bravo14

July 1, 2008

Think I may have solved it, won't know until later.

July 1, 2008

Thanks for all your help guys, still geting error message

Parse error: syntax error, unexpected '[' in /home/fhlinux147/y/yardleyheroes.co.uk/user/htdocs/fixtures.php on line 58

$result = mysql_query('SELECT * FROM `fixtures` ORDER BY `date`', $con) or die(mysql_error());
if(mysql_num_rows($result) < 1) 
{ 
   echo('<tr class="profiletext"><td colspan="7">Sorry, no results were found.</td></tr>'); 
} 
else 
{ 
   while($row = mysql_fetch_array($result))
    {
if ($row.[HomeAway].=="Home") {
echo ('<tr><td>'.$row[Date].'</td><td><b>Yardley</b></td>');
echo ('<td>'.$row[yardley_goals].'-'.$row[opposition_goals].'</td><td>'.$row[Opposition].'</td>');
echo ('<td></td><td><a href="http://www.yardleyheroes.co.uk/matchfines.php?match_id='.$row[match_id].'"View Fines</a></td></tr>');
       }

else{

echo ('<tr><td>'.$row[Date].'</td><td>'.$row[Opposition].'</td>');
echo ('<td>'.$row[opposition_goals].'-'.$row[yardley_goals].'</td><td>Yardley</td>');
echo ('<td></td><td><a href="http://www.yardleyheroes.co.uk/matchfines.php?'.$row[match_id].'<a></td></tr>');
        }
}
}
mysql_close($con);
?>

Line 58 is if ($row.[HomeAway].=="Home") {

June 30, 2008

Sorry guys being a bit thick where do I need to put the { at the moment I have done it as follows

if {($row[Home/Away]=="Home")

and I am getting an error message

Sorry about this

June 30, 2008

Thank you for your quick response

I have copied that text over so it now looks like

$result = mysql_query('SELECT FROM fixtures ORDER BY date', $con) or die(mysql_error());
if(mysql_num_rows($result) < 1) 
{ 
   echo('<tr class="profiletext"><td colspan="6">Sorry, no results were found.</td></tr>'); 
} 
else 
{ 
   while($row = mysql_fetch_array($result))
    {
if ($row[Home/Away]=="Home")
echo ('<tr><td>'.$row[Date].'</td><td><b>Yardley</b></td>');
echo ('<td>'.$row[yardley_goals].'-'.$row[opposition_goals].'</td><td>'.$row[Opposition].'</td>');
echo ('<td></td><td><a href="http://www.yardleyheroes.co.uk/matchfines.php?match_id='.$row[match_id].'"View Fines</a></td></tr>');
       }

else{

echo ('<tr><td>'.$row[Date].'</td><td>'.$row[Opposition]'</td>');
echo ('<td>'.$row[opposition_goals].'-'.$row[yardley_goals].'</td><td>Yardley</td>');
echo ('<td></td><td><a href="http://www.yardleyheroes.co.uk/matchfines.php?').$row[match_id].'<a></td></tr>');
        }
}
}
mysql_close($con);
?>

and get the following error message

Parse error: syntax error, unexpected T_ELSE in /home/fhlinux147/y/yardleyheroes.co.uk/user/htdocs/fixtures.php on line 64

Line 64 is else{

Any ideas?

Cheers Mark

June 30, 2008

Hi all

I am trying to display a list of fixtures using if statements.

If the Home/Away field is Home I want it to display the fixtures in one way and if it is Away I want it to show another way, I have had a stab at it but am getting an error message on the page at www.yardleyheroes.co.uk/fixtures.php

$result = mysql_query('SELECT FROM fixtures ORDER BY date', $con) or die(mysql_error());
if(mysql_num_rows($result) < 1) 
{ 
   echo('<tr class="profiletext"><td colspan="6">Sorry, no results were found.</td></tr>'); 
} 
else 
{ 
   while($row = mysql_fetch_array($result))
    {
if ($row[Home/Away]=="Home")
echo ('<tr><td>'.$row[Date].'</td><td><b>Yardley</b></td>');
echo ('<td>'.$row[yardley_goals].'-'.$row[opposition_goals].'</td><td>'.$row[Opposition].'</td>');
echo ('<td></td><td><a href="http://www.yardleyheroes.co.uk/matchfines.php?match_id='.$row[match_id].'"View Fines</a></td></tr>');

else{

echo ('<tr><td>'.$row[Date].'</td><td>'.$row[Opposition]'</td>');
echo ('<td>'.$row[opposition_goals].'-'.$row[yardley_goals].'</td><td>Yardley</td>');
echo ('<td></td><td><a href="http://www.yardleyheroes.co.uk/matchfines.php?').$row[match_id].'<a></td></tr>');
}
}
mysql_close($con);
?>

Cheers

Mark

Granting User Permissions · June 24, 2008

In addition to the previous post

db is cyb

Username is cybadmin

Granting User Permissions · June 24, 2008

Hello

I am trying to grant permissions on another database and used the template as laid out above

GRANT ALL permissions ON 'cyb@server213-171-219-90.livedns.org.uk' TO 'cybadmin.*';

and get the following error message

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'permissions ON 'cyb@server213-171-219-90.livedns.org.uk'TO 'cybadmin.*'' at line 1

MySQL client version: 5.0.22

does anybody have guidance

June 14, 2008

Thanks guys for your responses, it is a script I found from somewhere else, I am going to try and create a login using session, but this will be a first, so it will be interesting to see what I can come up with, does anyone have a starting point for me?

