How do I resolve this error?

Question

Please help me solve this problem:

if($_POST["keyword"]) {
$keyword       = $_POST["keyword"];
$keyword       = trim($keyword);
$keyword_array = explode(" ",$keyword);
$numberofwords = (integer)count($keyword_array);
require("server.php");
$link = open_koneksi();
$tbl_name      = "author";
$query         = "SELECT COUNT(*) as num FROM $tbl_name WHERE " ;
   for ($x = 0; $x<= $numberofwords; $x++) {
      $query .= "author LIKE '%$keyword_array[$x]%'";
      if ( $x <  $numberofwords ) {
         $query .= " AND ";
      }
   }
   echo("<SCRIPT>document.location.href='?p=result';</SCRIPT>");

}

If the code in the program segment is executed, there will be a warning as follows:

Notice: Undefined offset: 1 in C:\xampp\htdocs\bijang\result.php on line 111

Location of faults refer to this code:

$query .= "author LIKE '%$keyword_array[$x]%'";

How do I fix this?

score 3 · Answer 1 · answered Jan 12 '13 at 16:51

3

Your problem is probably this:

  for ($x = 0; $x<= $numberofwords; $x++) {
                 ##

You are counting indexes 0 till 1, because the previous count() gave you 1. But that's the total number of indexes, the last index is still 0.

Change it into:

  for ($x = 0; $x < $numberofwords; $x++) {
                  #

Better yet, just use an foreach:

  foreach ($keyword_array as $x => $kw) {

That counts the indexes in $x implicitly.
You probably still have to change your last entry detection for the AND fillers. (Commonly you first make an array of substrings, then implode() with the filler " AND ".)

YadaYada: Also take care about unfiltered input. Use the database escaping function for text strings. It's heaps easier to use PDO and prepared statements though.

answered Jan 12 '13 at 16:51

mario

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Thanks Mario.. after I tried to follow your advice, the next error message: Notice: Undefined variable: numberofwords in ... – Ado Bijang Jan 12 '13 at 17:09
You still need that count variable. – mario Jan 12 '13 at 18:33
See also [PHP: “Notice: Undefined variable” and “Notice: Undefined index”](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) – mario Jan 12 '13 at 18:34

score 0 · Answer 2 · edited Jan 12 '13 at 17:15

0

Cause the $keyword_array array don't have the $x key.

edited Jan 12 '13 at 17:15

bensiu

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answered Jan 12 '13 at 16:52

Ddo

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score 0 · Answer 3 · answered Jan 13 '13 at 06:30

0

turns out, to overcome, just one word:

error_reporting(0);

the question, whether the notice was included as part of the error?

answered Jan 13 '13 at 06:30

Ado Bijang

1
1

How do I resolve this error?

3 Answers3