Can anyone spot why is this code not working?

[layman]

I inserted the whole code section. Now I don`t get it...

 

The error message I got:

INSERT INTO student_course (SID, CID, GRADE, COMMENTS) VALUES ('1', '1', '', '')

Fatal error: Using $this when not in object context in C:\xampp\htdocs\cwk\Student_Course_Insert_Tan.php on line 62

 

else {
                $sid = $_POST['sid'];
                $cid = $_POST['cid'];
                $grade = $_POST['grade'];
                $comments = $_POST['comments'];
                $db1 = new student_course();
                $db1->openDB();
                $sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS) VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";
                echo $sql;
                $result = mysql_query($sql, $this->conn);                                               //line 62

                
                echo "Success. Number of rows affected: <strong>{$numofrows}<strong>";
                $db1->closeDB(); 

[layman]

And the function looks exactly like this:

 

function insert_student_course($sid, $cid, $grade, $comments) {
        $esc_grade = mysql_real_escape_string($grade, $this->conn);             
        $esc_comments = mysql_real_escape_string($comments, $this->conn);        
        $sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS)          
        VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";

        echo $sql;

        $result = mysql_query($sql, $this->conn);
        if (!$result) {
            die("SQL Insertion error: " . mysql_error());
        } else {
            $numofrows = mysql_affected_rows($this->conn);
            return $numofrows;
        }
    }

[cyberRobot]

Seems kind of weird that the "$this" reference stopped working after adding the echo statement. Did you change anything else?

 

 

I'm just guessing, but does it work if you change the "$this->" references to "$db1->"?

[layman]

I haven`t changed anything else.

I checked with    $this->$db1  , but no difference, same error message.

[cyberRobot]

So you changed the "$this->conn" to "$db1->conn"?

 

 

Is the function inside a class object?

[layman]

Yes, I have changed, its    $db1->conn    now.

The function in inside the              class student_course.

 

[cyberRobot]

The function in inside the              class student_course.

 

Hmmm...with it being in a class "$this->conn" should work.

[cyberRobot]

With this script, are you planning to open multiple databases? If not, the database connection reference ($this->con) isn't required. If the reference isn't there, PHP defaults to the last opened database.

 

For example see:

http://php.net/manual/en/function.mysql-query.php

[layman]

This is the class student_course {

 

The function

function insert_student_course($sid, $cid, $grade, $comments) {
        $esc_grade = mysql_real_escape_string($grade, $this->conn);
        $esc_comments = mysql_real_escape_string($comments, $this->conn);
        $sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS)
        VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";

        echo $sql;

        $result = mysql_query($sql, $this->conn);
        if (!$result) {
            die("SQL Insertion error: " . mysql_error());
        } else {
            $numofrows = mysql_affected_rows($this->conn);
            return $numofrows;
        }
    }

 

and the code

else {
                $sid = $_POST['sid'];
                $cid = $_POST['cid'];
                $grade = $_POST['grade'];
                $comments = $_POST['comments'];
                $db1 = new student_course();
                $db1->openDB();
                $sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS) VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";
                echo $sql;
                $result = mysql_query($sql, $db1->conn);

                echo "Success. Number of rows affected: <strong>{$numofrows}<strong>";
                $db1->closeDB();
            }

 

cyberRobot, I really appreciate your help!

I haven`t a clue anymore, what`s wrong...

[layman]

No, I have 1 database, so there are no multiple databases.

Just checking the link you sent.

[layman]

If not, the database connection reference ($this->con) isn't required. If the reference isn't there, PHP defaults to the last opened database.

Should I try to take it out then?

[cyberRobot]

Should I try to take it out then?

 

 

I would if you don't plan to open multiple database. It's one less thing you need to type.

[layman]

I took it out, and guess what happened!

 

Not fatal error anymore, just this:

INSERT INTO student_course (SID, CID, GRADE, COMMENTS) VALUES ('1', '1', '', '')Success. Number of rows affected:

 

But in the table, the row was not inserted...

