no showing values in text box

franknu · April 9, 2007

ok i have a form where the user type in user name and password to update the a form

the problem is that it is not showing values in the form when the user login

the only one that work is where it says user name and password i dont understand why

i will insert full code so u guys can see it


$User_Name=strtolower($_POST['User_Name']);
$Password=strtolower($_POST['Password']);

$usercheck = mysql_query("SELECT * FROM business_info where User_Name='$User_Name' AND Password='$Password'",$db);
$userrow = mysql_fetch_array($usercheck);
$num_rows = mysql_num_rows($usercheck);

$query = "SELECT * FROM business_info where  BusinessName='$BusinessName' AND User_Name='$User_Name' AND Password='$Password'";
$result = mysql_query($query);

while ($row = mysql_fetch_assoc($result)) {


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



if ($num_rows!=1)

{
print "<p><b>username and/or password not found. Try
again?</b></p>";
}
else

{

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

   echo"<td valign='top'>";
   echo"<table border='0'>";
        echo"<tr>";
       echo" <td bgcolor='cococo'>";
echo"<a href='index.php'>";
//100

echo"Home";
echo"<font color='cococo'>ddddddddddddddddddddd</font>
</a>";

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

echo"<tr>";


          echo "<td>welcome $BusinessName,  ";
echo"<font color='ffffff'>dddddddddddddddddddfffffffffffffffffffdd</font> </td>";
           echo "</tr>";

echo"          </table>
        <table background='../images/fondo2.jpg'  border='1' valign='top'>";

          echo"<tr>";
         echo" <td valign='top'>";
echo "<table border='1'background='../images/fondo2.jpg' valign='top'>";

echo"<tr>";
      echo"      <td>";
	echo"  <a href= 'http://entrepreneur.com'>Pictures Uploads/change</a>";

echo"  </td>
          </tr>

          <tr>
            <td>";
	echo"  <a href= 'http://entrepreneur.com'>Advertise your website</a>";
	echo"  </td>
          </tr><tr>
           <td></td></tr>
          <tr> <td>";
echo"<a href='http://69.89.27.201:2082/horde/index.php/'>Email</a>
	 </td></tr>
              <tr> <td></td> </tr>

<tr>
   <td>";

echo"<a href='http://www.nasdaq.com'>Stock Market</a></td>";
   echo" </tr>
            </table>
</td>
          <td  valign='top'>
<center><table border='1'>
<tr>"; //145
  echo"<td><a href=\'update.php?update=true\'>update</a></td>";

echo"<td><a href=\'keyword.php?keyword=true\'>Add Key words</a></td>";

                echo"<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 ("Problem with the query: <pre>$sql</pre><br>" . mysql_error());
$res2 = mysql_fetch_assoc($result);







{

$query="UPDATE  business_info  SET BusinessName= ('$BusinessName'), `Slogan`=('$_POST[$Slogan]'),
               Business_Address = ('$_POST[business_Address]'), Tel=('$_POST[Tel]'), Website= ('$_POST[Website]'),
               Email = ('$_POST[Email]'), Fax= ('$_POST[Fax]'), `type`='$type',
               make = '$make', Categories = ('$_POST[Categories]'), Keyword = ('$_POST[Keyword]'),  Picture1 = ('$_POST[Picture1]'),
               Headline = ('$_POST[Headline]'), Slogan2 = ('$_POST[slogan2]'), Description1 = ('$_POST[Description1]'),
               Description2 = ('$POST[Description2]'), Description3 =  ('$_POST[Description3]'),
               Picture2 = ('$_POST[Picture2]'), Picture3 = ('$_POST[Picture3]'),
`Password`= '$Password' WHERE User_Name = '$User_Name'";


$result = mysql_query($query) or die ("Problem with the query: <pre>$query</pre><br>" . mysql_error());

