PHP in_array not working true

Question

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."

Don't use mysql_* functions anymore. They're deprecated for a long time and get finally removed with PHP7 in the next few months. Use PDO or mysqli instead. — TiMESPLiNTER
– TiMESPLiNTER, Commented May 20, 2015 at 7:40
Thanks. Are you believe the mysql_* is the reason of the wrong answer? — Majid
– Majid, Commented May 20, 2015 at 8:18
Dear @deceze, I test mysqli_* and sorrily the answer is the same. It means there is no deference between mysl and mysqli in my localhost xampp. — Majid
– Majid, Commented May 20, 2015 at 8:33
@Majid The difference is that mysqli will work in the future, and mysql_query is already producing tons of warnings in PHP 5.5. — tadman
– tadman, Commented May 21, 2015 at 15:51

kamlesh.bar · Accepted Answer · 2015-05-20 07:46:23Z

2

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

edited May 20, 2015 at 7:46

answered May 20, 2015 at 7:41

kamlesh.bar

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1 Comment

Majid Over a year ago

Thanks for your answer. But I tried it before and it did not work. Because I believe there is no deference between your offered code and my code .

Majid · Accepted Answer · 2015-05-21 15:43:31Z

1

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.

answered May 21, 2015 at 15:43

Majid

192 bronze badges

1 Comment

Majid Over a year ago

I forgot to say my database is created by using a simple english words list. After replacing it with a .csv format list by importing to my database and using while(($words_row = mysql_fetch_assoc($words))) { if((in_array($word, $words_row)==false)) { my broblem was solved.

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