simple if problem

[hpesoj_m]

hi everyone...is there something wrong with these line

$query = mysql_query("SELECT emaildb, passworddb, fnamedb, lnamedb, statdb FROM $table WHERE emaildb = '$loginEmaildata'");	
$row = mysql_fetch_row($query);
   if(mysql_num_rows($query)==1)
   {
      if($loginPassworddata==$row[2]){
	  return "statsNum=" . $row[4]; 
	  }else{
	  return "statsNum=invalid";
	  } 
   } else{
      return "statsNum=error";	
   }

coz if ill omit some lines and change it to these...

$query = mysql_query("SELECT emaildb, passworddb, fnamedb, lnamedb, statdb FROM $table WHERE emaildb = '$loginEmaildata'");
$row = mysql_fetch_row($query);
   if(mysql_num_rows($query)==1)
   {
      return "statsNum=" . $row[2];  

   } else{
      return "statsNum=error";	
   }

... it works...my problem actually is that as soon as the comparison of the login username exist in the database, i want the script to compare its password whether it matches or not. coz if ever it doesnt it will give a different error other than the "error" word that i have for login details that doesnt exist in the database

thanks in advance!

[halojoy]

      if($loginPassworddata==$row[2]){ 
      return "statsNum=" . $row[4];  

      }else{ 
      return "statsNum=invalid"; 
      }

you should try to track
that
$loginPassworddata and $row[2] BOTH exists, and get their correct values

Bug tracking you can do like this,
with 'echo variable value' , right BEFORE this IF
When you get it to work, you take away this echo lines.

echo "loginPassworddata: $loginPassworddata";
echo "<br>";
echo "row2: $row[2]";
echo "<br>";
     if($loginPassworddata==$row[2]){ 
      return "statsNum=" . $row[4];  

      }else{ 
      return "statsNum=invalid"; 
      }

I do this very often, to make sure all variables used in
if ( $variable1 == $variable2 )
ARE alright

🙂

[chintupintu03]

Hello hpesoj_m,

Following is the part of your query : ( i have copied and pasted here )

---

$query = mysql_query("SELECT emaildb, passworddb, fnamedb, lnamedb, statdb FROM $table WHERE emaildb = '$loginEmaildata'");

$row = mysql_fetch_row($query);
if(mysql_num_rows($query)==1)
{
if($loginPassworddata==$row[2]){

---

As I have understood, your variable ( $loginPassworddata ) contains the password entered by the user. This you want to compare with the table value fetched from the query.

Please check, what I think is you have to $row[1] value instead of $row[2] value.

From your query $row[2] refers to fnamedb AND NOT passworddb value. ( the subscripts begin with 0 )

regards,