php error plz help

**Erez** · Jan 4, 2005 09:05

hey im working on a registration page and for some rison it doesnt work , plz help .

Code:

 <?php
  include "notice.html";
  include "connection.php";



 if (!isset($_POST["insert_button"]))

  {

     ?>
   <FORM action="<?= $_SERVER["notice.php"]; ?>" method="POST">
      <TABLE border=0 cellpadding=7 cellspacing=7>
         <TR>
            <br><br><br><br>
            <TD><font color="#ffbc8c83">account:</font></TD>
            <TD><INPUT type="TEXT" name="account" size=20 maxlength=20 value="<?= $_POST["account"]; ?>"></TD>
         </TR>
         <TR>
            <TD><font color="#ffbc8c83">password:</font></TD>
            <TD><INPUT type="password" name="password" size=20 maxlength=20 value="<?= $_POST["password"]; ?>"></TD>
         </TR>
         <TR>
            <TD><font color="#ffbc8c83">e-mail:</font></TD>
            <TD><INPUT type="TEXT" name="e-mail" size=30 maxlength=50 value="<?= $_POST["e-mail"]; ?>"></TD>
         </TR>
         <TR>
            <TD><font color="#ffbc8c83">age:</font></TD>
            <TD><INPUT type="TEXT" name="age" size=2 maxlength=2 value="<?= $_POST["age"]; ?>"></TD>
         </TR>
         <TR>
            <TD align="right"><INPUT type="SUBMIT" name="insert_button" value="register"></TD>
            <TD align="left"><INPUT type="RESET" value="reset"></TD>
         </TR>
      </TABLE>
   </FORM>
   <?
  }

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

  {
     $account=$_POST["account"];
     $password=$_POST["password"];
     $e_mail=$_POST["e-mail"];
     $age=$_POST["age"];


     $grade_insert_errors="";

     if ($account=="")
        $grade_insert_errors.="<br>account name isnt good.<BR>";

     if ($password=="")
        $grade_insert_errors.="password just isnt right.<BR>";

     if ($e_mail=="")
        $grade_insert_errors.="type a working e-mail plz.<BR>";

     if ($age=="")
        $grade_insert_errors.="age isnt good.<BR>";

     if ($grade_insert_errors=="")

     {

        $account=$account;
        $password=$password;
        $e_mail=$e_mail;
        $age=$age;

        $grade_insert_query="INSERT INTO ".$reg." (account".(($account!="")?",account":"")."".(($password!="")?",pass":"")."".(($e_mail!="")?",E-mail":"")."".(($age!="")?",age":"").")  VALUES (";
        $insert_query.=($account!="")?",'$account'":"";
        $insert_query.=($password!="")?",'$password'":"";
        $insert_query.=($e_mail!="")?",'$e-mail'":"";
        $insert_query.=($age!="")?",'$age'":"";
        $insert_query.=")";
        $result=@mysql_query ($grade_insert_query,$grade_con);

        if (mysql_affected_rows()==1)

        {
           echo "<BR><B><font color=orange>account made sucsesfully</font></B><BR>";

        }
        else

        {
           echo "<BR><br><B><font color=orange>account could not be made</font></B><BR>";
           unset($_POST["insert_button"]);

        }
     }
     else

     {
        echo "<BR><br><FONT color=\"#ffbc8c83\">".$grade_insert_errors."</FONT><BR>";
        unset($_POST["insert_button"]);

     }
  }
?>

i belive i have an error here :

Code:

$grade_insert_query="INSERT INTO ".$reg." (account".(($account!="")?",account":"")."".(($password!="")?",pass":"")."".(($e_mail!="")?",E-mail":"")."".(($age!="")?",age":"").")  VALUES (";

OR

Code:

$result=@mysql_query ($grade_insert_query,$grade_con);

if (mysql_affected_rows()==1)

how do i know it doesnt work ? it doesnt show me "account made sucssesfully"

**swdev** · Jan 4, 2005 09:17

From a quick look, it seems that you haven't defined $reg - which should be the name of the table you want to insert your values into.

Also, I would test the value of $result, if false, then display the results of mysql_error(), and echo the value of $grade_insert_query before you run the query.

**Erez** · Jan 4, 2005 09:24

is this ok ?:

Code:

  $grade_insert_query="INSERT INTO ".$grade_reg_table." (account".(($account!="")?",account":"")."".(($password!="")?",pass":"")."".(($e_mail!="")?",E-mail":"")."".(($age!="")?",age":"").")  VALUES (";
        $insert_query.=($account!="")?",'$account'":"";
        $insert_query.=($password!="")?",'$password'":"";
        $insert_query.=($e_mail!="")?",'$e-mail'":"";
        $insert_query.=($age!="")?",'$age'":"";
        $insert_query.=")";
        $result=@mysql_query ($grade_insert_query,$grade_con);

**swdev** · Jan 4, 2005 09:30

Try it and see

But once again, have you defined $grade_reg_table anywhere?

Also, try my suggestions about echo the SQL command before executing it - makes it easier to spot errors and also check the $result variable.

