parser error

franknu · November 22, 2006

can anyone help me out here, i am getting a parser error in my member page i dont see why i am getting it maybe someone can tell me why

here is the code:

[code=php:0]

<?php
$host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";

$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());

$BusinessName = (isset($_GET['BusinessName']) ? $_GET['BusinessName'] : '');
$Slogan = (isset($_GET['Slogan']) ? $_GET['Slogan']:'');
$Business_Address = (isset($_GET['Business_Address']) ? $_GET['Business_Address']:'');
$Tel = (isset($_GET['Tel']) ? $_GET['Tel']:'');
$Website = (isset($_GET['Website']) ? $_GET['Website']:'');
$Email = (isset($_GET['Email']) ? $_GET['Email']:'');
$Member_Status = (isset($_GET['Member_Status']) ? $_GET['Member_Status']:'');
$Fax =(isset($_GET['Fax']) ? $_GET['Fax']:'');
$type = (isset($_GET['type']) ? $_GET['type']:'');
$make = (isset($_GET['make']) ? $_GET['make']:'');
$Categories = (isset($_GET['Categories']) ? $_GET['Categories']:'');
$Keyword = (isset($_GET['Keyword']) ? $_GET['Keyword']:'');
$Picture1 = (isset($_GET['Picture1']) ? $_GET['Picture1']:'');
$Headline = (isset($_GET['Headline']) ? $_GET['Headline']:'');
$Slogan2 = (isset($_GET['Slogan2']) ? $_GET['Slogan2']:'');
$Description1 = (isset($_GET['Description1']) ? $_GET['Description1']:'');
$Description2 = (isset($_GET['Description2']) ? $_GET['Description2']:'');
$Description3= (isset($_GET['Description3']) ? $_GET['Description3']:'');
$Contact2 = (isset($_GET['Contact2']) ? $_GET['Contact2']:'');
$Picture2 = (isset($_GET['Picture2']) ? $_GET['Picture2']:'');
$Picture3 = (isset($_GET['Picture3']) ? $_GET['Picture3']:'');
$Picture4 = (isset($_GET['Picture4']) ? $_GET['Picture4']:'');
$User_Name = (isset($_GET['User_Name']) ? $_GET['User_Name']:'');
$Password = (isset($_GET['Password']) ? $_GET['Password']: '');
$user = (isset($_POST['User_Name']) ? $_GET['User_Name']:'');

if (isset($_POST['submit'])) {

foreach ($_POST as $key => $val) {
$$key = mysql_real_escape_string($val);
}

$q = "SELECT `Password` FROM business_info WHERE User_Name = '$User_Name'";
$sql = mysql_query($q) or die(mysql_error() . "<br />$q");
if (mysql_num_rows($sql) == 1)
{
$confPass = mysql_result($sql, 0, 'Password');

if ($Password == $confPass)
{

$query = "SELECT * FROM `business_info` WHERE `User_Name` ='$User_Name', AND
`Password`='$Password',"

}
}
echo "<table >";
echo "<tr>";
echo" <td valign='top'> ";

echo" <table>";
echo "<tr> ";

echo" <td>";
echo"Home";
echo"</td>";

echo"</tr>";
echo"<tr>";
echo"<td>";

echo"welcome $BusinessName";
echo"</td>";
echo"</tr>";
echo"</table>";
v echo"<table>";
echo"<tr> ";
echo"<td >";
echo"<table >";
echo "<tr> ";
echo"<td>";
echo"Advertise your website";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Email Marketin";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Add An email ";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Messeges";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"<td valign='top'> ";
echo"<table>";
echo" <tr>";
echo"<td>";
echo"Update my site";
echo"</td>";
echo"<td>";
echo"Add Key words";
echo"</td>";
echo"<td>";
echo"CCBILL";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"<td valign='top'>";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"</tr>";
echo"</table>";

}

} else {
// wrong user name and password
echo "Wrong User Name and Password Try Again:";
}
}
?>
[/code]

my parser error : Parse error: parse error in c:\program files\easyphp1-8\home\townsfinder\authorization\update_page.php on line 65

this is what i have on line 65

62: if ($Password == $confPass)
63: {

64 $query = "SELECT * FROM `business_info` WHERE `User_Name` ='$User_Name', AND
`Password`='$Password',"

65 echo "<table >";
66 echo "<tr>";

mb81 · November 22, 2006

You need a semicolon at the end of line 64.

franknu · November 22, 2006

I made some changes to make it easy; but i am still gettin a parser error

here is the code:

<?php
$host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";

$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
$User_Name= ($_POST['User_Name']);
$Password=($_POST['Password']);

if( User_Name=="$User_Name" || Password="$Password")
{

$query = "SELECT * FROM business_info WHERE `User_Name`== '$User_Name' AND `Password=='$Password'" ;
$result = mysql_query($query) or die (mysql_error());
$row = mysql_fetch_assoc($result);

