Help ! Password problem unexpected $end

Discussion in 'PHP' started by johnnh, May 4, 2007.

  1. #1
    Would be greatfull if anyone can advise. I am trying to put a secure logon that opens a connection and then populates a form once a password has been submitted but i keep getting 'Parse error: parse error, unexpected $end in update.php on line 56'

    Can someone advise before i loose all hope!

    One virtual pint available, or probably a barrel

    Regards

    <html>
    <head>
    <title>Update</title>
    </head>
    <body>


    <?php if (!isset($PHP_AUTH_USER)) {

    // If empty, send header causing dialog box to appear

    header('WWW-Authenticate: Basic realm="My Private Stuff"');
    header('HTTP/1.0 401 Unauthorized');
    echo 'Authorization Required.';
    exit;

    } else if (isset($PHP_AUTH_USER)) {

    if (($PHP_AUTH_USER != "a") || ($PHP_AUTH_PW != "ab")) {

    header('WWW-Authenticate: Basic realm="My Private Stuff"');
    header('HTTP/1.0 401 Unauthorized');
    echo 'Authorization Required.';
    exit;

    } else {


    //connect to database
    $id = $HTTP_GET_VARS['pageid'];

    $dbConnection = @mysql_connect ("localhost","root","") OR die ('could not connect:'.mysql_error());

    // select the database to use

    @mysql_select_db ("accessibility") OR die ('could not connect to database:'.mysql_error());
    $SQL = "INSERT INTO subscribers (Firstname, Surname, Email)";

    $result = mysql_query ("SELECT * FROM content WHERE id ='$id'") or die ("could not retrieve");

    $update = mysql_query ("UPDATE content SET WHERE id ='$id'") or die ("could not retrieve");

    $database_record = mysql_fetch_row($result);
    //$database_record = $database_record[1]; // mysql_fetch_row returns an array. we only want the content so we just set it excluseively.

    }?>

    <form action="updated.php" method="post">
    <?php echo $id;?></br>
    content a<input type="text" name="contenta" value="<?php Echo $database_record [1]; ?>"><br>
    CONTENT B<input type="text" name="contentb" value="<?php Echo $database_record [2]; ?>"><br>
    Image a<input type="text" name="iamgea" value="<?php Echo $database_record [3]; ?>"><br>
    Image b<input type="text" name="imageb" value="<?php Echo $database_record [4]; ?>"><br>
    script<input type="text" name="script" value="<?php Echo $database_record [5]; ?>"><br>
    Links<input type="text" name="links" value="<?php Echo $database_record [6]; ?>"><br>
    <input type="submit" value="Update">
    </form>

    </body>
    </html>

    ?were did it all go wrong
     
    johnnh, May 4, 2007 IP
  2. sky22

    sky22 Guest

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    #2
    I think it's because
    } else if (isset($PHP_AUTH_USER)) {
    PHP:
    is missing the end } this should work:

    <html>
    <head>
    <title>Update</title>
    </head>
    <body>
    
    
    <?php
    if (!isset($PHP_AUTH_USER)) {
    
    	// If empty, send header causing dialog box to appear
    
    	header('WWW-Authenticate: Basic realm="My Private Stuff"');
    	header('HTTP/1.0 401 Unauthorized');
    	echo 'Authorization Required.';
    	exit;
    
    } else if (isset($PHP_AUTH_USER)) {
    
    	if (($PHP_AUTH_USER != "a") || ($PHP_AUTH_PW != "ab")) {
    
    		header('WWW-Authenticate: Basic realm="My Private Stuff"');
    		header('HTTP/1.0 401 Unauthorized');
    		echo 'Authorization Required.';
    		exit;
    
    	} else {
    
    
    		//connect to database
    		$id = $HTTP_GET_VARS['pageid'];
    
    		$dbConnection = @mysql_connect ("localhost","root","") OR die ('could not connect:'.mysql_error());
    
    		// select the database to use
    
    		@mysql_select_db ("accessibility") OR die ('could not connect to database:'.mysql_error());
    		$SQL = "INSERT INTO subscribers (Firstname, Surname, Email)";
    
    		$result = mysql_query ("SELECT * FROM content WHERE id ='$id'") or die ("could not retrieve");
    
    		$update = mysql_query ("UPDATE content SET WHERE id ='$id'") or die ("could not retrieve");
    
    		$database_record = mysql_fetch_row($result);
    		//$database_record = $database_record[1]; // mysql_fetch_row returns an array. we only want the content so we just set it excluseively.
    	}
    }?>
    
    <form action="updated.php" method="post">
    <?php echo $id;?></br>
    content a<input type="text" name="contenta" value="<?php Echo $database_record [1]; ?>"><br>
    CONTENT B<input type="text" name="contentb" value="<?php Echo $database_record [2]; ?>"><br>
    Image a<input type="text" name="iamgea" value="<?php Echo $database_record [3]; ?>"><br>
    Image b<input type="text" name="imageb" value="<?php Echo $database_record [4]; ?>"><br>
    script<input type="text" name="script" value="<?php Echo $database_record [5]; ?>"><br>
    Links<input type="text" name="links" value="<?php Echo $database_record [6]; ?>"><br>
    <input type="submit" value="Update">
    </form>
    
    </body>
    </html> 
    PHP:
     
    sky22, May 4, 2007 IP
  3. johnnh

    johnnh Peon

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    #3
    Thanks for taking a look

    Unfortunately im still gettin parse error even with the code that you provided?
     
    johnnh, May 5, 2007 IP