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Querystring $end error

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Querystring $end error

 
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  #1
Oct 27th, 2009
This is my code - which displays different articles and produces articles as http://www.mysite.com/testpage.php?test=1

PHP Syntax (Toggle Plain Text) 
  1. <html>
  2. <body>
  3. <?
  4. $i=$_GET['day'];
  5. switch ($i) {
  6. case 1:
  7.     include("2.php");
  8.     break;
  9. case 2:
  10.     include("3.php");
  11.     break;
  12. case 3:
  13.     include("4.php");
  14.     break;
  15. default:
  16.     //echo "page not found!!!!";
  17. }
  18. ?>
  19. <!--
  20. <? if($_GET['day']==1){ ?>
  21.  <?php include("2.php") ?>
  22. <? }else if($_GET['day']==2){ ?>
  23.  <?php include("3.php") ?>
  24. <? }else if($_GET['day']==3){ ?>
  25.  <?php include("4.php") ?>
  26.  <? }else if($_GET['day']==4){ ?>
  27.  <?php include("5.php") ?>
  28.  <? }else if($_GET['day']==5){ ?>
  29.  <?php include("6.php") ?>
  30.  <? }else if($_GET['day']==6){ ?>
  31.  <?php include("7.php") ?>
  32.  <? }else if($_GET['day']==5){ ?>
  33.  <?php include("1.php") ?>
  34. <? }else if(!is_numeric($_GET['day']) || $_GET['day']<1){ ?>
  35. <!--invalid id (or page not found)--> -->
  36. <?php
  37. $d=date("D");
  38. switch ($d)
  39. {
  40. case "Mon":
  41.     include("2.php");
  42.   //echo "Today is Monday";
  43.   break;
  44. case "Tue":
  45.     include("3.php");
  46.   //echo "Today is Tuesday";
  47.   break;
  48. case "Wed":
  49.     include("4.php");
  50.   //echo "Today is Wednesday";
  51.   break;
  52. case "Thu":
  53.     include("5.php");
  54.   //echo "Today is Thursday";
  55.   break;
  56. case "Fri":
  57.     include("6.php");
  58.   //echo "Today is Friday";
  59.   break;
  60. case "Sat":
  61.     include("7.php");
  62.   //echo "Today is Saturday";
  63.   break;
  64. case "Sun":
  65.     include("1.php");
  66.   //echo "Today is Sunday";
  67.   break;
  68. default:
  69.   //echo "Wonder which day is this ?";
  70. }
  71. ?>
  72. </body>
  73. </html>

However, I get this error:
Parse error: syntax error, unexpected $end in C:\www\vhosts\mylocalhostsite\testing.php on line 73

I have tried for myself to fix it, but keep getting the $end error repeatedly, anyone else done this type of PHP and managed to find a fix for this?
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-1
  #2
Oct 27th, 2009
You most likely forget a closing }.
Google is your friend.

Use [code] tags.

If you have found a solution to your problem, please mark the thread as SOLVED.
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  #3
Oct 28th, 2009
else if(!is_numeric($_GET['day']) || $_GET['day']<1){ ?>

no correspond }
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  #4
Oct 28th, 2009
Why so many opening and closing tags for PHP?

Get rid of all the unneeded tags, indent your code and missing braces will be more easy to spot.
AJAX is not a programming language, scripting language or any other sort of language.
It is acheived by using JavaScript http functions.
So, AJAX = JavaScript.
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-1
  #5
Oct 28th, 2009
put the break in the default and forgot the closing brace of last else if()
Last edited by network18; Oct 28th, 2009 at 5:36 am.
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