Problems retrieving correct values from database.

[loveitandhateit]

OK, my PHP skills are progressing but here is a problem I cannot solve:

if ($studentid) { 
    $verifyStudent = mysql_db_query ($dbname, "SELECT StudentID from Students", $linktodb); 
    if (in_array ($studentid, mysql_fetch_array ($verifyStudent))) { 
        $showName = implode("", mysql_fetch_array (mysql_db_query ($dbname, "SELECT StudentName from Students where (StudentID='$studentid')", $linktodb))); 
        print "$showName"; 
    } else { 
        print "Student name not found!"; 
    } 
}

The value of $showName is JohnJohn instead of John. I checked the database and the value in StudentName is just John. Why does it out put the name twice?

[HalfaBee]

JFYI

From the manual.

See also mysql_connect() and mysql_query().

Note: This function has been deprecated since PHP 4.0.6. Do not use this function. Use mysql_select_db() and mysql_query() instead.

This whole section should be done with one query
HalfaBee

   $verifyStudent = mysql_db_query ($dbname, "SELECT StudentID,StudentName from Students where STudentID=$studentid", $linktodb); 

if( mysql_numrows( $verifyStddent)==0 ) 
     echo "Student not found";
   else
   {
        $result = mysql_fetch_array( $verifyStudent );
        echo $result[1];
    }
 

[loveitandhateit]

it doesn't work, gives me:
mysql_numrows(): supplied argument is not a valid MySQL result resource

[HalfaBee]

He He

I can't spell 🙂
mysql_numrows( $verifyStddent)==0
mysql_numrows( $verifyStudent)==0

HalfaBee

[loveitandhateit]

No... I corrected that problem... but I still get that error. Besides it doesn't find that student in the first place.

[HalfaBee]

How about the capital T in STudent in the query.

HalfaBee

[loveitandhateit]

Nope I corrected everything...

[HalfaBee]

Put the quotes around $studentID in the where bit.
Halfabee