PHP problem again??

Coolprogram · Sep 7, 2005

I keep getting this problem Parse error: parse error, unexpected T_CONSTANT_ENCAPSED_STRING in /home/paintman/public_html/signupck.php on line 51 here is the

Code:

$dbservertype='mysql';
$servername='localhost';
// username and password to log onto db server
$dbusername='paintman_Admin';
$dbpassword='';
// name of database
$dbname='paintman_signup';

connecttodb($servername,$dbname,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbuser,$dbpassword)
{
global $link;
$link=mysql_connect ("$servername","$dbuser","$dbpassword");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());
}
//////////////////////////////
?>
<!doctype html public "-//w3c//dtd html 3.2//en">

<html>

<head>
<title>Signing Up</title>
<meta name="GENERATOR" content="Arachnophilia 4.0">
<meta name="FORMATTER" content="Arachnophilia 4.0">
</head>

<body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080" alink="#ff0000">
<?
if(isset($todo) and $todo=="post"){

$status = "OK";
$msg="";

// if userid is less than 3 char then status is not ok
if (!isset($userid) or strlen($userid) <3){
$msg=$msg"Username should be =3 or more than 3 char length<br>";
$status= "notok";}					

if mysql_query("SELECT userid FROM plus_signup WHERE userid = '$userid'"))){
$msg=$msg."Userid already exists. Please try another one<BR>";
$status= "NOTOK";}					


if ( strlen($password) < 3 ){
$msg=$msg."Password must be more than 3 char legth<BR>";
$status= "NOTOK";}					

if ( $password <> $password2 ){
$msg=$msg."Both passwords are not matching<BR>";
$status= "NOTOK";}					


if ($agree<>"yes") {
$msg=$msg."You must agree to terms and conditions<BR>";
$status= "NOTOK";}	

if($status<>"OK"){
echo "<font face='Verdana' size='2' color=red>$msg</font><br><input type='button' value='Retry' onClick='history.go(-1)'>";
}else{
$query=mysql_query("insert into plus_signup(userid,password,email,name,sex) values('$userid','$password','$email','$name','$sex')");
echo "<font face='Verdana' size='2' color=green>Welcome, You have successfully signed up</font>";
}
}
?>


</body>

</html>

Can someone help me on this

Thank You,
-Coolprogram

user-7256 · Sep 7, 2005

Could you be so kind and comment line 51? Also, edit your post and change your [ code ] tag to a [ php ] tag.

Thanks.

EDIT: You probably forgot a semicolon at the end of the previous line. Also, I notice your code doesn't start with <?php

Crusader · Sep 7, 2005

It might be this.

$msg=$msg"Username should be =3 or more than 3 char length<br>";

$msg=$msg."Username should be =3 or more than 3 char length<br>";

Space things out lad! It's easier to look for errors that way.

Eric · Sep 7, 2005

Yes, Indeed. Space things out!!!! Much easier to read and to find any errors! Lol, anyway, I took your code, and not only cleaned it up... fixed your error (I believe)

PHP:

<?php

$dbservertype = 'mysql';
$servername = 'localhost';
$dbusername = 'paintman_Admin';
$dbpassword = '';
$dbname = 'paintman_signup';

connecttodb($servername, $dbname, $dbusername, $dbpassword);
function connecttodb($servername, $dbname, $dbuser, $dbpassword)
{
global $link;
$link = mysql_connect("$servername", "$dbuser", "$dbpassword");

if(!$link)
{
  die("Could not connect to MySQL");
}

mysql_select_db("$dbname", $link) or die ("Could not open db " .mysql_error());
}

?>
<!DOCTYPE HTML PUBLIC "-//w3c//dtd html 3.2//en">

<html>

<head>
<title>Signing Up</title>
<meta name="GENERATOR" content="Arachnophilia 4.0">
<meta name="FORMATTER" content="Arachnophilia 4.0">
</head>

<body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080" alink="#ff0000">
<?php

if(isset($todo) and $todo == "post")
{

$status = "OK";
$msg = "";

// UserID is less than 3 characters. Status is NOTOK
if (!isset($userid) or strlen($userid) <3)
{
$msg = "Username should be atleast 3 or more characters in length.<br>";
$status= "NOTOK";
}

if (mysql_query("SELECT userid FROM plus_signup WHERE userid = '$userid'"))
{
$msg = "Userid already exists. Please try another.<BR>";
$status = "NOTOK";
}

if ( strlen($password) < 3 )
{
$msg = "Password must be more than 3 characters in legth.<BR>";
$status = "NOTOK";
}

if ( $password <> $password2 )
{
$msg = "Passwords did not match.<BR>";
$status = "NOTOK";
}

if ($agree<>"yes")
{
$msg = "You must agree to terms and conditions.<BR>";
$status= "NOTOK";
}

if($status<>"OK")
{
echo "<font face='Verdana' size='2' color=red>$msg</font>
<br>
<input type='button' value='Retry' onClick='history.go(-1)'>";
}

else

{

$query = mysql_query("INSERT INTO plus_signup(userid, password, email, name, sex) VALUES('$userid','$password','$email','$name','$sex')") or die("Could not insert data " .mysql_error());

echo "<font face='Verdana' size='2' color=green>Welcome, You have successfully signed up.</font>";
}
}
?>

</body>
</html>

Coolprogram · Sep 8, 2005

There is another problem it says this Could not insert data Table 'paintman_signup.plus_signup' doesn't exist

slantednet · Sep 8, 2005

I guess the quick answer would be does the database paintman_signup exist and does it contain the table plus_signup?

Coolprogram · Sep 8, 2005

The db does but the table plus_signup does not.Okay i will see if i could fix it

slantednet · Sep 8, 2005

Yeah, either create the table, or check the database to see if you already havea table there and the name is just incorrect in teh script. I don't know if this is a roll-your-own app or something that you purchased/downloaded, so I can't be more help than that...

PHP problem again??

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