**Erez** · Jan 4, 2005 09:37

ok sorry im a noob :/ i have no idia how to define the reg table ... and i've done the part of posting the stuff it sandes to the table ...

**swdev** · Jan 4, 2005 09:40

We were all noobs once upon a time

What is the name of the table that you want to insert that data into? Whatever that name is, that is what you shoule set $grade_reg_table to, or just hard-code the table name into the query string.

e.g

PHP Code:


 $grade_reg_table = 'GradeTable';

would be good enough, if your table was called GradeTable.

**Erez** · Jan 4, 2005 09:42

ok here is what he posted me when i told him to post what he is going to send :

Code:

INSERT INTO reg (reg,Admin,yoman,frerez@hotmail.com,15) VALUES (

first i duno why he added the "reg" in the start ...
and reg is the name of the table
the rest is exacly the info i entered ....
and this is the code :

Code:

        if ($grade_insert_errors=="")

     {

        $account=$account;
        $password=$password;
        $e_mail=$e_mail;
        $age=$age;
        $grade_reg_table='reg';

        $grade_insert_query="INSERT INTO ".$grade_reg_table." (reg".(($account!="")?",$account":"")."".(($password!="")?",$password":"")."".(($e_mail!="")?",$e_mail":"")."".(($age!="")?",$age":"").")  VALUES (";
        $insert_query.=($account!="")?",'$account'":"";
        $insert_query.=($password!="")?",'$password'":"";
        $insert_query.=($e_mail!="")?",'$e-mail'":"";
        $insert_query.=($age!="")?",'$age'":"";
        $insert_query.=")";
        $result=@mysql_query ($grade_insert_query,$grade_con);

        if (mysql_affected_rows()==1)

        {
           echo "<BR><B><font color=orange>account made sucsesfully</font></B><BR>";

        }
        else

        {
           echo "<BR><br><B><font color=orange>account could not be made</font></B><BR>";
           unset($_POST["insert_button"]);
           echo "<BR><br><FONT color=\"#ffbc8c83\">".$grade_insert_query."</FONT><BR>";
           echo "<BR><br><FONT color=\"#ffbc8c83\">".$grade_con."</FONT><BR>";

        }

**swdev** · Jan 4, 2005 09:50

Ok - the syntax for an INSERT SQL statement is

INSERT INTO table_name (col1, ...) VALUES (val1, ...)

where

table_name is the name of the table you want to store the values
col1 ... are the names of the columns where you want the values stored
val1... are the values to be stored

so in your case, guessing at the column names, you want something that looks like

INSERT INTO GradeTable (AccountID, Password, EMail, Age) VALUES ('account', 'Admin', 'frerez@hotmail.com', 15);

Note very carfully which values are surrounded by single quotes.

Hope this helps

**Erez** · Jan 4, 2005 10:04

sound cool but is the result

Code:

$result=@mysql_query ($grade_insert_query,$grade_con);

i havent placed any $grade_con
its nothing ... and i got an error on this line:

Code:

$grade_insert_query=INSERT INTO reg (account,password,e-mail,age) VALUES ($account,$password,$e_mail,$age);

any idia what's wrong?

**Erez** · Jan 4, 2005 10:17

made some changes ... didnt help thought ... still error in the first line here :

Code:

        $grade_insert_query= INSERT INTO reg (account,password,e-mail,age) VALUES ('$account','$password','$e_mail',$age);
        $insert_query.=($account!="")?",'$account'":"";
        $insert_query.=($password!="")?",'$password'":"";
        $insert_query.=($e-mail!="")?",'$e_mail'":"";
        $insert_query.=($age!="")?",'$age'":"";
        $insert_query.=")";
        $result=@mysql_query ($grade_insert_query,$grade_con);

and btw to understand the uper info know this :

Code:

   ?>
   <FORM action="<?= $_SERVER["notice.php"]; ?>" method="POST">
      <TABLE border=0 cellpadding=7 cellspacing=7>
         <TR>
            <br><br><br><br>
            <TD><font color="#ffbc8c83">account:</font></TD>
            <TD><INPUT type="TEXT" name="account" size=20 maxlength=20 value="<?= $_POST["account"]; ?>"></TD>
         </TR>
         <TR>
            <TD><font color="#ffbc8c83">password:</font></TD>
            <TD><INPUT type="password" name="password" size=20 maxlength=20 value="<?= $_POST["password"]; ?>"></TD>
         </TR>
         <TR>
            <TD><font color="#ffbc8c83">e-mail:</font></TD>
            <TD><INPUT type="TEXT" name="e-mail" size=30 maxlength=50 value="<?= $_POST["e-mail"]; ?>"></TD>
         </TR>
         <TR>
            <TD><font color="#ffbc8c83">age:</font></TD>
            <TD><INPUT type="TEXT" name="age" size=2 maxlength=2 value="<?= $_POST["age"]; ?>"></TD>
         </TR>
         <TR>
            <TD align="right"><INPUT type="SUBMIT" name="insert_button" value="register"></TD>
            <TD align="left"><INPUT type="RESET" value="reset"></TD>
         </TR>
      </TABLE>
   </FORM>
   <?

