Not sure how to fix PHP error -

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i have created php script , keep on getting "wsod". have turned on error reporting not appear working. decided run script through online php checker , gave me error: "parse error: syntax error, unexpected 'index' (t_string) in code on line 6". not sure how solve , hoping able me.

<?php  if ($username && $userid) {     echo "you logged in <b>$username</b>."; } else {     $form = "<form action="index.php" method="post">         <table>           <tr>             <td>username:</td>             <td><input type="text" name="username"/></td>           </tr>           <tr>             <td>password:</td>             <td><input type="password" name="password"/></td>           </tr>           <tr>             <td><input type="submit" name="loginbutton" value="login"/></td>           </tr>           <tr>             <td><a href="register.php">register</a><a href="forgotpass.php">forgot password?</a></td>           </tr>         </table>         </form>";      if ($_post['loginbutton']) {         $user = $_post['username'];         $password = $_post['password'];          if ($username) {              if ($password) {                 require("connect.php");                 $password = md5 (md5("fkjhrfcdsj".$password."rehfkjrfd"));                 $query = mysql_query("select * members username ='$username'");                 $numrows = mysql_num_rows($query);                  if ($numrows == 1) {                     $row = mysql_fetch_assoc($query);                     $dbid = $row ['id'];                     $dbuser = $row['username'];                     $dbpass = $row['password'];                     $dbactive = $row['active'];                      if ($password == $dbpass) {                          if ($dbactive == 1) {                             $_session['userid'] = $dbid;                             $_session['username'] = $dbuser;                             echo "you have been logged in <b>$dbuser</b>."} else echo "please activate account login. $form";                     } else echo "you did not enter correct password. $form";                 } else echo "the username entered not found. $form";                 mysql_close();             } else echo "you must enter password. $form";         } else    echo "you muse enter username. $form";     } else echo "$form"; }  ?>

you have 

$form = "<form action="index.php" method="post">

convert to

$form = "<form action=\"index.php\" method=\"post\">

this because you're wanting include html double quotes inside php double quote breaking php string creation. escaping slashes resolve issue.

you this

$form = '<form action="index.php" method="post">'

and use single quotes define php string stops injecting variables later on without first breaking out of string.

so using php double quotes do

$form = "<form action=\"$anotherphpvarhere\" method=\"$phpvarhere\">"

however using single quotes you'd have this

$form = '<form action="'.$anotherphpvarhere.'" method="'.$phpvarhere.'">'

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