}








echo'<table bgcolor="ffffff">';

  echo'<tr><td> <table><tr> </tr>  </table>  </td></tr><tr><td></td>
  </tr><tr><td>  <table> <tr> <td><table bgcolor="ffffff">
<tr> <td>

<table> ';


//208


echo'<form action="'. $_SERVER['PHP_SELF'].'" method="post" enctype="multipart/form-data">';





echo'<tr>';

echo'<td> Busiess Info </td></tr><tr>
    <td>
    <table>
        <tr>
          <td>
          Business Name
          </td>
          <td>';
echo'<input type="text" NAME="BusinessName" value="'. stripslashes($_POST['BusinessName']).'" >';





echo '</td>';
echo'</tr> <tr> <td> Slogan </td>        <td>';

echo'<input type="text" NAME="Slogan" value="'. stripslashes($_POST['Slogan']).'" >';

echo"</td> <td> Website </td> <td>";

echo'<input type="text" name="Website" value="' . stripslashes($_POST['Website']). '" >';

echo' </td>';

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

          echo'<input type="text" name="Tel" value="' . stripslashes ($_POST['Tel']). '" >';

         echo"</td>";
          echo"<td>
          Key Words
          </td>
          <td>";
        echo'<input type="text" name="Keyword" value="' . stripslashes($_POST['Keyword']). '" >';
       echo"</td>
       </tr>
        <tr>
          <td>
          Fax
         </td>
          <td>";
          echo '<input type="text" name="Fax" value="' . stripslashes($_POST['Fax']). '">';
          echo"</td>
          <td>
         Address
          </td>
          <td>";

echo '<textarea rows="1" cols="1"  name="Business_Address">' . stripslashes($_POST['Business_Address']) . '</textarea>';





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

echo '<textarea rows="5" cols="65"  name="Categories">' . stripslashes($_POST['Categories']) . '</textarea>';


echo" </td></tr></table><table> <tr><td>Business  Webpage</td>
        </tr><tr><td>Headline</td> </tr><tr><td>";


echo '<input type="text" name="Headline" value="' . stripslashes($_POST['Headline']).'" >';
echo" </td>";
echo"  </tr>";
        echo"<tr>";
echo" <td>
Slogan
</td>
</tr>
<tr>";

   echo '<td> <input type="text" name="Slogan2" value="' . stripslashes($_POST['Slogan2']). '" >';
echo'</td></tr><tr> <td>   About US </td> </tr><tr><td>';



echo '<textarea rows="5" cols="65"  name="Description1">' . stripslashes($_POST['Description1']) . '</textarea>';

          echo'</td>
       </tr>
        <tr> <td> Products / Services
          </td>
       </tr>
        <tr>
          <td>';

echo '<textarea rows="5" cols="65"  name="Description2">' . stripslashes($_POST['Description2']) . '</textarea>';



    
          echo'</td>
        </tr>
        <tr>
          <td>
         Areas Served
         </td>
        </tr>
        <tr>
         <td>';

echo '<textarea rows="5" cols="65"  name="Description3">' . stripslashes($_POST['Description3']) . '</textarea>';



         echo'</td>
        </tr>

       <tr>
          <td>

        <table>';

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

   echo" Picture1";

echo' <input name="Picture1" type="file"    value="' . stripslashes($_POST['Picture1']).'" >';

echo'</td>';
echo'<td> ';




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

echo"<tr>
<td>";

   echo" Picture2";
	  echo" <input name='Picture2' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";


echo"<tr>
<td>";

   echo" Picture3";
	  echo" <input name='Picture3' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";
echo"<tr>
<td>";

   echo" Picture4";
	  echo" <input name='Picture4' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";
echo"<tr>
<td>";

   echo" Picture5";
	  echo" <input name='Picture5' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";
echo"<tr>
<td>";

   echo" Picture6";
	  echo" <input name='Picture6' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";
echo"<tr>
<td>";

   echo" Video";
	  echo" <input name='video' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";
echo"<tr>
<td>";

