Very New php/mysql

phpnewbie · #1 26-Jan-2005, 10:04

I must admit that I am treading foreign waters when working with mysql & php, but in researching many other errors that I have been receiving while "trying to correct some errors in the code below, I decide to register, as I am stumped.

Parse error: parse error, unexpected T_VARIABLE on line 47

is the most recent error that I am receiving, and I am not sure what to do. I have read several posts, and manuals, but I am lost. I do apologize for the, what is probably an elementary problem, but your help is most appreciated.
Here is the code:

PHP Code:


$result=mysql("$DBName","SELECT * FROM Category ORDER BY Category");
fontFace("Arial","Select a category:<br><br>");
echo "<ul>";   
while (($row=mysql_fetch_row($result)) 
//line 47 error is next line
$Cat=$row[0];
$CatID=$row[1];
fontFace("Arial","<li><a href='$Relative/items.php?CA=$CatID&UID=$UID'>$Cat</a></li>");
        }
echo "</ul>";

JdS · #2 26-Jan-2005, 10:25

Hello phpnewbie,

It's simply a typo I think, you missed the (opening) curly bracket in the while() structure.

PLUS, you seem to have an extra (opening) bracket in this line: while (($row=mysql_fetch_row($result))

phpnewbie · #3 26-Jan-2005, 10:41

Thanks You, I have corrected that one (I think) but I am now seeing

Parse error: parse error, unexpected $ on line 56

And the code stops at line 55, or at least that is how it looks.. Here is the code from line 46 to the end:

PHP Code:


{
while ($row=mysql_fetch_row($result)) 
$Cat=$row[0];
$CatID=$row[1];
fontFace("Arial","<li><a href='$Relative/items.php?CA=$CatID&UID=$UID'>$Cat</a></li>");
        
echo "</ul>";

commonFooter($Relative,$UID);
?>

Again I apologize for my ignorance.... this is all a learning exerience for me.

Quote:

Originally Posted by JdS

Hello phpnewbie,

It's simply a typo I think, you missed the (opening) curly bracket in the while() structure.

PLUS, you seem to have an extra (opening) bracket in this line: while (($row=mysql_fetch_row($result))

JdS · #4 26-Jan-2005, 10:59

Try replacing that bit with this instead...

PHP Code:


<?

// comment this out ---> {
while ($row=mysql_fetch_row($result))
{
  $Cat=$row[0];
  $CatID=$row[1];
  fontFace("Arial","<li><a href='$Relative/items.php?CA=$CatID&UID=$UID'>$Cat</a></li>");
}
echo "</ul>";

commonFooter($Relative,$UID);
?>

phpnewbie · #5 26-Jan-2005, 12:32

I have made the requested changes and a different group errors have displayed. Here are the errors that I am receiving, and my Code. Sorry about the length, I was trying NOT to post all of it, but I am stuck, I have read most of the MySql manual, but not know the intricacies of the program, I am not Quite sure what I'm looking for.

??:

Errors:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource on line 11

Warning: Cannot add header information - headers already sent by (output started at xx.php:11) on line 40

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource on line 48

Here is the code starting at line 9

PHP Code:


$result=mysql("$DBName","SELECT CartItemsID,Date FROM CartItems");
while ($row=mysql_fetch_row($result))
{
$CII=$row[0];
$CDa=$row[1];
$pieces=explode(":",$CDa);
$DCHK=$pieces[1];
if ($DCHK < $old) {
mysql("$DBName","DELETE FROM CartItems WHERE CartItemsID = '$CII'");
    }
}


if ($UID != "") {
$result=mysql("$DBName","SELECT * FROM Users WHERE User='$UID'");
$num=mysql_num_rows($result);
if ($num == "0") {
$dt=date("YmdHis");
$UID="$dt$REMOTE_ADDR";
$date=date("z");
mysql("$DBName","INSERT INTO Users VALUES ('$UID','$date')");
Header("Location: $PHP_SELF?UID=$UID");
}
}

if ($UID == "") {
$dt=date("YmdHis");
$UID="$dt$REMOTE_ADDR";
$date=date("z");
mysql("$DBName","INSERT INTO Users VALUES ('$UID','$date')");
Header("Location: $PHP_SELF?UID=$UID");
}
commonHeader("$Company","Select a category");

$result=mysql("$DBName","SELECT * FROM Category ORDER BY Category");
fontFace("Arial","Select a category:<br><br>");
echo "<ul>";
 
while ($row=mysql_fetch_row($result)) { 
  $Cat=$row[0]; 
  $CatID=$row[1]; 
  fontFace("Arial","<li><a href='$Relative/items.php?CA=$CatID&UID=$UID'>$Cat</a></li>"); 
} 
echo "</ul>"; 

commonFooter($Relative,$UID); 
?>

firebird · #6 28-Jan-2005, 14:29

By taking a quick look at your code, it seems to me that you should replace all mysql() with mysql_db_query.

So

$result=mysql("$DBName","SELECT CartItemsID,Date FROM CartItems");

would become

$result=mysql_db_query("$DBName","SELECT CartItemsID,Date FROM CartItems");

etc.

Hope that's it,

Dave.

phpnewbie · #7 28-Jan-2005, 14:54

Thanks Dave, I have printed out your suggestion, but future use, but that doesn't seem to solve my issue. I started another thread, as this thread answered the 1st few questions I had. Thank you for your input

Quote:

Originally Posted by Dave

By taking a quick look at your code, it seems to me that you should replace all mysql() with mysql_db_query.

So

$result=mysql("$DBName","SELECT CartItemsID,Date FROM CartItems");

would become

$result=mysql_db_query("$DBName","SELECT CartItemsID,Date FROM CartItems");

etc.

Hope that's it,

Dave.

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Very New php/mysql

different parse