Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in

ukrspp21 · #1 13-Aug-2003, 03:34

Dear All

I am trying to create a form in which users can update their prefrences, Address, Phone no., contact prefrence etc. The code I am using is below. The code is very scrapy but I have been trying many things. I am trying to check whether a field in a form contains any data if so update the db. I am trying to stop the update function from empting my db.

I have a feeling that I am accessing an array variable incorrectly, but am at a lose as to how to fix. Any help would be appreciated.

Best Regards
Phil

Error:
Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in /home/sites/site112/web/user_prefs9.php on line 66

PHP Code:


<?php

require_once('incl_funs.php');
  session_start();
  do_header('Prefrences Updated', 'Prefs<br/>Updated', 4, 1);
  check_valid_user();
  //create short names
      $username = $HTTP_SESSION_VARS['valid_user'];
    $salutation =$HTTP_POST_VARS['salutation'];
    $cname = $HTTP_POST_VARS['cname'];
    $fname = $HTTP_POST_VARS['fname'];
    $company = $HTTP_POST_VARS['company'];
    $jobtitle = $HTTP_POST_VARS['jobtitle'];
    $phoneno = $HTTP_POST_VARS['phoneno'];
    $mobileno = $HTTP_POST_VARS['mobileno'];
    $faxno = $HTTP_POST_VARS['faxno'];
    $email = $HTTP_POST_VARS['email'];
    $address1 = $HTTP_POST_VARS['address1'];
    $address2 = $HTTP_POST_VARS['address2'];
    $town = $HTTP_POST_VARS['town'];
    $city = $HTTP_POST_VARS['city'];
    $county = $HTTP_POST_VARS['county'];
    $postcode = $HTTP_POST_VARS['postcode'];
    $country = $HTTP_POST_VARS['country'];
    $contact_method = $HTTP_POST_VARS['contact_method'];
    
  function clean(&$value)
 {
 $value=trim($value);
 addslashes($value);
 ucwords($value);
  };
 
 array_walk($_POST, 'clean');
  valid_email($email);
  $details = array(
         'salutation'=>$salutation,
         'cname'=>$cname,
         'fname'=>$fname,
         'company'=>$company,
         'jobtitle'=>$jobtitle,
         'phoneno'=>$phoneno,
         'mobileno'=>$mobileno,
         'faxno'=>$faxno,
         'address1'=>$address1,
         'address2'=>$address2,
         'town'=>$town,
         'city'=>$city,
         'county'=>$county,
         'postcode'=>$postcode,
         'country'=>$country,
         'contact_method'=>$contact_method
 );    
    
   
   do_db_connect();
   $db_conn = mysql_pconnect();
   mysql_select_db('username', $db_conn);
   
   reset($details);
   $no_items = count($details);
   while($element = each($details))
   {
       if(!empty($element['value']))
   {
//error is here   
$result = mysql_query("update userdata set'$element['key']' = '$element['value']' where username = '$username'", $db_conn);
   }
   } 
   
    if(!empty($email))   {
       
$result1 = mysql_query( "update users
                            set email = ('$email')
                            where username = ('$username')", $db_conn);
                            }
                            
    if (!$result && !$result1)
     {echo '<p class="body">Your details could not be updated please try again.</p>';}
    else
     {echo '<p class="body" align="center">Your details were updated</p>';}
    
do_footer();
?>

JdS · #2 13-Aug-2003, 05:13

I didn't go through all your code, so this may eliminate the error or it may not:

PHP Code:


<?php
// replace this:
// $result = mysql_query("update userdata set'$element['key']' = '$element['value']' where username = '$username'", $db_conn);

// with
$result = mysql_query( "update `userdata` set `".$element['key']."` = '".$element['value']."' where username = '$username'", $db_conn );
?>

By the way, didn't you ask something similar a few days ago, or was that someone else? :-)

ukrspp21 · #3 13-Aug-2003, 05:23

Yeah sorry I did.

Please accept my appolgises I had totally forgotten that I had done. I am kind of getting brain melt down.

Thank you for your help again.
Phil

dexthageek · #4 21-Aug-2003, 08:03

I am having a similar problem with my code on update.
I am receiving Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' whenever I attempt to execute the script.

PHP Code:


<?php 

include 'accesscontrol.php';
include_once 'common.php'; 
include_once 'db.php';

 dbConnect("test_db"); 
  $sql = "SELECT * FROM auth_users WHERE 
  UserID = '$uid' AND Password = '$pwd'"; 
  $result = mysql_query($sql); 

   if (!$result) 
     { 
      error('A database error occurred while checking your '. 
      'login details.\\n If This error persists, please '. 
      'contact admin@example.com.');
     }


$date = mysql_result($result,0,'Date_Created');
$userid = mysql_result($result,0,'UserID');
$firstname = mysql_result($result,0,'FName');
$lastname = mysql_result($result,0,'LName');
$email = mysql_result($result,0,'email');
$user_rank =  mysql_result($result,0,'rank');