$BusinessName= ($row['BusinessName']);
$Keyword =($row['Keyword']);
$Picture1 = ($row['Picture1']);
$Headline = ($row['Headline']);
$Slogan2 = ($row['Slogan2']);
$Description1 =($row['Description1']);
$Description2 = ($row['Description2']);
$Description3= ($row['Description3']);
$Contact2 = ($row['Contact2']);
$Picture2 = ($row['Picture2']);
$Picture3 = ($row['Picture3']);
$Business_Address=($row['Business_Address']);
$make=($row['make']);
$type=($row['type']);
$Tel=($row['Tel']);
$Website=($row['Website']);

echo "<table >";
echo "<tr>";
echo" <td valign='top'> ";

echo" <table>";
echo "<tr> ";

echo" <td>";
echo"Home";
echo"</td>";

echo"</tr>";
echo"<tr>";
echo"<td>";

echo"welcome $BusinessName";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"<table>";
echo"<tr> ";
echo"<td >";
echo"<table >";
echo "<tr> ";
echo"<td>";
echo"Advertise your website";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Email Marketin";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Add An email ";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Messeges";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"<td valign='top'> ";
echo"<table>";
echo" <tr>";
echo"<td>";
echo"Update my site";
echo"</td>";
echo"<td>";
echo"Add Key words";
echo"</td>";
echo"<td>";
echo"CCBILL";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"<td valign='top'>";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"</tr>";
echo"</table>";
}
else

{
// wrong user name and password
echo "Wrong User Name and Password Try Again:";
}

?>

My error say

Parse error: parse error in c:\program files\easyphp1-8\home\townsfinder\authorization\update_page.php on line 22

this is what i have on line 22

20.$User_Name= ($_POST['User_Name']);
21.$Password=($_POST['Password']);

22. if( User_Name=="$User_Name" || Password="$Password")
23.{

franknu · November 22, 2006

I made some changes to make it easy; but i am still gettin a parser error

here is the code:
[code=php:0]
<?php
$host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";

$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
$User_Name= ($_POST['User_Name']);
$Password=($_POST['Password']);

if( User_Name=="$User_Name" || Password="$Password")
{

$query = "SELECT * FROM business_info WHERE `User_Name`== '$User_Name' AND `Password=='$Password'" ;
$result = mysql_query($query) or die (mysql_error());
$row = mysql_fetch_assoc($result);

$BusinessName= ($row['BusinessName']);
$Keyword =($row['Keyword']);
$Picture1 = ($row['Picture1']);
$Headline = ($row['Headline']);
$Slogan2 = ($row['Slogan2']);
$Description1 =($row['Description1']);
$Description2 = ($row['Description2']);
$Description3= ($row['Description3']);
$Contact2 = ($row['Contact2']);
$Picture2 = ($row['Picture2']);
$Picture3 = ($row['Picture3']);
$Business_Address=($row['Business_Address']);
$make=($row['make']);
$type=($row['type']);
$Tel=($row['Tel']);
$Website=($row['Website']);

echo "<table >";
echo "<tr>";
echo" <td valign='top'> ";

echo" <table>";
echo "<tr> ";

echo" <td>";
echo"Home";
echo"</td>";

echo"</tr>";
echo"<tr>";
echo"<td>";

echo"welcome $BusinessName";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"<table>";
echo"<tr> ";
echo"<td >";
echo"<table >";
echo "<tr> ";
echo"<td>";
echo"Advertise your website";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Email Marketin";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Add An email ";
echo"</td>";
echo"</tr>";
echo"<tr> ";
echo"<td>";
echo"Messeges";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"<td valign='top'> ";
echo"<table>";
echo" <tr>";
echo"<td>";
echo"Update my site";
echo"</td>";
echo"<td>";
echo"Add Key words";
echo"</td>";
echo"<td>";
echo"CCBILL";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"<td valign='top'>";
echo"</td>";
echo"</tr>";
echo"</table>";
echo"</td>";
echo"</tr>";
echo"</table>";
}
else

{
// wrong user name and password
echo "Wrong User Name and Password Try Again:";
}

?>
[/code]

[code=php:0]

My error say

Parse error: parse error in c:\program files\easyphp1-8\home\townsfinder\authorization\update_page.php on line 22

this is what i have on line 22

20.$User_Name= ($_POST['User_Name']);
21.$Password=($_POST['Password']);

22. if( User_Name=="$User_Name" || Password="$Password")
[/code]

kenrbnsn · November 22, 2006

Your line 22 is syntacally wrong. What are you attempting to compare?

Do you want
[code]<?php
if( $User_Name=="User_Name" || $Password=="Password")
?>[/code]

Ken

franknu · November 23, 2006

well, i am trying to do is: if there is a user_name and password that match in the database display what isin the tables

franknu · November 23, 2006

ok, now the problem that i am havin is that it is not going through the database,or tha what it seems like because whenever the user typed passward and user name it taken the to the end of the scripts

this is the error display

garry27 · November 23, 2006

you need to do something like this:

[code]
$email = ( $_GET['email'] );
$pwd = ( $_GET['pwd'] );

// validate login
$sql = mysql_query("select * from Business_Info where User_Name='$userName' and User_Password='$pwd'");
$num_rows = mysql_num_rows($sql);
if ($num_rows >0 )
{
//a record exists and user is valid- but not validated
}
[/code]

read up on sessions to find out how to validate users.

roopurt18 · November 23, 2006

You can leave the parentheses off of statements like this:
$Keyword =($row['Keyword']);

You must always put $ in front of variable names.

When posting your code, can you please enclose it inside of [ code ] [ / code ] tags, but without the spaces in the tags.

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