what do you think, why that?

[vinnie]

it only displays the name, but the last name and the address seems not
to exist! why?

<?php
@ $db=mysql_pconnect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db('list');
$query="select * from index";
$result=mysql_query($query);
$num_results=mysql_num_rows($result);
if ($num_results == 0)
{
echo'<br><br>nothing to dispaly';
}
else
{
echo'<br><br>here are the results: '.$num_results;
}
for ($i<0; $i<$num_results; $i++)
{
$row=mysql_fetch_array($result);
echo '<br>name: ';
echo htmlspecialchars(stripslashes($row['name']));
echo '<br>last name: ';
echo htmlspecialchars(stripslashes($row['last']));
}
?>

[Floortje]

vinnie wrote:

it only displays the name, but the last name and the address seems not
to exist! why?
echo '<br>name: ';
echo htmlspecialchars(stripslashes($row['name']));
echo '<br>last name: ';
echo htmlspecialchars(stripslashes($row['last']));

Because you tell the script to display the name, not the adres

try var_dump($row); to see what's in the row

[Denis Gerina]

vinnie wrote:

it only displays the name, but the last name and the address seems not
to exist! why?

<?php
@ $db=mysql_pconnect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db('list');
$query="select * from index";
$result=mysql_query($query);
$num_results=mysql_num_rows($result);
if ($num_results == 0)
{
echo'<br><br>nothing to dispaly';
}
else
{
echo'<br><br>here are the results: '.$num_results;
}
for ($i<0; $i<$num_results; $i++)
{
$row=mysql_fetch_array($result);
echo '<br>name: ';
echo htmlspecialchars(stripslashes($row['name']));
echo '<br>last name: ';
echo htmlspecialchars(stripslashes($row['last']));
}
?>

You don't seem to be printing address anywhere. As for last name, check
the name of the column in your table (index).

And, to avoid notices, the code

for ($i<0; $i<$num_results; $i++)

should be

for ($i=0; $i<$num_results; $i++)
Although I would probably write something like

while ($row = mysql_fetch_array($result))
{
....
}

[mypetprogrammer]

On Jun 4, 5:53 pm, Denis Gerina <[e-mail]:

vinnie wrote:

it only displays the name, but the last name and the address seems not
to exist! why?

<?php
@ $db=mysql_pconnect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db('list');
$query="select * from index";
$result=mysql_query($query);
$num_results=mysql_num_rows($result);
if ($num_results == 0)
{
echo'<br><br>nothing to dispaly';
}
else
{
echo'<br><br>here are the results: '.$num_results;
}
for ($i<0; $i<$num_results; $i++)
{
$row=mysql_fetch_array($result);
echo '<br>name: ';
echo htmlspecialchars(stripslashes($row['name']));
echo '<br>last name: ';
echo htmlspecialchars(stripslashes($row['last']));
}
?>

You don't seem to be printing address anywhere. As for last name, check
the name of the column in your table (index).

And, to avoid notices, the code

for ($i<0; $i<$num_results; $i++)

should be

for ($i=0; $i<$num_results; $i++)

Although I would probably write something like

while ($row = mysql_fetch_array($result))
{
...

}

See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_array($result)) != null)
{

So as not to be operating on a null on the last iteration of the loop.
Agree with the while, though.

~A!

[Denis Gerina]

my*************@gmail.com wrote:

On Jun 4, 5:53 pm, Denis Gerina <[e-mail]:

>vinnie wrote:

>>it only displays the name, but the last name and the address seems not
to exist! why?
<?php
@ $db=mysql_pconnect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db('list');
$query="select * from index";
$result=mysql_query($query);
$num_results=mysql_num_rows($result);
if ($num_results == 0)
{
echo'<br><br>nothing to dispaly';
}
else
{
echo'<br><br>here are the results: '.$num_results;
}
for ($i<0; $i<$num_results; $i++)
{
$row=mysql_fetch_array($result);
echo '<br>name: ';
echo htmlspecialchars(stripslashes($row['name']));
echo '<br>last name: ';
echo htmlspecialchars(stripslashes($row['last']));
}
?>

You don't seem to be printing address anywhere. As for last name, check
the name of the column in your table (index).

And, to avoid notices, the code

for ($i<0; $i<$num_results; $i++)

should be

for ($i=0; $i<$num_results; $i++)

Although I would probably write something like

while ($row = mysql_fetch_array($result))
{
...

}

See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_array($result)) != null)
{

So as not to be operating on a null on the last iteration of the loop.
Agree with the while, though.

~A!

Interesting. Do explain, please, in light of the docs.
http://www.php.net/mysql_fetch_array

Returns an array of strings that corresponds to the fetched row, or
FALSE if there are no more rows.

[Jerry Stuckle]

my*************@gmail.com wrote:

On Jun 4, 5:53 pm, Denis Gerina <[e-mail]:

>vinnie wrote:

>>it only displays the name, but the last name and the address seems not
to exist! why?
<?php
@ $db=mysql_pconnect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db('list');
$query="select * from index";
$result=mysql_query($query);
$num_results=mysql_num_rows($result);
if ($num_results == 0)
{
echo'<br><br>nothing to dispaly';
}
else
{
echo'<br><br>here are the results: '.$num_results;
}
for ($i<0; $i<$num_results; $i++)
{
$row=mysql_fetch_array($result);
echo '<br>name: ';
echo htmlspecialchars(stripslashes($row['name']));
echo '<br>last name: ';
echo htmlspecialchars(stripslashes($row['last']));
}
?>

You don't seem to be printing address anywhere. As for last name, check
the name of the column in your table (index).

And, to avoid notices, the code

for ($i<0; $i<$num_results; $i++)

should be

for ($i=0; $i<$num_results; $i++)

Although I would probably write something like

while ($row = mysql_fetch_array($result))
{
...

}

See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_array($result)) != null)
{

So as not to be operating on a null on the last iteration of the loop.
Agree with the while, though.

~A!

Denis's code is correct. There will be no out of bounds error in Denis's
code.

mysql_fetch_array() is a function which returns either an array or
false. It does not return null. Denis's code will continue to fetch
rows as long as the exist; after the last row has been fetched, the next
call to mysql_fetch_array() returns false, stopping the loop.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================