I am trying to echo two different parts of HTML code based on the results from the if statement, which are again taken from the database. The number of results is not definite and it will vary based on the user input. It all works well except echoing the HTML. Here is the code:
<?php
$query = mysql_query("SELECT dt, soba1, soba2, soba3, soba4, soba5, soba6, soba7, soba8, soba9, soba10 FROM calendar_table WHERE (dt BETWEEN '$dateFrom' AND '$dateTo')");
while($row = mysql_fetch_assoc($query))
{
$date = $row['dt'];
$soba1 = $row['soba1'];
$soba2 = $row['soba2'];
$soba3 = $row['soba3'];
$soba4 = $row['soba4'];
$soba5 = $row['soba5'];
$soba6 = $row['soba6'];
$soba7 = $row['soba7'];
$soba8 = $row['soba8'];
$soba9 = $row['soba9'];
$soba10 = $row['soba10'];
echo "
<div class='column'>
<p class='dateBox'> $date </p>
".
( ($soba1 != NULL) ?
echo "<p class='status'> <input type='text' name='persons'> </p>"
:
echo "<p class='status'> $soba1 </p>"
."
<p class='status'> $soba2 </p>
<p class='status'> $soba3 </p>
<p class='status'> $soba4 </p>
<p class='status'> $soba5 </p>
<p class='status'> $soba6 </p>
<p class='status'> $soba7 </p>
<p class='status'> $soba8 </p>
<p class='status'> $soba9 </p>
<p class='status'> $soba10 </p>
</div>";
};
?>
Any suggestions on how this can be writen differently, or modified so that it works. THe error that i am getting is Parse error: syntax error, unexpected '=' in C:\xampp\htdocs\plazahotel\cro\confirm.php on line 137