**swdev** · Jan 4, 2005 10:20

try this code

PHP Code:


  if ('' == $grade_insert_errors)
  {
    // make sure strings are quoted and escaped
    $account = '\''. mysql_real_escape_string($account) . '\'';
    $password = '\''. mysql_real_escape_string($password) . '\'';
    $e_mail = '\''. mysql_real_escape_string($e_mail) . '\'';
    $age = $age;
    $grade_reg_table = 'reg';
  
    $grade_insert_query = 'INSERT INTO '
                      . $grade_reg_table
                   . ' (account, password, email, age)'
                      . ' VALUES ('
                      . $account
                      . ', ' . $password
                      . ', ' . $email
                      . ', ' . $age
                        . ')'
                        ;
  
    // START DIAGNOSTIC - REMOVE WHEN WORKING
    echo 'Executing ' . $grade_insert_query . '<br />';
    // END DIAGNOSTIC - REMOVE WHEN WORKING
  
    $result = @mysql_query ($grade_insert_query,$grade_con);
    if (false == $result)
    {
      echo ' Failed to execute ' . $grade_insert_query . ' due to ' . mysql_error() . '<br />';
    }
    else
    {
      echo 'Account created <br />';
    }

**Erez** · Jan 4, 2005 10:30

i got this msg :

Code:

Executing INSERT INTO reg (account, password, e-mail, age) VALUES ('Admin', 'yoman', 'frerez@hotmail.com', 15)
Failed to execute INSERT INTO reg (account, password, e-mail, age) VALUES ('Admin', 'yoman', 'frerez@hotmail.com', 15) due to

err was the line suppuse to end like this ? "due to" ..... nothing else was writen m8 ...

**swdev** · Jan 4, 2005 10:42

ok - change these 2 lines

PHP Code:


     $result = @mysql_query ($grade_insert_query,$grade_con);
     if (false == $result)

to

PHP Code:


     $result = mysql_query ($grade_insert_query,$grade_con);
 // START OF DIAGNOSTIC - REMOVE WHEN WORKING
 echo 'Result = ' . $result . '<br />';
 // END OF DIAGNOSTIC - REMOVE WHEN WORKING
 
     if (false === $result)

The error line should display the MySQL error as well as the query that caused the error

**Erez** · Jan 4, 2005 10:46

ok now this is wierd:

Code:

Executing INSERT INTO reg (account, password, e-mail, age) VALUES ('Admin', 'yoman', 'frerez@hotmail.com', 15)

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/www/somatnt.superihost.com/notice.php on line 90
Result = 
Account created

**swdev** · Jan 4, 2005 10:52

Ah I think I see what the problem might be

$grade_con

Where is the connection defined / set-up?

Also, you should change the line

PHP Code:


 if (false === $result)

back to

PHP Code:


 if (false == $result)

**Erez** · Jan 4, 2005 10:54

u mean the file that tells me wich host and db to connect to ?

**swdev** · Jan 4, 2005 10:55

Yes

**Erez** · Jan 4, 2005 10:56

this ?:

Code:

<?php

$db_host = "superihost.com";
$username = "somatnt_yosoma";
$password = "******";
$DB_name = "somatnt_yosoma";

echo "<HTML><BODY>";

$chan = mysql_connect ($db_host, $username, $password);
mysql_select_db ($DB_name, $chan);

?>

i hope u dont mind i changed the pass .... (its not really ****)
u can see in my first post i've added it (connection.php)

**swdev** · Jan 4, 2005 11:00

No - I'm glad you **** out the password.

As you can see from that code, the MySQL link identifier is called $chan, so this is what you shuold use in the mysql_query function call.

Try it and post the results

**Erez** · Jan 4, 2005 11:04

here is the resolt ... (long):

Code:

Executing INSERT INTO reg (account, password, e-mail, age) VALUES ('Admin', 'yoman', 'frerez@hotmail.com', 15)
Result = 
Failed to execute INSERT INTO reg (account, password, e-mail, age) VALUES ('Admin', 'yoman', 'frerez@hotmail.com', 15) due to You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '-mail, age) VALUES ('Admin', 'yoman', 'frerez@hotmail.com',

**Erez** · Jan 4, 2005 11:10

btw ... the fields in my table are :
account , password , e-mail , age

**Erez** · Jan 4, 2005 11:19

should i change the field name "e-mail" to somthing else?

**swdev** · Jan 4, 2005 11:20

Ah - ok

change this line

PHP Code:


                  . ' (account, password, email, age)'

to

PHP Code:


                  . ' (account, password, `e-mail`, age)'

i.e use backticks around the e-mail column name

**Erez** · Jan 4, 2005 11:22

cant be done ... made an error on that line ...

**swdev** · Jan 4, 2005 11:24

What was the error? It worked ok for me

Also is it possible to change the name of the e-mail field by removing the '-'?

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