from HRData ORDER BY name ASC";
$db->setQuery($query);
$query = $db->loadObjectList();
if ($query)
{
?>
<table border="1">
<thead>
<tr>
<th>Name </th>
<th>Hire Date </th>
<th>Birthday </th>
<th>Pay Rate </th>
<th>hourlypay </th>
</tr>
</thead>
<?php
foreach ($query as $res)
{
print "<tr>";
print "<td>" . $res->name . "</td>";
print "<td>" . $res->hiredate . "</td>";
print "<td>" . $res->bday . "</td>";
print "<td>" . $res->payrate . "</td>";
print "<td>" . $res->hourlypay . "</td>";
print "</tr>";
}
}
EDIT
This is the syntax I am trying to adapt, but I keep getting a 500 Error
$option = array();
$option['driver'] = 'mssql';
$option['host'] = 'IP Address';
$option['user'] = 'UserName';
$option['password'] = 'Password';
$option['database'] = 'DB';
$option['prefix'] = '';
$db = JDatabase::getInstance($option);
$query = $db->getQuery(true);
$query = "SELECT name, MAX(Pay) As PayYTD FROM HRINFO";
$db->setQuery($query);
$query = $db->loadObjectList();
if ($query)
{
?>
<table border="1">
<thead>
<tr>
<th>Name </th>
<th>YTD Pay </th>
</tr>
</thead>
<?php
foreach ($query as $res)
{
print "<tr>";
print "<td>" . $res->name . "</td>";
print "<td>" . "$" . round($res->PayYTD) . "</td>";
print "</tr>";
}
}
<br><br><br>
//Query
$query = $db->getQuery(true);
$query = "SELECT name, hiredate, bday, payrate, hourlypay from HRData ORDER BY name ASC";
$db->setQuery($query);
$query = $db->loadObjectList();
if ($query)
{
?>
<table border="1">
<thead>
<tr>
<th>Name </th>
<th>Hire Date </th>
<th>Birthday </th>
<th>Pay Rate </th>
<th>hourlypay </th>
</tr>
</thead>
<?php
foreach ($query as $res)
{
print "<tr>";
print "<td>" . $res->name . "</td>";
print "<td>" . $res->hiredate . "</td>";
print "<td>" . $res->bday . "</td>";
print "<td>" . $res->payrate . "</td>";
print "<td>" . $res->hourlypay . "</td>";
print "</tr>";
}
}

Answer 1

The problem you're having is that you are using the calls incorrectly.
$query = $db->getQuery(true);
$query = "SELECT name, MAX(Pay) As PayYTD FROM HRINFO";
$db->setQuery($query);
The first line will create an object, it doesn't matter which. The object will be in $query.
The second line will immediately destroy the object and assign a string to $query (this is incorrect).
The third line expects an object as a parameter to setQuery, but unfortunately it is a string! Error.
If you want this to work correctly, then you need to use the object in $query correctly.
I'm not a Joomla expert, so I link you to a page for how to do this correctly: https://docs.joomla.org/Selecting_data_using_JDatabase

Answer 2

Put the output in a variable.
$StringOut = '';
$StringOut .= '<table border="1">
<thead>
<tr>
<th>Name </th>
<th>Hire Date </th>
<th>Birthday </th>
<th>Pay Rate </th>
<th>hourlypay </th>
</tr>
</thead>';
foreach ($query as $res)
{
$StringOut .= "<tr>";
$StringOut .= "<td>" . $res->name . "</td>";
$StringOut .= "<td>" . $res->hiredate . "</td>";
$StringOut .= "<td>" . $res->bday . "</td>";
$StringOut .= "<td>" . $res->payrate . "</td>";
$StringOut .= "<td>" . $res->hourlypay . "</td>";
$StringOut .= "</tr>";
}
$StringOut .= '</table>';
#Do other logic to retrieve first table.
#You could put it in a different variable if you like. And print when and whereever you wish.
echo $StringOut;
#Optionally close connection, done.