PHP function call

Question

I want to create function display() to be called to display a sorted list of all previous names and marks entered. Name parameter should be set as optional. I am unable to call the function in the proper place. Note that the code is working without the function. I want to it to work by the function display(); Please help. Thanks

Here's my code:

<?php

session_start();

if(isset($_POST['lname']) && isset($_POST['marks'])){
    if ($_POST['marks']<=100){
        $_SESSION['info'][] = array($_POST['lname'] => $_POST['marks']);
    }elseif (isset($_SESSION['info'])) {
        $marks = array();
        foreach($_SESSION['info'] as $key =>$values){
            foreach($values as $name=>$mark){
                $marks[$key] = $mark;
            }
        }
        display();

        function display(){ 
            arsort($marks);
            foreach($marks as $key => $value){
                foreach($_SESSION['info'][$key] as $name => $marks){
                    printf("Name: %s &nbsp; Marks: %d &nbsp Grade: %s <br>", $name, $marks, CalcGrade($marks) );
                }
            }
        } 
    }
}

Sathya Baman · Accepted Answer · 2015-04-18 09:12:20Z

3

Try like this

  session_start();

 if(isset($_POST['lname']) && isset($_POST['marks'])){        
if ($_POST['marks']<=100){
  $_SESSION['info'][] = array($_POST['lname'] => $_POST['marks']);

 }elseif (isset($_SESSION['info'])) {
  $marks = array();
  foreach($_SESSION['info'] as $key =>$values){
       foreach($values as $name=>$mark){
            $marks[$key] = $mark;
   }
 }
 display($marks);

function display($marks){ 
arsort($marks);
 foreach($marks as $key => $value){
 foreach($_SESSION['info'][$key] as $name => $marks){
 printf("Name: %s &nbsp; Marks: %d &nbsp Grade: %s <br>", $name, $marks, CalcGrade($marks) );
      }
    }
  } 
}
  }

answered Apr 18, 2015 at 9:12

Sathya Baman

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Comments

mariobgr · Accepted Answer · 2015-04-18 09:10:21Z

2

As each function works for itself, you must either pass the variables you work with, or get it from the global namespace.

function display() {

    global $marks; 

    arsort($marks);

    ......
}

Or

function display($marks) {

    arsort($marks);

    ......
}

answered Apr 18, 2015 at 9:10

mariobgr

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2 Comments

yetin Over a year ago

Its not working bro.. Fatal error: Call to undefined function display() in line 16 Here's my full code pastebin.com/ZsmfL0S3

mariobgr Over a year ago

First of all, you shoud pass the variable $marks to the function on line 16 of your code... but is it just me, or the closing brackets are much more than the opening ?

lakshman · Accepted Answer · 2015-04-18 09:08:53Z

0

Define $marks in the function..in PHP functions can't get the value outside it by your above written codes.

answered Apr 18, 2015 at 9:08

lakshman

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