You can't start a variable name with a number.
-Rasmus
On Fri, 26 Oct 2001, jtjohnston wrote:
> Friday night and I can't get my head around this. OK, so I don't get
> this. (From A Reformer Perler?? - unitil now)
>
> I was getting
>
> "parse error, expecting `T_VARIABLE' or `'$'' in ..."
>
> until I exchanged the naming of these:
>
> $1a = 0; $1b = 0; $1c = 0; $1d = 0; $1e = 0; $1f = 0;
>
> (It won't accept $1a as a variable.) ... for these:
>
> $Onea = 0; $Oneb = 0; $Onec = 0; $Oned = 0; $Onee = 0; $Onef = 0;
>
> (Which is Absolute silliness, but I didn't create PHP !!)
>
> Now I cannot do:
>
> $Onea = $Onea + $survey->1a;
>
> without receiving:
>
> "parse error, expecting `T_STRING' or `T_VARIABLE' or `'{'' or `'$''"
>
> Can someone show me how to do this please? What the ***k am I doing
> wrong now? A post & reply is always appreciated,
>
> John
>
> --------------snip----------------------
>
> $myconnection = mysql_pconnect("localhost","","");
> mysql_select_db("",$myconnection);
>
>
> $1a = 0; $1b = 0; $1c = 0; $1d = 0; $1e = 0; $1f = 0;
>
> $rows = mysql_query("select * from bookmark_unit4 ORDER by id desc");
> while ($survey = mysql_fetch_object($rows))
> {
> //This does not work. I'm trying to ++ $1a witht he value of $survey->1a
>
> // $1a = $1a + $survey->1a;
> // $1b = $1b + $survey->1b;
> // $1c = $1c + $survey->1c;
> // $1d = $1d + $survey->1d; $1e = $1e + $survey->1e; $1f = $1f +
> $survey->1f;
> }
>
>
>
>
>
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