PHP if statement in html-table outputs error - php

Question

I am new to HTML5 and PHP, I am trying to output a specific value in table data, If the database-retrieved-value is per condition.
My code:
<table class="scroll">
<thead style="background-color: #99E1D9; color: #705D56;">
<tr>
<th>ID</th>
<th>Name Client</th>
<th>Last Update</th>
<th style="padding-left: 30%;">Status</th>
</tr>
</thead>
<tbody id="hoverTable">
<?php
$connection = mysql_connect('localhost', 'root', '');
mysql_select_db('patientdb');
$query = "SELECT id, name, date FROM clients";
$result = mysql_query($query);
//status waarden uit
$status = "SELECT status FROM clients";
$status_ = mysql_query($status);
while($row = mysql_fetch_array($result)){ //Loop through results
echo "<tr>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['date'] . "</td>
<td style='padding-left: 30%;'>" .
if ($status_ > 60){echo "red";
} elseif ($status_ > 50){echo "yellow";
} else{echo "green";}
. "</td>
</tr>";
}
mysql_close();
?>
</tbody>
</table>
Error output
Parse error: syntax error, unexpected T_IF in
/test/composition/login/portal/portal.php
on line 204
What is the right way to solve this?
EDIT
my current code:
<table class="scroll">
<thead style="background-color: #99E1D9; color: #705D56;">
<tr>
<th>Naam Client</th>
<th>Laatste Update</th>
<th style="margin-left: 40%; padding-left: 0%;">Status</th>
</tr>
</thead>
<tbody id="hoverTable" style="font-size: 11pt;">
<?php
$connection = mysql_connect('localhost', 'root', '');
mysql_select_db('patientdb');
$query = "SELECT id, naam, datum FROM clients";
$result = mysql_query($query);
$query2 = "SELECT status FROM clients";
$result2 = mysql_query($query2);
if (!empty ($result2)) {
while ($row2 = mysql_fetch_assoc($result2)) {
echo $row2['status'] . "<br />";
}
}
while($row = mysql_fetch_array($result)){ //Loop through results
echo "<tr>
<td>" . $row['id'] . "</td>
<td>" . $row['naam'] . "</td>
<td>" . $row['datum'] . "</td>
<td style='padding-left: 30%;'>";
if ($results2 > 60 && $results2 < 70) {
echo "red";
} elseif ($results2 > 50 && $results2 < 60) {
echo "yellow";
} else {
echo "green";
}
echo "</td>
</tr>";
}
mysql_close();
?>
</tbody>
</table>
Output the right data. but partly outside and partly inside the table.

Answer 1

You will have to remove the if statement out of the echo to get rid of the error Try this:
<table class="scroll">
<thead style="background-color: #99E1D9; color: #705D56;">
<tr>
<th>ID</th>
<th>Name Client</th>
<th>Last Update</th>
<th style="padding-left: 30%;">Status</th>
</tr>
</thead>
<tbody id="hoverTable">
<?php
$connection = mysql_connect('localhost', 'root', '');
mysql_select_db('patientdb');
$query = "SELECT id, name, date FROM clients";
$result = mysql_query($query);
//status waarden uit
$status = "SELECT status FROM clients";
$status_ = mysql_query($status);
while($row = mysql_fetch_array($result)){ //Loop through results
echo "<tr>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['date'] . "</td>
<td style='padding-left: 30%;'>";
if ($status_ > 60) {
echo "red";
} elseif ($status_ > 50) {
echo "yellow";
} else {
echo "green";
}
echo "</td>
</tr>";
}
mysql_close();
?>
</tbody>
</table>

Answer 2

You can't have an if statement (or any other statement, for that matter) in the middle of another statement like echo. If you want to concatenate different strings depending on a variable, you can use the conditional (AKA "ternary") operator.
echo "<tr>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['date'] . "</td>
<td style='padding-left: 30%;'>" .
$status_ > 60 ? "red" : ($status_ > 50 ? "yellow" : "green" )
. "</td>
</tr>";

Answer 3

Try:
$status = "green";
if ($status > 50)
{
$status="yellow";
}
elseif($status>60)
{
$status="red";
}
echo "<tr>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['date'] . "</td>
<td style='padding-left: 30%;'>" .$status. "</td>
</tr>";
You can't append to a string a conditional statement, assign to a variable first for example (like I posted)

Answer 4

This part isn't at the right place:
if ($status_ > 60){echo "red";
} elseif ($status_ > 50){echo "yellow";
} else{echo "green";}
should be:
echo "<tr>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['date'] . "</td>
<td style='padding-left: 30%;'>";
if ($status_ > 60){
echo "red";
} elseif ($status_ > 50){
echo "yellow";
} else{
echo "green";
}
echo "</td></tr>";

Answer 5

Surely status_ would not come back with a number, but an array.
$status_ = mysql_query($status);
Without knowing what data is coming back, it is difficult to help.
mysql_query