Error message: Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE),

[dgunning]

I'm getting the above error message and I can't figure out what the problem is. Here's some of my code:

PHP Code:

$db = mysql_connect("dgunning.db.8219830.hostedresource.com", "*****", "******");
mysql_select_db("dgunning", $db) or die(mysql_errno() . ": " . mysql_error() . "<br>");
$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET['stnum']) AND op.direction = htmlspecialchars($_GET['dir']) AND 
        op.street = htmlspecialchars($_GET['street']);" 


I'm getting the error on the $sql= line. I'm pretty new to using php, so would really appreciate any help. thanks!

[elhippie]

got it, read it on another form, hopefully it works

$db = mysql_connect("dgunning.db.8219830.hostedresource.com", "*****", "******");
mysql_select_db("dgunning", $db) or die(mysql_errno() . ": " . mysql_error() . "<br>");
$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch
WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET[stnum]) AND op.direction = htmlspecialchars($_GET[dir]) AND
op.street = htmlspecialchars($_GET[street]);"


the fix is with the ($_GET[]) functions, you gotta take away the ' ' around them

[dgunning]

Thanks for the response. After making the change you suggest, I now get this error message:

Parse error: syntax error, unexpected '$result' (T_VARIABLE)

[elhippie]

are you accessing another page using include (); cause its saying that $result is in the wrong spot. or your missing a ;, i'd have to see more code to help you

[dgunning]

Originally Posted by elhippie

are you accessing another page using include (); cause its saying that $result is in the wrong spot. or your missing a ;, i'd have to see more code to help you

$result is in the next line of code, it's:

PHP Code:

$result = safe_query($sql); 


and here's safe query:

PHP Code:

function safe_query ($query = "")
{
    if (empty($query)) {return FALSE; }
    $result = mysql_query($query)
        or die("ack! query failed: "
            ."<li>errorno=".mysql_errno()
            ."<li>error=".mysql_error()
            ."<li>query=".$query
        );
    return $result;
} 


[elhippie]

Originally Posted by dgunning

$result is in the next line of code, it's:

PHP Code:

$result = safe_query($sql); 


and here's safe query:

PHP Code:

function safe_query ($query = "")
{
    if (empty($query)) {return FALSE; }
    $result = mysql_query($query)
        or die("ack! query failed: "
            ."<li>errorno=".mysql_errno()
            ."<li>error=".mysql_error()
            ."<li>query=".$query
        );
    return $result;
} 


the error is here

PHP Code:

$db = mysql_connect("dgunning.db.8219830.hostedresource.com", "*****", "******");
mysql_select_db("dgunning", $db) or die(mysql_errno() . ": " . mysql_error() . "<br>");
$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET['stnum']) AND op.direction = htmlspecialchars($_GET['dir']) AND 
        op.street = htmlspecialchars($_GET['street'])";
 #HERE IS THE ERROR SWITCH THE ';' and the " around 


[dgunning]

Now I'm back to the original error message. Here's how my code stands now, this is from the beginning of this file.

PHP Code:

include('basic.php');
// database connection stuff
$db = mysql_connect("dgunning.db.8219830.hostedresource.com", "*****", "*****");
mysql_select_db("dgunning", $db) or die(mysql_errno() . ": " . mysql_error() . "<br>");


$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET['stnum']) AND op.direction = htmlspecialchars($_GET['dir']) AND 
        op.street = htmlspecialchars($_GET['street'])";
        
$result = safe_query($sql); 


Thanks for your efforts, any other suggestions?

[elhippie]

Originally Posted by dgunning

Now I'm back to the original error message. Here's how my code stands now, this is from the beginning of this file.

PHP Code:

include('basic.php');
// database connection stuff
$db = mysql_connect("dgunning.db.8219830.hostedresource.com", "*****", "*****");
mysql_select_db("dgunning", $db) or die(mysql_errno() . ": " . mysql_error() . "<br>");


$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET'stnum]) AND op.direction = htmlspecialchars($_GET[dir]) AND 
        op.street = htmlspecialchars($_GET[street])";
        
$result = safe_query($sql); 


Thanks for your efforts, any other suggestions?

lol im srry i copied the first php code remove the ' from the $_GET functions

PHP Code:

include('basic.php');
// database connection stuff
$db = mysql_connect("dgunning.db.8219830.hostedresource.com", "*****", "*****");
mysql_select_db("dgunning", $db) or die(mysql_errno() . ": " . mysql_error() . "<br>");


$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET[stnum]) AND op.direction = htmlspecialchars($_GET[dir]) AND 
        op.street = htmlspecialchars($_GET[street])";
        
$result = safe_query($sql); 


[tangoforce]

PHP Code:

$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = htmlspecialchars($_GET['stnum']) AND op.direction = htmlspecialchars($_GET['dir']) AND 
        op.street = htmlspecialchars($_GET['street'])"; 


Should be:

PHP Code:

$sql = "SELECT op.name, op.stnum, op.direction, op.street, arch.architect, op.yearbuilt, op.url, op.id FROM op, arch 
        WHERE op.archno = arch.archno AND op.stnum = '$_GET[stnum]' AND op.direction = '$_GET[dir]' AND 
        op.street = '$_GET[street]'"; 


Inside a "double quoted" string, you do not use single['quotes'] in your arrays. This will cause the error you have been complaining of. Also as far as I'm aware, mysql doesn't have that htmlspecialchars() function. In php again, inside your "double quoted" string, you cannot use a function. It IS a PHP function so you must use it in PHP and not inside a string:

PHP Code:

$_GET['street'] = htmlspecialchars($_GET['street']); 


[dgunning]

That did it, the query is executing now. I have some additional issues later on in the code, but I'll see if I can get them working myself before asking for any additional help. Thanks a lot!