Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'

[Mutley]

New error for me, line 9 which is:

[code]7.          {
8.          if(isset($id)){
9.          $1 = $_POST['1'];
10.        $2 = $_POST['2'];
11.        $3 = $_POST['3'];[/code]

Error:
Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$' on line 9.

[Daniel0]

I do not think you can start a variable with a number.

[wildteen88]

That is correct you cannot use a number as the variable name. Use a non numeric character first if want to use a number as a variable eg:
$_1 = 'blah';
$_2 = 'blah';

[zq29]

[quote author=Daniel0 link=topic=108663.msg437383#msg437383 date=1158687465]
I do not think you can start a variable with a number.
[/quote]
Correct, variable names can only start with a letter or an underscore.

[b]EDIT:[/b] Beaten to it :P

[Mutley]

Thanks.

My script seems to not be updating my content, does this look right now I've changed it?

[code]if(isset($id)){
$_1 = $_POST['1'];
$_2 = $_POST['2'];
$_3 = $_POST['3'];
$_4 = $_POST['4'];
$_5 = $_POST['5'];
$_6 = $_POST['6'];
$_7 = $_POST['7'];
$_8 = $_POST['8'];
$_9 = $_POST['9'];
$_10 = $_POST['10'];
$_11 = $_POST['11'];
$_12 = $_POST['12'];
$_13 = $_POST['13'];
$_14 = $_POST['14'];
$_15 = $_POST['15'];
$_16 = $_POST['16'];
$_17 = $_POST['17'];
$_18 = $_POST['18'];
$_19 = $_POST['19'];

mysql_select_db("rufc");

$sql = "UPDATE scores SET 1='$_1', 2='$_2', 3='$_3', 4='$_4', 5='$_5', 6='$_6', 7='$_7', 8='$_8', 9='$_9', 10='$_10', 11='$_11', 12='$_12', 13='$_13', 14='$_14', 15='$_15', 16='$_16', 17='$_17', 18='$_18', 19='$_19' WHERE score_id = '".$id."'";
mysql_query($sql);[/code]

The SQL echos on the page and looks fine but makes no change to the database. It has other fields in rows not just 1, 2, 3, 4 etc, would that be the problem? Because I'm not updating them with anything? I just want the fields listed above (the numbers) to be updated.

[wildteen88]

Do you have fields named 1 throught to 19 in your scores table? Also prehaps see if there is an error with your query too, by adding or die(mysql_errro()) after mysql_query($sql) so it becomes:
[code=php:0]mysql_query($sql) or die('Unable to perform query<br />' . mysql_error());[/code]

[.josh]

you also might wanna check and see if $id really is set.  try

echo $id;

before your if statement

[Mutley]

ID is fine (first line which is "18" below), I got an error though:
[code]
18
Unable to perform query
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 '1='', 2='', 3='', 4='', 5='', 6='', 7='', 8='', 9='', 10='', 11='', 12='', 13=''' at line 1[/code]

I deliberatly put no data in by the way, so don't worry that nothing = anything. Thanks for the help guys!

[.josh]

try putting backticks around your field names.  `1` = '$_1', `2` = '$_2", etc...

[Mutley]

Cool, it worked, thanks.

What do backticks` do?

[zq29]

Backticks generally allow you to use reserved words, par example:
[code]SELECT * FROM `table` WHERE `where`='England' ORDER BY `id` ASC LIMIT 10[/code]

[.josh]

since you are using numbers for fields, sql gets confused when you try to use them in your query. it sees 1='$_1' and tries to look at it like a condition/comparison, rather than an assignment.  therefore you use the backticks to tell sql that the number is really a field name and that you want to assign a value to it.

backticks are also used for what SA mentioned: if you use a reserved word for a table or field name, sql will get confused and try to parse the string under the function of the reserved word. So if you wish to use a reserved word as a table or field name, you need to use backticks so sql knows to treat it as a table/field name and not a reserved word. 

You should know that it is not good practice to use reserved words or even numbers as table and field names.  It's lazy and non-intuitive and begs for trouble to happen down the road...just like right now.