Queston regarding echoing contents of a table in MySQL

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June 21st, 2002, 07:24 PM

MuuTuwon

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Queston regarding echoing contents of a table in MySQL

I'm not really sure if I have this in the right place. Heh, with all the boards for help I have a feeling there might have been one better suited for this queston that I might have missed. If that's the case, sorry in advance. Also this is probably really simple because I'm a beginner, but nevertheless it has me stumped.

My problem I think is fairly simple, I'm trying to echo a single field in every row of a table into an html table. So basicly, I'd like it to look like this:

<tr>
<td>field1, row1</td>
<td>field1, row2</td>
<td>field1, row3</rd>
</tr>
<tr>
ect..

My queston is how do I get it to put in the <tr>'s and make a new row in the html table every 3 times like that? the way I've always pulled stuff out of a database with with the while loop, but if I had something like..

while ($row = mysql_fetch_array($result) {
ECHO "<td>$row[field1]</td>";
}

Then I just get a straight table down the page of my info. So how do I make it throw <tr>'s into there every 3 records? and still echo everything that's in the table when I don't know how much is there?

June 21st, 2002, 07:42 PM

Matt

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RE: Queston regarding echoing contents of a table in MySQL

Here's one approach:

php Code:

Original - php Code
<? $num_rows = mysql_num_rows($result); echo "<TABLE>n"; for($i = 0; $i < $num_rows; $i++) { if($i % 3 == 0) { echo "<TR>n"; } $row = mysql_fetch_array($result); echo "<TD>".$row['field1']."</TD>n"; if($i % 3 == 2) { echo "</TR>n"; } } if(($i-1) % 3 != 2) { echo "</TR>n"; } echo "</TABLE>n"; ?>

<?
$num_rows = mysql_num_rows($result);
echo "<TABLE>n";
for($i = 0; $i < $num_rows; $i++) {
  if($i % 3 == 0) {
    echo "<TR>n";
  }
  $row = mysql_fetch_array($result);
  echo "<TD>".$row['field1']."</TD>n";
  if($i % 3 == 2) {
    echo "</TR>n";
  }
}
if(($i-1) % 3 != 2) {
  echo "</TR>n";
}
echo "</TABLE>n";
?>

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