Text Input to Variable then Variable to Array

[belbin09]

I have a form that asks for the users name and student number. Then in a separate script I going to test it against a txt file that looks like this: 

 

Jim Smith, 400424565

Sarah Hillier, 534712479

Jonathan Quinlan, 764134296

Keith Roberts, 123456789

Sarah Hillier, 200343656

Chloe Butler, 678123987

 

I am new to php so I am not sure where I went wrong. I keep getting warnings:

 

Notice: Undefined index: snumber in /Applications/XAMPP/xamppfiles/htdocs/assignment1/enrolled.php on line 9

Warning: in_array() expects parameter 2 to be array, string given in /Applications/XAMPP/xamppfiles/htdocs/assignment1/enrolled.php on line 21

 

enrolled.php

  <?php

  $students = "student.txt"; // text file for students and student number

  //converting a string into a variable

  $name = $_POST["name"];
  $number = $_POST["snumber"];


//open student file and explode into an array

  $sfile = fopen($students, 'r+') or die ("Student file does not exist");

  while ($sline = fgets ($sfile)) {
    $list = explode(",", $sline);

    //test array against text input

    if (in_array($name, $list[0])) {

      if (in_array($number, $list[1])) {

        echo $name;
        echo $number;

      } // end of inner if

    } //end of outer if

  } // end of while



  fclose($sfile)



  ?>

[Barand]

Are you sure your input form has an input with name "snumber"?

 

The second message is self-explanatory - $list[0] and $list[1] are not arrays. However, $list is an array.

[Barand]

P.S.

 

You should also note that $sline will contain the trailing linefeed character. You need to trim() the line before exploding otherwise $line[1] will also have the linefeed and will not, therefore, match the number.

[phpmillion]

Didn't you post this on other forums yesterday? I believe I answered your question.

[belbin09]

I figured out what was wrong with the text name snumber. I forgot the equal sign on the html side of the form. 

 

Made the correction from list[1] to list. No errors. However without list[1] meaning the name how will it be compared to the txt file for name and number?

[Barand]

You can use

if ($name == $list[0]

or

if (in_array($name, $list))

Similar for the number using $list[1]. Don't forget the trim();

[belbin09]

Didn't you post this on other forums yesterday? I believe I answered your question.

 

You did. Where I rewrote my code I wasn't sure if the same answer applied. 

[belbin09]

You can use

if ($name == $list[0]

or

if (in_array($name, $list))

Similar for the number using $list[1]. Don't forget the trim();

 

 

Thank you. My code is now running

[belbin09]

I was having someone help me but they are confusing me more than anything. So I do apologize for the questions. I have the code reading the text file perfectly. However as soon as I include the else statement it outputs it 6 times which I know is because its within the while statement. However when I leave it out of the while statement and create a new if statement it will always show the form

if ($name != $list[0] && $number != $list[1]) {

    include 'index.php';
  }

  while ($sline = fgets ($sfile)) {
    $list = explode(",",trim ($sline));

    //test array against text input

    if ($name == $list[0] && $number == $list[1]) {

      echo "Welcome $name $number";

    } else {

      include 'index.php';

    }

  } // end of while

  fclose($sfile)

  ?>

[Barand]

Try something like this

 

found = 0
while (get next line) {
    if number and name match {
        found  = 1
        echo welcome message
        break;
    }
}

if not found {
    include form
}

[belbin09]

Try something like this

 

found = 0
while (get next line) {
    if number and name match {
        found  = 1
        echo welcome message
        break;
    }
}

if not found {
    include form
}

 

That worked brilliantly! Thank you!