check if value exist in database array not working

Question

need to check if the value exist in array , the array created from database table, it just not working with me keep getting error says "in_array() expects parameter 2 to be array, string given" can someone please help ?

PHP code

<?php 
    $lang='en';
    $query_rsLanguages = "SELECT * FROM languages";
    $rsLanguages = mysql_query($query_rsLanguages);
    $languages_array = array();
    while($row = mysql_fetch_array($rsLanguages)){ 
      $languages_array[] = "\"".$row['language_sign']."\""; 
    }
    $languages_string = implode(",", $languages_array);
    if (in_array($lang, $languages_string)) {
        echo 'found' ;
        }
?>

Rhumborl · Accepted Answer · 2015-01-11 12:42:37Z

1

You are imploding $languages_array to a string and passing that to in_array, which is incorrect and unnecessary. As the error says you need to pass an array to search in. You just need to build the array of languages from your result and check that:

<?php 
    $lang='en';
    $query_rsLanguages = "SELECT * FROM languages";
    $rsLanguages = mysql_query($query_rsLanguages);
    $languages_array = array();
    while($row = mysql_fetch_array($rsLanguages)){ 
      $languages_array[] = $row['language_sign'];
    }

    if (in_array($lang, $languages_array)) {
        echo 'found' ;
    }
?>

answered Jan 11, 2015 at 12:42

Rhumborl

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user4275254 · Accepted Answer · 2015-01-11 12:42:55Z

1

This will check if the String $lang exists inside your $languages_array or not

if (in_array($lang, $languages_array)) {
        echo 'found' ;
}

answered Jan 11, 2015 at 12:42

user4275254

Comments

Utkarsh Dixit · Accepted Answer · 2015-01-11 12:43:10Z

1

The above error is saying that the second parameter of the in_array should be an array, by using implode you are making it a string use the code below.

<?php 
    $lang='en';
    $query_rsLanguages = "SELECT * FROM languages";
    $rsLanguages = mysql_query($query_rsLanguages);
    $languages_array = array();
    while($row = mysql_fetch_array($rsLanguages)){ 
      $languages_array[] = $row['language_sign'];
    }

    if (in_array($lang, $languages_array)) {
        echo 'found' ;
    }
?>

Hope this helps you

answered Jan 11, 2015 at 12:43

Utkarsh Dixit

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