如何将数据从一个mySQL表显示到php中的多个表中？

Question

I have on mySQL table with data and I want to display it in multiple tables. When I try to do it with the code I have now, the data doesn't show up in the second table. Is my code wrong?

$id=$_GET['id'];
$sql = "SELECT * FROM buildings WHERE room = '{$id}' AND number = '81'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
echo "<center><h2>Room $id</h2></center> <br> <b>General Information<br>     
<table><tr>
<th>Room Type</th>
<th>Department</th>
<th>College</th>
<th>Primary Owner</th></tr>";

// output data of each row

while($row = mysqli_fetch_assoc($result)) {
echo "<tr>
<td>".$row["type"]."</td>
<td>".$row["Department"]."</td>
<td>".$row["College"]."</td>
<td>".$row["Primary"]."</td>
</tr>";
}
  echo "</table>";
}

//CONTACT INFO//
if (mysqli_num_rows($result) > 0) {
echo "<p><b>Contact Information<br><table><tr>
<th>Reservations</th>
<th>CTL Contact Name</th>
<th>CTL Contact Email</th>  
<th>CTL Contact Phone</th>
<th>Department Contact Name</th>
<th>Department Contact Email</th>
<th>Department Contact Phone</th>
<th>IT Contact Name</th>
<th>IT Contact Email</th>
<th>IT Contact Phone</th>
</tr>";

// output data of each row

while($row = mysqli_fetch_assoc($result)) {
echo "<tr>
<td>".$row["reservations"]."</td>
<td>".$row["contact"]."</td>  
<td>".$row["contactphone"]."</td>
<td>".$row["deptcontact"]."</td>
<td>".$row["deptcontactemail"]."</td>
<td>".$row["deptcontactphone"]."</td>
<td>".$row["itcontact"]."</td>
<td>".$row["itcontactemail"]."</td>
<td>".$row["itcontactphone"]."</td>
</tr>";
}
  echo "</table>";
}
else{
echo("0 results"); 
}

This is what I get: http://i57.tinypic.com/2heijyc.png

Answer 1

Read the manual of mysqli_fetch_assoc

You've already called while($row = mysqli_fetch_assoc($result))

This will loop through the entire result array and get each row. mysqli_fetch_assoc returns null when there are no more rows. So the next time you run mysqli_fetch_assoc($result) it will be null.

What you need to do is use mysqli_fetch_all at the top to get an array of the result set. Then you can loop through this array as needed.

---

As I stated in my comment, this query is subject to SQL injections because you use $id which is sourced by $_GET directly in the query. Read up on prepared statements or if it is an integer, use intval($id) .