?>

 <form action="<?=$_SERVER['PHP_SELF']?>" method="post">  
       <table width="300" align="center">
        <tr>
         <td width="100" align="right"><div id="storetext">Account Created: </div></td>
         <td width="100" align="left"><div id="storetext"><?=$date?></div></td>
        </tr>
        <tr>
         <td width="100" align="right"><div id="storetext">UserID: </div></td>
         <td width="100" align="left"><input type="text" name="update_userid" size="25" value="<?=$userid?>"></td> 
        </tr>
        <tr>
         <td width="100" align="right"><div id="storetext">First Name: </div></td>
         <td width="100" align="left"><input type="text" value="<? echo "$firstname" ?>" name="update_firstname" size="25"></td> 
        </tr>
        <tr>
         <td width="100" align="right"><div id="storetext">Last Name: </div></td>
         <td width="100" align="left"><input type="text" name="update_lastname" size="25" value="<? echo "$lastname" ?>"></td> 
        </tr>
        <tr>
         <td width="100" align="right"><div id="storetext">Email: </div></td>
         <td width="100" align="left"><input type="text" name="update_email" size="25" value="<? echo "$email" ?>"></td> 
        </tr>
        <tr>
         <td width="100" align="right"><div id="storetext">Rank: </div></td>
         <td width="100" align="left">
           <?
              if ($userpriv == "admin")
               {
                 ?>
                    <input type="text" name="rank" size="25" value="<? echo "$user_rank" ?>">
                 <?
               }
              else
               {
                 ?>
                  <div id="storetext"><? echo "$user_rank" ?></div>
                 <?
               }
           ?>
         </td> 
        </tr>
        <tr>
         <td colspan="2"><hr noshade><input type="Submit" value="Update"><input type="reset" value="Reset"></td>
        </tr>
       </table>
      </form>

<?php

//Assign Values from form to variables


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

//Testing to make sure variables are being passed

echo 'Testing to make  sure variables are being passed';
echo '<br>';
echo $_POST['update_userid'];
echo '<br>';
echo $update_userid;
echo '<br>';
echo $update_lastname;
echo '<br>';
echo $update_email;

dbConnect("test_db"); 
  //Error is on the line below
  $sql = "UPDATE auth_users SET UserID = $_POST['update_userid'] Fname = $_POST['update_firstname'] LName = $_POST['update_lastname'] email = $_POST['update_email'] WHERE UserID='$uid' AND Password='$pwd'";
  $result = mysql_query($sql); 

   if (!$result) 
     { 
      error('A database error occurred while checking your '. 
      'login details.\\n If This error persists, please '. 
      'contact [email]admin@example.com[/email].');
     }



 }
else
 {
  echo 'There is still a problem with your script';
 }
 
?>

If I remove my sql statement from my code the rest of the page works fine and it displays all variables they should be.

NOTE: The script is no where near final, there are other things I am gonna change, I just need this error to disappear.

Thanks in Advance

Mike

ukrspp21 · #5 21-Aug-2003, 08:11

Surround your variables like

PHP Code:


'".$var."'

it should work.

dexthageek · #6 21-Aug-2003, 08:30

Thanks, that worked on removing the error.

However now my script is throwing my DB Login Error

Here is the code for the alert its throwing up.

PHP Code:


if (!$result) 
     { 
      error('A database error occurred while checking your '. 
      'login details.\\n If This error persists, please '. 
      'contact [email]example@example.com[/email].');
     }

So that tells me my script is having a problem writing to my database.

Have any ideas based on the code I posted in my previous post.

Thanks in Advance

Mike

ukrspp21 · #7 21-Aug-2003, 08:36

Which line? first or second.

Also seperate your field groups with commas

ukrspp21 · #8 21-Aug-2003, 08:38

Also do you know anything about working with dates with Mysql and Php?

dexthageek · #9 21-Aug-2003, 08:46

thanks for the help, I think that took care of my problem.

What are you trying to do with dates?

Do you want to manipulate a TimeStamp?

Mike

ukrspp21 · #10 21-Aug-2003, 08:54

Essentially I am trying to display FTSE 100 Stock prices on my page.

I have decided to create a Mysql table which I will update daily. The primary key for this table is the date, currently a datetime field.

So I want to take todays date in php and pass it to Mysql in the form of a Sql query to return the prices.

I also what to select the previous days values to perform a change calculation.

I have a test page I am playing with at the mo, I just can't get it to return values, no errors just no matches. Thus I must assume that date format is incorrect!!

PHP Code:


$day2 = date('j m Y');
$day1 = date('Y').' '.date('m').' '.(date('j')-1);

echo $day2;
echo $day1;

$q1 = ("select * from stock where (date = '".$day1."')");
$db_conn = mysql_pconnect('localhost', , );
mysql_select_db('username', $db_conn);
$r1 = mysql_query($q1, $db_conn);
$date1 = mysql_fetch_array($r1);

echo $date1['ftse'];

Any suggestions?

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