parse error

[pinehead18]

$db = mysql_select_db("pinehead", $con) or die(mysql_error());
$sql= "select * from awating where id='$id'";
$result = mysql_db_query($con, $sql);
$row = mysql_fetch_array($result);

$sql = 'insert into articles values ("' . $id . '", "' . $row["author"] . '", "' . $row["subject"] . '", "' . $row["date"] . '", "' . $row["1"] . '")';
mysql_db_query($con, $sql,);
print mysql_error($con);
$sql = "delete from subarts where subid='$subid'";
mysql_db_query($con, $sql,);
print "Added $id";

This is my code..

$row = mysql_fetch_array($result);

$sql = 'insert into articles values ("' . $id . '", "' . $row["author"] . '", "' . $row["subject"] . '", "' . $row["date"] . '", "' . $row["1"] . '")'; this is the lines i'm getting a parse error on.

Parse error: parse error, unexpected ')' in /home/pinehead/public_html/add.php on line 15

that is the error i'm getting.

Sorry to be so blunt, just really ticked off at this, and on top of all that i dont' know if the code even works hehe.

Thank you for your help

-PiNeHeAd

[RWJackson2006]

$sql = 'insert into articles values ("' . $id . '", "' . $row["author"] . '", "' . $row["subject"] . '", "' . $row["date"] . '", "' . $row["1"] . '")';
Try:

$sql = 'insert into articles values ("' . $id . '", "' . $row["author"] . '", "' . $row["subject"] . '", "' . $row["date"] . '", "' . $row["1"] . '");

[RWJackson2006]

Did that work?

[pinehead18]

Parse error: parse error, unexpected T_VARIABLE in /home/pinehead/public_html/add.php on line 17

nope this is the error i receive now.

[jpmoriarty]

• Put php tags around your code so we can read it properly

• dont think you should have a final , in these lines: mysql_db_query($con, $sql,);

[pinehead18]

Parse error: parse error, unexpected T_VARIABLE in /home/pinehead/public_html/add.php on line 14

that is the error i'm receiving now.. and let me show you line 14

$sql = "delete from subarts where subid='$id'"; that is line 14

and here is the rest of my code..

<?php
$id = $_GET[id];
$con = mysql_connect("localhost", "pinehead", "blowme") or die(mysql_error());
$db = mysql_select_db("pinehead", $con) or die(mysql_error());
$sql = "select * from awating where id='$id'";
$result = mysql_db_query($con, $sql);
$row = mysql_fetch_array($result);

$sql = 'insert into articles values ("' . $id . '", "' . $row["author"] . '", "' . $row["subject"] . '", "' . $row["date"] . '", "' . $row["1"] . '");
mysql_db_query($con, $sql);
print mysql_error($con);
$sql = "delete from subarts where subid='$id'";
mysql_db_query($con, $sql);
print "Added $id";
?>

Thank you for all your help....

-PiNeHeAd

[jpmoriarty]

please put php tags in

[pinehead18]

php tags as in <?php and ?> ?

[jpmoriarty]

no why would i care about that?

php as in the [ php] tags around the text so that it formats it correctly. or use the "PHP" button above

[pinehead18]

hmm i have no idea what your talking about you mean like the

tags?

[jpmoriarty]

YES!! put your code in that and we'll spot the error easily

[pinehead18]

Parse error: parse error, unexpected T_VARIABLE in /home/pinehead/public_html/add.php on line 14

that is the error i'm receiving now.. and let me show you line 14

$sql = "delete from subarts where subid='$id'"; that is line 14 

and here is the rest of my code..

<?php 
$id = $_GET[id]; 
$con = mysql_connect("localhost", "pinehead", "blowme") or die(mysql_error()); 
$db = mysql_select_db("pinehead", $con) or die(mysql_error()); 
$sql = "select * from awating where id='$id'"; 
$result = mysql_db_query($con, $sql); 
$row = mysql_fetch_array($result); 

$sql = 'insert into articles values ("' . $id . '", "' . $row["author"] . '", "' . $row["subject"] . '", "' . $row["date"] . '", "' . $row["1"] . '"); 
mysql_db_query($con, $sql); 
print mysql_error($con); 
$sql = "delete from subarts where subid='$id'"; 
mysql_db_query($con, $sql); 
print "Added $id"; 
?> 

Thank you for all your help....

-PiNeHeAd

[jpmoriarty]

and straight away you can spot by the colours that you're missing a closing ' on the sql line.

i dont know why you changed from the " to ' on that line, but change it to this and you should be fine.

$sql = "insert into articles values ('$id','". $row['author']."','" . $row['subject'] . "', '" . $row['date'] ."', '" . $row['1'] . "')"; 

[pinehead18]

This is my error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/pinehead/public_html/add.php on line 9
Unknown database 'Resource id #2'Added

This is the code on line 9

  $row = mysql_fetch_array($result);

This is the entire code all together.

<?php
	$id = $_GET[id]; 
$con = mysql_connect("localhost", "pinehead", "blowme") or die(mysql_error());
$db = mysql_select_db("pinehead", $con) or die(mysql_error()); 
 $sql = "select * from awating where id='$id'";
  $result = mysql_db_query($con, $sql);
  $row = mysql_fetch_array($result);

 $sql = "insert into articles values ('$id','". $row['author']."','" . $row['subject'] . "', '" . $row['date'] ."', '" . $row['1'] . "')";
  mysql_db_query($con, $sql);
  print mysql_error($con);
  $sql = "delete from subarts where subid='$id'";
  mysql_db_query($con, $sql);
  print "Added $id";
?>

Thanks again 🙁 🙂

[jpmoriarty]

it'll do that when it gets no results from the query. ie your query is returning no results.

you sure your table isn't called "awaiting"? You've got it in as "awating".

Just a little bit of self-investigation can usually avoid most of these problems...

[pinehead18]

no, actually the table is awating ...

[jpmoriarty]

well whatever, but you're getting no results.

maybe the $_GET[id] is the problem, since i think the [id] should have quotes in it:

$_GET['id']