June 13, 2008

Thanks for your help guys.

Craygo just out of curiosity as I am relatively new to PHP/MySQL where in the code below would I put the code you suggested?

$result = mysql_query('SELECT * FROM `reports` ORDER by report_id LIMIT 0, 30 ', $con);
while($row = mysql_fetch_array($result))

{
echo ('<tr class="profiletext"><td>'.$row[date].'</td>');

June 13, 2008

Hi all

I have a forum on my site however as the users do not have to log in, I am having a lot of messages on there that I do not want. I want to try and create a registration process and log in script for registered users if they want to add a post to the forum.

Does anybody have any ideas or sources?

Thanks again

Mark

June 13, 2008

Hi All

I have some dates in a mysql database and it is the format of yyyy-mm-dd and I wanto to display it in a differnet format on the web pages.

Any assistance would be much appreciated.

Mark

June 12, 2008

Hi all

I have probable made a very simple error hopefully someone can help

I am trying to insert a record into a database using PHP

the form is built using the following code

<?php
echo ('<form action="insert_fine.php"><table width="65%" border="0">');
echo ('<tr>');
echo ('<td width="30%" class="profiletext">Name</td>');
echo ('<td><select name="name">');
$con = mysql_connect('IPaddress', 'database', 'password', true);
if(!$con)
{
      die('Could not connect: ' . mysql_error());
}
mysql_select_db('yardleyheroes', $con);
$res=mysql_query('SELECT * FROM `profiles` WHERE `profiletype` =\'player\'ORDER BY `name`');
if(mysql_num_rows($res)==0) 
echo('<option>there is no data in table..</option>');
else
{
while($row=mysql_fetch_assoc($res))
{
echo('<option value='.$row[profile_id].'>'.$row[name].'</option>');
}
}

echo ('</select></td>');
echo ('</tr>');
echo ('<tr>');
echo ('<td class="profiletext">Game</td><td><select name="match" width="145">');
$fixtures=mysql_query('SELECT * FROM `fixtures` ORDER BY `date`');
if(mysql_num_rows($fixtures)==0) 
echo('<option>there is no data in table..</option>');
else
{
while($matchrow=mysql_fetch_assoc($fixtures))
{
echo('<option value='.$matchrow[match_id].'>'.$matchrow.'|'.$matchrow[Opposition].'</option>');
}
}
echo ('</select></td></tr>');
echo ('<tr>');
echo ('<td class="profiletext">Amount</td>');
echo ('<td><input type="text" name="amount"></td>');
echo ('</tr>');
echo ('<tr>');
echo ('<td class="profiletext">Reason</td>');
echo ('<td><textarea name="reason" cols="50" rows="4"></textarea></td>');
echo ('</tr>');
echo ('<tr>');
echo ('<td> </td>');
echo ('<td><input type="submit" name="Submit" value="Submit"></td>');
echo ('</tr>');
echo ('</table>');
echo('</form>');
?>

And the process code is

<?php
$con = mysql_connect('IPaddress', 'username', 'password', true);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db('yardleyheroes', $con);$sql="INSERT INTO fines (profile_id, fine_amount, reason)
VALUES
('$_POST[name]','$_POST[amount]','$_POST[reason]')";if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo ('1 record added');
echo ('<table><tr><td>Name</td><td>Amount</td><td>Reason</td></tr>');
echo ('<tr><td>'.$_POST[name].'</td><td>'.$_POST[amount].'</td><td>'.$_POST[reason].'</td></tr></table>');
mysql_close($con)

?>

When processing the Insert, the code says that the record has been produced, but the data from the form does not get created.

Thanks in advance

Granting User Permissions · June 8, 2008

The error message I am getting here is

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'permissions to 'yardleyheroes.*' on 'yardleyheroes@server213-171-218-147.livedns' at line 1

Granting User Permissions · June 8, 2008

I am trying to grant permission to a user on a mysql database

GRANT ALL permissions TO 'yardleyheroes.*' ON 'yardleyheroes@server213-171-218-147.livedns.org.uk'

Can anybody give me any guidance

June 7, 2008

Hi folks

Newbie to this forum, but hoping there is someone there who can help

I am trying to display the results of a MySQL query on a page using PHP, the query works in phpmyadmin

see code below

<?php

$con = mysql_connect('213.171.219.88', 'yardleyheroes', 'heroes', true);

if(!$con)

{

die('Could not connect: ' . mysql_error());

}

mysql_select_db('yardleyheroes', $con);

$result = mysql_query('SELECT Sum(fines.fine_amount) AS SumOffine_amount, profiles.name FROM fines INNER JOIN profiles ON fines.profile_id = profiles.profile_id GROUP BY profiles.name ORDER BY Sum(fines.fine_amount); LIMIT 0, 30 ', $con);

if(mysql_num_rows($result) < 1)

{

echo('<tr><td colspan="2">Sorry, no results were found.</td></tr>');

}

else

{

while($row = mysql_fetch_array($result))

{

echo('<tr><td valign="top">' . $row['name'] . '</td>');

echo('<td> £' . $row['SumOffine_amount'] .'.00</td></tr>');

}

mysql_close($con);

?>

I get the following message on the page

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/fhlinux147/y/yardleyheroes.co.uk/user/htdocs/fines.php on line 52

Line 52 is if(mysql_num_rows($result) < 1)

Any ideas, the page can be seen at www.yardleyheroes.co.uk/fines.php

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bravo14

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