如何在同一PHP代码中执行两个查询？

Question

I'm new to PHP programming. I want to execute 2 queries at the same PHP code. The first query gave me expected results. But the second query seems to be not working. But it does not show me any error message. Can someone help me?

 <?php
                  mysql_connect('localhost','root','');
                  mysql_select_db('health');
                  $query1="SELECT * FROM messages,doctor WHERE doctor.regNo=messages.regNo AND messages.ID='".$id."';"; 
                  $result1=mysql_query($query1) or die(mysql_error());
                  $row=mysql_fetch_array($result1);
                  echo '<table cellspacing="10">';
                  echo '<tr><td><b>From: </b></td><td>Dr. '.$row['firstName'].' '.$row['lastName'].'</td></tr>';
                  echo '<tr><td><b>Category: </b></td><td>'.$row['category'].'</td></tr>';
                  echo '</table>';
                  echo '<p style="padding-left:13px;">'.$row['reply'].'</p>';
                  echo '</br><b style="padding-left:13px;">Your question</b></br>';
                  echo '<p style="padding-left:13px;">'.$row['question'].'</p>';

                  $query2="UPDATE health.messages SET read='Y' WHERE messages.ID='".$id."';"; 
                  $result2=mysql_query($query2);
                  mysql_close();
             ?>

Answer 1

You are using reserved MySQL words (such as READ) as column names.

To fix this, change the line

$query2="UPDATE health.messages SET read='Y' WHERE messages.ID='".$id."';"; 

to

$query2="UPDATE health.messages SET `read` = 'Y' WHERE messages.ID='".$id."';"; 

I would also suggest that you heed the warnings from everyone else about using depreceated MySQL functions and SQL injection issues.

You should also try to be more unique with naming columns that don't collide with reserved MySQL keywords. Try using a consistent prefix (e.g. messageRead or isRead)