   echo"Sound";
	  echo" <input name='sound' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";

echo"<tr>
<td>";

   echo"You Tube";
	  echo" <input name='tube' type='file' value=''></td>
<td> <img src='dfdfd.jpg'></td>";
echo"</tr>";



echo"
  <tr>





          <td bgcolor='c0c0c0'>";

echo"<br>

<br>
              


User Name";


echo"<input type='text' name='User_Name' value='$User_Name'>";


               echo" Password";


echo" <input type='text' name='Password' value='$Password'>";


echo"</td>
             </tr></tr>
</table>


                 </table>
      </td>
</tr>
  <tr>
    <td>";

    echo"<input type='submit'  value='Update my database' name='submit' />";
      echo"  <input type='reset' value='Reset fields' />";

kenrbnsn · April 9, 2007

Please edit you post and surround you code with the

tags. This will make your post much easier to read.

Ken

neel_basu · April 9, 2007

Add This at the top of your code

echo "<pre>";

print_r($_POST);

echo "</pre>";

And Show the Output

franknu · April 9, 2007

i got this

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING

franknu · April 9, 2007

the add things is the value= passoword ans user name is showing the values

but not the others

franknu · April 9, 2007

hey

this

print_r($_POST);

echo "";

gave me on my display

Array ( [user_Name] => franknu [Password] => abc123 [submit] => LogMeIn )

neel_basu · April 9, 2007

SECURITY ISSUE I WANNA HIGHLIGHT THE ATTENTION OF THE MODERATORS AND ADMINISTRATORS OF PHPFREAKS IN THIS POST OF MINE

The posts is Sometime parsing some HTML Codes as HTML but its not Safe to PHPFREAKS. PLEASE Fix This. e.g. Using htmlspecialchars() while Inserting the values In the Database.

IN My above I wrote <pre></pre>But It parsed It as a HTML Code

To ~franknu Use This

echo "<pre>";
print_r($_POST);
echo "</pre>";

---------------------

<pre>This Line Is In pre

Another Line in pre

Look Although I've Used pre code blocks here

Its not showing</pre>Another line out of the pre

franknu · April 9, 2007

ok that is the new display

Array

(

[user_Name] => franknu

[Password] => abc123

[submit] => LogMeIn

)

if i type all the info it will work fine

my problems is when i log out and then try to log in again.

not info show in the value except the one that say User_Name and Password

neel_basu · April 9, 2007

Post the code for log Out

franknu · April 9, 2007

i dont have any code for log out

i did some exactly just like the user_Name AND Password and nothing

neel_basu · April 9, 2007

my problems is when i log out and then try to log in again.

So how you are logging out

franknu · April 9, 2007

just closing the window or browser i mean it updates all the info with not problem in the database, it is just when i need to log in again let say to make new changes

kenrbnsn · April 9, 2007

SECURITY ISSUE I WANNA HIGHLIGHT THE ATTENTION OF THE MODERATORS AND ADMINISTRATORS OF PHPFREAKS IN THIS POST OF MINE

That's what the

tags do which is why you should ALWAYS use them when entering code snippets.

Ken

neel_basu · April 9, 2007

Thanks

So what it shows when you try to login again.

franknu · April 9, 2007

something i think my be interesting it is when i do the post dispaly

it is not showing all the values

it show say something like this

Array

(

[businessName] => tyrtytr

[slogan] =>

[Website] =>

[Tel] =>

[Keyword] =>

[Fax] =>

[business_Address] =>

[Categories] =>

[Headline] =>

[slogan2] =>

[Description1] =>

[Description2] =>

[Description3] =>

[user_Name] => franknu

[Password] => abc123

[submit] => Update my database

)

franknu · April 9, 2007

when i try to login again it shows not value in the text box

neel_basu · April 9, 2007

Cause these Fields Are Blank ??

