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
@ $db=mysql_pconn ect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db ('list');
$query="select * from index";
$result=mysql_q uery($query);
$num_results=my sql_num_rows($r esult);
if ($num_results == 0)
{
echo'<br><br>no thing to dispaly';
}
else
{
echo'<br><br>he re are the results: '.$num_results;
}
for ($i<0; $i<$num_results ; $i++)
{
$row=mysql_fetc h_array($result );
echo '<br>name: ';
echo htmlspecialchar s(stripslashes( $row['name']));
echo '<br>last name: ';
echo htmlspecialchar s(stripslashes( $row['last']));
}
?>

[Floortje]

Re: what do you think, why that?

vinnie wrote:

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

Re: what do you think, why that?

vinnie wrote:

it only displays the name, but the last name and the address seems not
to exist! why?
>
<?php
@ $db=mysql_pconn ect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db ('list');
$query="select * from index";
$result=mysql_q uery($query);
$num_results=my sql_num_rows($r esult);
if ($num_results == 0)
{
echo'<br><br>no thing to dispaly';
}
else
{
echo'<br><br>he re are the results: '.$num_results;
}
for ($i<0; $i<$num_results ; $i++)
{
$row=mysql_fetc h_array($result );
echo '<br>name: ';
echo htmlspecialchar s(stripslashes( $row['name']));
echo '<br>last name: ';
echo htmlspecialchar s(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 ($row = mysql_fetch_arr ay($result))
{
....
}

[mypetprogrammer@gmail.com]

Re: what do you think, why that?

On Jun 4, 5:53 pm, Denis Gerina <denisREMOVET.. .@cced.bawrote:

vinnie wrote:

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

>

<?php
@ $db=mysql_pconn ect('host', 'UserID', 'PWD');
if (!$db)
{
echo 'conneciton eror';
exit;
}
else
{
echo 'connection on!';
}
mysql_select_db ('list');
$query="select * from index";
$result=mysql_q uery($query);
$num_results=my sql_num_rows($r esult);
if ($num_results == 0)
{
echo'<br><br>no thing to dispaly';
}
else
{
echo'<br><br>he re are the results: '.$num_results;
}
for ($i<0; $i<$num_results ; $i++)
{
$row=mysql_fetc h_array($result );
echo '<br>name: ';
echo htmlspecialchar s(stripslashes( $row['name']));
echo '<br>last name: ';
echo htmlspecialchar s(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_arr ay($result))
{
...
>
}

See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_arr ay($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]

Re: what do you think, why that?

mypetprogrammer @gmail.com wrote:

On Jun 4, 5:53 pm, Denis Gerina <denisREMOVET.. .@cced.bawrote:

>vinnie wrote:

>>it only displays the name, but the last name and the address seems not
>>to exist! why?
>><?php
>>@ $db=mysql_pconn ect('host', 'UserID', 'PWD');
>>if (!$db)
>> {
>> echo 'conneciton eror';
>> exit;
>> }
>>else
>> {
>> echo 'connection on!';
>> }
>>mysql_select_ db('list');
>>$query="selec t * from index";
>>$result=mysql _query($query);
>>$num_results= mysql_num_rows( $result);
>>if ($num_results == 0)
>>{
>> echo'<br><br>no thing to dispaly';
>>}
>>else
>>{
>> echo'<br><br>he re are the results: '.$num_results;
>>}
>> for ($i<0; $i<$num_results ; $i++)
>>{
>> $row=mysql_fetc h_array($result );
>> echo '<br>name: ';
>> echo htmlspecialchar s(stripslashes( $row['name']));
>> echo '<br>last name: ';
>> echo htmlspecialchar s(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_arr ay($result))
>{
>...
>>
>}

>
See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_arr ay($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]

Re: what do you think, why that?

mypetprogrammer @gmail.com wrote:

On Jun 4, 5:53 pm, Denis Gerina <denisREMOVET.. .@cced.bawrote:

>vinnie wrote:

>>it only displays the name, but the last name and the address seems not
>>to exist! why?
>><?php
>>@ $db=mysql_pconn ect('host', 'UserID', 'PWD');
>>if (!$db)
>> {
>> echo 'conneciton eror';
>> exit;
>> }
>>else
>> {
>> echo 'connection on!';
>> }
>>mysql_select_ db('list');
>>$query="selec t * from index";
>>$result=mysql _query($query);
>>$num_results= mysql_num_rows( $result);
>>if ($num_results == 0)
>>{
>> echo'<br><br>no thing to dispaly';
>>}
>>else
>>{
>> echo'<br><br>he re are the results: '.$num_results;
>>}
>> for ($i<0; $i<$num_results ; $i++)
>>{
>> $row=mysql_fetc h_array($result );
>> echo '<br>name: ';
>> echo htmlspecialchar s(stripslashes( $row['name']));
>> echo '<br>last name: ';
>> echo htmlspecialchar s(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_arr ay($result))
>{
>...
>>
>}

>
See, and I would avoid the inevitable out of bounds on that with:
while( ($row = mysql_fetch_arr ay($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_arr ay() 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_arr ay() returns false, stopping the loop.

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===