parser error on table

[franknu]

well i was wondering if i have to echo tables when i have an if statement for examples

 

if ($num_rows!=1)
{
print "<p><b>username and/or password not found. Try
again?</b></p>";
}
else

{
echo " Welcome";



<table border="0">
<tr>

   <td valign='top'>
   <table border='0'>

 

this is giving me an error where the tables start however if i add echo"<tables"; it wont give me an error

 

Parse error: syntax error, unexpected '<' in /home/townsfin/public_html/authorization/Update_Page.php on line 98

 

[papaface]

You have HTML in PHP, you cant do that. Should be:

if ($num_rows!=1)
{
print "<p><b>username and/or password not found. Try
again?</b></p>";
}
else

{
echo " Welcome";


echo '
<table border="0">
<tr>

   <td valign='top'>
   <table border='0'>';

[franknu]

i did it that way and now i am getting an

 

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/townsfin/public_html/authorization/Update_Page.php on line 101

[papaface]

This is a simple problem.

You used 'top' instead of "top".

Should be:

if ($num_rows!=1)
{
print "<p><b>username and/or password not found. Try
again?</b></p>";
}
else

{
echo " Welcome";


echo '
<table border="0">
<tr>

   <td valign="top">
   <table border="0">';

 

[franknu]

ok, i made the changes the problem that i am having now is that  it is not going into to the database.

 

this is my code

 

<td>
<a href=\"keyword.php?keyword=true\">Add Key words</a>
			</td>
                <td>
			<a href="http://www.ccbill.com">CCBILL</a>
			</td>
              </tr>

            </table>

</center>



$sql = "SELECT * FROM business_info WHERE     

`BusinessName`= "$BusinessName"
               AND `User_Name`="$User_Name"";

$result = mysql_query($query) or die (mysql_error());
$res2 = mysql_fetch_assoc($result); 



$BusinessName

if(isset($_POST["submit"]))

{
$query="UPDATE  business_info  SET `BusinessName`= "{$BusinessName}", `Slogan`="{$Slogan}",
               `Business_Address`="{$Business_Address}", `Tel`="{$Tel}", `Website`="{$Website}",
               `Email`="{$Email}", `Fax`="{$Fax}", `type`="{$type}",
               `make`="{$make}", `Categories`="{$Categories}", `Keyword`="{$Keyword}", `Picture1`="{$Picture1}",
               `Headline`="{$Headline}", `Slogan2`="{$Slogan2}", `Description1`="{$Description1}",
               `Description2`="{$Description2}", `Description3`= "{$Description3}",
               `Picture2`="{$Picture2}", `Picture3`="{$Picture3}",
               `Password`="{$Password}" WHERE `User_Name`="{$User_Name}"";


$result = mysql_query($query) or die (mysql_error());


}




<table bgcolor="ffffff">
  <tr>
    <td>
        <table>
          <tr>

                     </tr>
        </table>
      </td>
  </tr>
  <tr>
    <td></td>
  </tr>
  <tr>
    <td>
    <table> 
        <tr>

	  <td>

	<table bgcolor="ffffff ">
<tr>
  <td>
<table>
<form action="<?php echo $_SERVER["PHP_SELF"] method="Post" ?>";> 






<tr>
    <td>
    Busiess Info
   </td>
  </tr>
  <tr>
    <td>
    <table>
        <tr>
          <td>
          Business Name
          </td>
          <td>






<input type="text" name="BusinessName"  Value="$BusinessName">

 

here is what i am displaying

 

please help

 

$sql = "SELECT * FROM business_info WHERE `BusinessName`= "$BusinessName" AND `User_Name`="$User_Name""; $result = mysql_query($query) or die (mysql_error()); $res2 = mysql_fetch_assoc($result); $BusinessName if(isset($_POST["submit"])) { $query="UPDATE business_info SET `BusinessName`= "{$BusinessName}", `Slogan`="{$Slogan}", `Business_Address`="{$Business_Address}", `Tel`="{$Tel}", `Website`="{$Website}", `Email`="{$Email}", `Fax`="{$Fax}", `type`="{$type}", `make`="{$make}", `Categories`="{$Categories}", `Keyword`="{$Keyword}", `Picture1`="{$Picture1}", `Headline`="{$Headline}", `Slogan2`="{$Slogan2}", `Description1`="{$Description1}", `Description2`="{$Description2}", `Description3`= "{$Description3}", `Picture2`="{$Picture2}", `Picture3`="{$Picture3}", `Password`="{$Password}" WHERE `User_Name`="{$User_Name}""; $result = mysql_query($query) or die (mysql_error()); } 

[franknu]

help please