PHP Table Format

Question

I've put together the following code which creates a table upon a selection being made with a drop down menu.

echo "<table>";
$result=mysql_query($query);
while($rows=mysql_fetch_array($result)){
echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";
echo "<tr>";
echo "<td>".$rows['findname']."</td>";
echo "<td>".$rows['finddescription']."</td>";
echo "</tr>";
}
echo "</table>";

The problem I've got is that for each record that is returned the 'headers' are repeated. The live page can be found here. I just wonder whether someone could perhaps take a look at this please and tell me where I've gone wrong.

Apologies for the really simple question, but I've been looking at this for a while and I just can't find the answer. I think it just needs a fresh pair of eyes to look at it.

Chirag · Accepted Answer · 2012-03-17 13:55:32Z

4

Try this, you just need to get headers out of while loop

echo "<table>";
echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";

$result=mysql_query($query);
while($rows=mysql_fetch_array($result)){

echo "<tr>";
echo "<td>".$rows['findname']."</td>";
echo "<td>".$rows['finddescription']."</td>";
echo "</tr>";
}
echo "</table>";

answered Mar 17, 2012 at 13:55

Chirag

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Comments

QuantumBlack · Accepted Answer · 2012-03-17 13:56:08Z

4

The answer is obvious, you are repeating the output of the headers in the loop. Move the

while($rows=mysql_fetch_array($result)){

after the first

echo "</tr>";

answered Mar 17, 2012 at 13:56

QuantumBlack

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Comments

cmbuckley · Accepted Answer · 2012-03-17 17:54:26Z

4

You need to put the headers outside of the while loop:

echo "<table>";
echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";

$result = mysql_query($query);

while ($rows = mysql_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $rows['findname'] . "</td>";
    echo "<td>" . $rows['finddescription'] . "</td>";
    echo "</tr>";
}

echo "</table>";

edited Mar 17, 2012 at 17:54

cmbuckley

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answered Mar 17, 2012 at 13:56

James Keightley

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Comments

Kaustubh Karkare · Accepted Answer · 2012-03-17 13:56:29Z

3

This should work:

echo "<table>";
echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";
$result=mysql_query($query);
while($rows=mysql_fetch_array($result)){
    echo "<tr>";
    echo "<td>".$rows['findname']."</td>";
    echo "<td>".$rows['finddescription']."</td>";
    echo "</tr>";
    }
echo "</table>";

answered Mar 17, 2012 at 13:56

Kaustubh Karkare

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Comments

user229044 · Accepted Answer · 2012-03-17 13:55:15Z

3

Your headers are being repeated because you are writing them inside the loop, for each row returned by the query. You simply need to move your headers outside the loop so they're only written once, before the rows returned by the query start printing:

echo "<table>";
echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";

$result=mysql_query($query);
while($rows=mysql_fetch_array($result)){
  echo "<tr>";
  echo "<td>".$rows['findname']."</td>";
  echo "<td>".$rows['finddescription']."</td>";
  echo "</tr>";
}
echo "</table>";

answered Mar 17, 2012 at 13:55

user229044♦

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Comments

test30 · Accepted Answer · 2012-03-17 13:55:50Z

3

headers are repeating becasue they are in while loop, it should work well

echo "<table>";
echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";
$result=mysql_query($query);
while($rows=mysql_fetch_array($result)){

echo "<tr>";
echo "<td>".$rows['findname']."</td>";
echo "<td>".$rows['finddescription']."</td>";
echo "</tr>";
}
echo "</table>";

answered Mar 17, 2012 at 13:55

test30

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Comments

iwishiwasacodemonkey · Accepted Answer · 2012-03-17 13:56:49Z

2

Change to:

echo "<tr>";
echo "<th>Find Name:</th>";
echo "<th>Find Description:</th>";
echo "</tr>";
echo "<tr>";
while($rows=mysql_fetch_array($result)){
echo "<td>".$rows['findname']."</td>";
echo "<td>".$rows['finddescription']."</td>";
echo "</tr>";
}
echo "</table>";

answered Mar 17, 2012 at 13:56

iwishiwasacodemonkey

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