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1Possible duplicate of [PHP Parse/Syntax Errors; and How to solve them?](http://stackoverflow.com/questions/18050071/php-parse-syntax-errors-and-how-to-solve-them) – Charlotte Dunois May 16 '16 at 19:55
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Look at the line before it. Is that how you end an echo statement? It's not how anyone else ends an echo statement. – kainaw May 16 '16 at 19:55
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1Welcome to StackOverflow! Please read [How do I ask a good question?](http://stackoverflow.com/help/how-to-ask) – Pedro Lobito May 16 '16 at 19:58
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1Link only questions (and answers) are likely to be down voted. Please post code directly (with proper formatting). – BM5k May 16 '16 at 20:00
1 Answers
0
The referenced code contains a number of issues (an undefined variable $no
on line 218, and what looks like a mysql query being run within an echo
to name a few...), however, without elaborating on the particulars of each issue, I would suggest the following to simplify your code and make it easier to debug: First build your output within a variable and then echo it to the browser after executing both queries, like so (edit begins on line 203 of your code):
$output = '';
while($row=mysql_fetch_array($guest))
{
$output.= '<tr>';
$output.= '<td align=\"center\">'.$row["ID_guest"].'</td>';
$output.= '<td align=\"center\">'.$row["First_name"].' '.$row["Last_name"].'</td>';
$output.= '<td align=\"center\">'.$row["Check_In_Date"].'</td>';
$output.= '<td>'.$row["Check_Out_Date"].'</td>';
$output.= '<td>'.$row["Request_Date"].'</td>';
$output.= '<td align=\"center\">'.$row["No_Of_Room"].'</td>';
$output.= '<td align=\"center\">'.$row["Category_Name"].'</td>';
$output.= '<td align=\"center\"><select>';
$roomsql = "SELECT...";
$roomresult = mysql_query($roomsql);
while($rooms=mysql_fetch_array($roomresult))
{
$output.= '<option>'.$rooms['Room_No'].'</option>';
}
$output.= '</select></td>';
$output.= '</tr>';
}
echo($output);

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