[cyberRobot]

Could you try posting your current code again (function to insert the row and the code that calls the function)?

[layman]

This is the code. Hope you can spot the mistake!

 

<?php
        $db1 = new student_course();
        $db1->openDB();

        $sql = "SELECT sid FROM STUDENTS";
        $result1 = $db1->getResult($sql);

        $sql = "SELECT cid FROM COURSES";
        $result2 = $db1->getResult($sql);

       if (!$_POST) {
        ?>
            <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
                Choose student ID: <select name = "sid">
                <?php
                while ($row = mysql_fetch_array($result1))
                echo "<option value='{$row['sid']}'>{$row['sid']}</option>";
             
                ?>
            </select><br/>
                Choose course ID: <select name = "cid">
                <?php
                while ($row = mysql_fetch_array($result2))
                    echo "<option value ='{$row['cid']}'>{$row['cid']} </option>";
                ?>
            </select><br/>

            Enter grade:<input type="text" name="grade" /><br />
            Enter comment:<input type="text" name="comments" /><br />

            <input type="submit" value="Save" />  

        </form>

        <?php
            } else {
                $sid = $_POST['sid'];
                $cid = $_POST['cid'];
                $grade = $_POST['grade'];
                $comments = $_POST['comments'];
                $db1 = new student_course();
                $db1->openDB();
                $sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS) VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";
                echo $sql;
                $result = mysql_query($sql);

                echo "Success. Number of rows affected: <strong>{$numofrows}<strong>";
                $db1->closeDB();
            }
        ?>

 

<?php

require("db_config.php");

class student_course {

    private $conn;

    function insert_student_course($sid, $cid, $grade, $comments) {
        $esc_grade = mysql_real_escape_string($grade, $this->conn);             
        $esc_comments = mysql_real_escape_string($comments, $this->conn);       
        $sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS)          
        VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";
        $result = mysql_query($sql, $this->conn);
        if (!$result) {
            die("SQL Insertion error: " . mysql_error());
        } else {
            $numofrows = mysql_affected_rows($this->conn);
            return $numofrows;
            
        }
    }

  
   function getResult($sql) {
     $getResult = mysql_query($sql, $this->conn);
     if (!getResult) {
          die("SQL Insertion error: " . mysql_error());
     } else {
          return $getResult;
     }
}
   

    function openDB() {
        $this->conn = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
        if (!$this->conn) {
            die("SQL Connection error: " . mysql_error());
        }
        $db_selected = mysql_select_db(DB_NAME, $this->conn);
        if (!$db_selected) {
            die("SQL Selection error: " . mysql_error());
        }
    }

[layman]

I managed to sort of insert SOME data to my studentcourse!!!

I have changed the table. The message I get back is:

 

INSERT INTO studentcourses (SID, CID, GRADE, COMMENTS) VALUES ('5', '3', '', '')Success. Number of rows affected:

 

It inserts data to the SID and CID but NOTHING to the grade & comments.

 

Any idea why? I think I`m so close!

[cyberRobot]

Grade and comments don't work due to the incorrect variable names. ($grade vs $esc_grade)

 

...
$grade = $_POST['grade'];
$comments = $_POST['comments'];
...
$sql = "INSERT INTO student_course (SID, CID, GRADE, COMMENTS) VALUES ('{$sid}', '{$cid}', '{$esc_grade}', '{$esc_comments}')";

 

 

Of course those variable names would work if keep the two mysql_real_escape_string() statements.

 

...
$esc_grade = mysql_real_escape_string($grade, $this->conn);             
$esc_comments = mysql_real_escape_string($comments, $this->conn);       
...

[layman]

Dear cyberRobot,

 

Yeah, got them mixed up, corrected it, and voila! IT IS WORKING!!!

Thank you very much for your time and answers, I really appreciate, what you did for me!

 

You know I only started to learn PHP a month ago, I know, I am an amateur..., but with your help, I have learned a lot recently!

 

Thanks again!

 

All the very best to you,

layman

[cyberRobot]

No problem, glad everything worked out in the end.