[slogan]

[Website]

[Tel]

[Keyword]

[Fax]

[business_Address]

[Categories]

[Headline]

[slogan2]

[Description1]

[Description2]

[Description3]

this fields were not filled Up in the HTML form

franknu · April 9, 2007

well it should show the data that is in the database when i login

but it is not even showing them. i mean if put values in there i will go right into the database...

and it will show it for example

Array

(

[businessName] => tyrtytr

[slogan] => rtytry

[Website] => dfdfdf

[Tel] => dfd

[Keyword] => fddf

[Fax] => df

[business_Address] => df

[Categories] => fdf

[Headline] => df

[slogan2] => dfd

[Description1] => df

[Description2] => dfd

[Description3] => dfd

[user_Name] => franknu

[Password] => abc123

[submit] => Update my database

)

that is what shows if i typew in values and hit update the database bottom but let say i close the window and try to log in again the it wont show me the values that are in the databse exept for the user_name and password

neel_basu · April 9, 2007

Honestly I was being Confused. I'vent go through all your long codes.first of all let me make sure wheather what I am thinking is correct or not ??

Your script Is for updating the database. Its first updating the database and the select * from table to show all the update Values ??

am I right ??

and Please Clarify Whats your problem.

franknu · April 9, 2007

yes that is right,...

and my problem is that after i update all the info and then i login again it is not show any values which means that if i hit the update bottom i will delete all the info that is on the database. The odd Thing is that it shows the value for User_Name and Password..

it is just not showing it for all the other stuff

boo_lolly · April 9, 2007

you're not going to be able to logout successfully by simply closing your browser. every browser handles it differently. my suggestion would be to logout by using a link that will unset the session data. and on each page, you should be checking to see if the user is logged in by using something like:

if(!isset($_SESSION['username'] || !isset($_SESSION['password'])){ die('you do not have access to this page. please login'); }

of course, this is only after you've verified their username and password with your sql query and logged them in successfully.

neel_basu · April 9, 2007

Replace

if ($num_rows!=1)

{
print "<p><b>username and/or password not found. Try
again?</b></p>";
}

By

if ($num_rows == 0 || $num_rows > 1)

{
print "<p><b>username and/or password not found. Try
again?</b></p>";
}

neel_basu · April 9, 2007

{

$query="UPDATE  business_info  SET BusinessName= ('$BusinessName'), `Slogan`=('$_POST[$Slogan]'),
               Business_Address = ('$_POST[business_Address]'), Tel=('$_POST[Tel]'), Website= ('$_POST[Website]'),
               Email = ('$_POST[Email]'), Fax= ('$_POST[Fax]'), `type`='$type',
               make = '$make', Categories = ('$_POST[Categories]'), Keyword = ('$_POST[Keyword]'),  Picture1 = ('$_POST[Picture1]'),
               Headline = ('$_POST[Headline]'), Slogan2 = ('$_POST[slogan2]'), Description1 = ('$_POST[Description1]'),
               Description2 = ('$POST[Description2]'), Description3 =  ('$_POST[Description3]'),
               Picture2 = ('$_POST[Picture2]'), Picture3 = ('$_POST[Picture3]'),
`Password`= '$Password' WHERE User_Name = '$User_Name'";


$result = mysql_query($query) or die ("Problem with the query: <pre>$query</pre><br>" . mysql_error());

}

This '{' and '}' Is for what ??

I didn't find any if or else or while or any other construct before this code block

franknu · April 9, 2007

ok. i Deleted the

{

}

i am getting the same result

however that is not where i think the problem is because this update the database and it is doing its job

it is when i login that it is not showing the values from the database

neel_basu · April 9, 2007

I am going through your script line by line. and Posting the errors I find.

Look Over here

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

Closing Braces Of else e.g. '}' of else not found

Use this instead of teh previous one

$usercheck = mysql_query("SELECT * FROM business_info where User_Name='$User_Name' AND Password='$Password'",$db)or die(mysql_error());

and

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

Sign In

no showing values in text box

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