Parse error: parse error, unexpected T_STRING

Richardknox · #1 18-Nov-2006, 19:49

Hello,

I am slowly learning to use PHP for the website that I am putting together. So, please understand that I am a newbee to PHP, and MySql.

I am also creating this script from other bits and pieces from other scripts, so that might be part of my problem.

What I am attempting to accomplish here, is to update a variable by adding one to it, and then sending it back to the database to be updated there, before going to the website that the user selected.

This is the error that I am getting...

Parse error: parse error, unexpected T_STRING in /home/content/R/i/c/RichardAKnox/html/ace.php on line 43

This is my script:

PHP Code:


<?php
//Connect To Database
$hostname="mysql247.secureserver.net";
$username="RichardAKnox";
$password="password";
$dbname="maindatabase";
$usertable="wcdatabase";

$dblink = mysql_connect($hostname,$username, $password) OR DIE ("Unable to connect to database! Please try again later.");

mysql_select_db($dbname, $dblink);


$id=$_GET['id'];

$query = "SELECT * FROM wcdatabase WHERE id = ($id) LIMIT 0, 30 ";


$result = mysql_query($query,$dblink);

$row = mysql_fetch_assoc($result);

$viewed = $row["viewed"];
                         
$row["viewed"]++;
    
$viewed = $row["viewed"];

Line 43 error ---->>>  $query = UPDATE wcdatabase SET viewed = $viewed;




// Redirect to the link URL

Header("Location: $link");
exit();

?>

I used to have a few echo statements in this script to verify that certain aspects of the script worked. But, as soon as I added the Line 43 portion, then the script stopped working, and gave me the errors listed.

Thanks for any help,

Richard

TreyAU21 · #2 21-Dec-2006, 10:15

The query that you want to run should be stored as a string...

PHP Code:


// Yours
$query = UPDATE wcdatabase SET viewed = $viewed;

//CORRECT WAY(S)... Notice the quotations.
$query = "UPDATE wcdatabase SET viewed = $viewed";
// OR
$query = "UPDATE wcdatabase SET viewed = " . $viewed;

Look at how you have the SELECT query above in quotations... you need the same thing for the update query... including actually calling the mysql_query() function call to actually query the db.

Curtiss · #3 19-May-2007, 12:27

Im getting an error like that also, and can not figure out why i am also new to PHP and Mysql. The line that i am having the error on is the one that says VALUES ('".$_POST['username']."', '".$_POST['password']."')";
thanks for any help

PHP Code:


<title>login</title>

</head>

<body><?php 
// Connects to your Database 
mysql_connect("host", "user", "pass") or die(mysql_error()); 
mysql_select_db("EGAlive") or die(mysql_error()); 

//This code runs if the form has been submitted
if (isset($_POST['submit'])) { 

//This makes sure they did not leave any fields blank
if (!$_POST['username'] | !$_POST['pass'] | !$_POST['pass2'] ) {
die('You did not complete all of the required fields');
}

// checks if the username is in use
if (!get_magic_quotes_gpc()) {
$_POST['username'] = addslashes($_POST['username']);
}
$usercheck = $_POST['username'];
$check = mysql_query("SELECT username FROM users WHERE username = '$usercheck'") 
or die(mysql_error());
$check2 = mysql_num_rows($check);

//if the name exists it gives an error
if ($check2 != 0) {
die('Sorry, the username '.$_POST['username'].' is already in use.');
}

// this makes sure both passwords entered match
if ($_POST['pass'] != $_POST['pass2']) {
die('Your passwords did not match. ');
}

// here we encrypt the password and add slashes if needed
$_POST['pass'] = md5($_POST['pass']);
if (!get_magic_quotes_gpc()) {
$_POST['pass'] = addslashes($_POST['pass']);
$_POST['username'] = addslashes($_POST['username']);
}

// now we insert it into the database $insert = "INSERT INTO users (username, password)
VALUES ('".$_POST['username']."', '".$_POST['password']."')";[/color]
$add_member = mysql_query($insert);
?>

<!-- Now we let them know if their registration was successful -->
<h1>Registered</h1>
<p>Thank you, you have registered - you may now login</a>.</p>
<?php 
} 
else 
{    
?>

<!-- This is what they see before they have registered -->
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<table border="0">
<tr><td>Username:</td><td>
<input type="text" name="username" maxlength="60">
</td></tr>
<tr><td>Password:</td><td>
<input type="password" name="pass" maxlength="10">
</td></tr>
<tr><td>Confirm Password:</td><td>
<input type="password" name="pass2" maxlength="10">
</td></tr>
<tr><th colspan=2><input type="submit" name="submit" value="Register"></th></tr> </table>
</form>

<?php 
} 
?>



</body> 
</html>

TurboPT · #4 19-May-2007, 14:42

Bring the assignment to $insert to the next line.

If you look back at the formatted code in your post, you should see the problem, it looks like it is part of the comment.

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