CODEWITHSUNDEEP
phpmysql
Majid
Majid

Reputation: 19

PHP in_array not working true

I have a mysql_query to select * from an 'english' list database and mysql_fetch_assoc returns an array. I try to search word 'flick' (which actually exists in the database) by using in_array() if 'flick' is found, it shouldn't be shown but it is shown. I think in_array function does not find the word 'flick'. Please look at code below:

<?php
error_reporting(E_ALL);
require 'db.php';

function spellcheck($word)
{
$output = array();
$word   = mysql_real_escape_string($word);

$words  = mysql_query("SELECT `word` FROM `english` WHERE LEFT(`word`, 1) = '" .
 substr($word, 0, 1) . "'");

 while(($words_row = mysql_fetch_assoc($words)) && (in_array($word, $words_row)==false))
  {
    similar_text($word, $words_row['word'], $percent);
    if($percent > 82)
    {
      $output[] = $words_row['word'];
    }
  }

  return (empty($output)) ? false : $output;
}

  if (isset($_GET['word']) && trim($_GET['word']) != null)
  {
    $word = $_GET['word'];
    $spellcheck = spellcheck($word);

    if ($spellcheck !== false)
    {
      echo '<pre>' . print_r($spellcheck, true) . '</pre>'; 
    } else {
      echo '<p>' . $word . ' spelled correctly, or no suggestions founds.</p>';
    }
  }
?>
<form action="" method="GET">
Check single word spelling:
<input type="text" name="word" />
<input type="submit" value="Check" /> 
</form>

The code returns:

Array (
    [0] => flick 
    [1] => flicks
)

But it should be:

 "spelled correctly, or no suggestions founds."

Upvotes: 0

Views: 363

Answers (2)

Majid
Majid

Reputation: 19

After 2 days working on my problem I found the answer. the mistake is in the output of the query by mysql_fetch_assoc. Actually it returns an associative array but every key is added a space (' ') after that. So, the result is not like abcdefg. The result is like a b c d e f g. It means when I search a special word in the associative array, the in_array() function returns false. Because for example the word 'flick' is not equal to 'flick ' and there is a space after the keys in array. I used trim() function and solved my problem:

while ($rows = mysql_fetch_assoc($query))
  {
    foreach($rows as $key)
    {
      $key = trim($key);
      $array[] = $key; 
    } 
  }
    if (in_array($word, $array))
    {
      echo "The word is spelled correctly";
    } else {
      foreach($array as $key)
      {
        similar_text($word, $key, $percent);
        if ($percent > 82)
        {
          $output[] = $key;  
        }
      }      
    }

tank you for your paying attention to my answer.

Upvotes: 1

kamlesh.bar
kamlesh.bar

Reputation: 1804

replace this line

 while(($words_row = mysql_fetch_assoc($words)) && (in_array($word, $words_row)==false))

with

 while(($words_row = mysql_fetch_assoc($words))) {
    if((in_array($word, $words_row)==false)) {

and at bottom close if statement

Upvotes: 2

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