Using an array with a MySQL Query

Question

I am using a SELECT query to create an array of values. I am then using another select query, but I wan't to exclude the original array values from being returned. But I keep getting an error with the second query. Am I creating the array correctly? Is my MySQL sytax correct?

My PHP:

// Grabs all the users the logged in user is already friends with or following
$already_following_query= "SELECT recipient FROM relations WHERE sender= '".$user_id."'    
AND status= '1' OR status= '2'";
$already_following_result= mysqli_query($connect, $already_following_query)
    or die('Error with already following query');               
$already_following_array= mysqli_fetch_array($already_following_result);

$suggestions_query= "SELECT * FROM users WHERE user_id NOT IN   
'".$already_following_array."' AND user_id != '".$user_id."'";
$suggestions_result= mysqli_query($connect, $suggestions_query)
or die('Error with suggestions query');

Would be useful to include the error you get when you say an error is hindering your progress.\ — Rottingham
– Rottingham, Commented Jan 14, 2014 at 19:06
Also, if $already_following_array is an array, you will need to use implode() — Dutchie432
– Dutchie432, Commented Jan 14, 2014 at 19:07
You may want to do a bit of research on how mysqli_fetch_array works, check your SQL syntax (you're missing "()"'s around your IN values), and research how to join an array to a string of commas. There are a few issues here to fix. — Andrew McGivery
– Andrew McGivery, Commented Jan 14, 2014 at 19:07
it will help you ouput $suggestions_query along with the error message. Simply putting $already_following_ARRAY in the query string will not generate the intended query.. — aconrad
– aconrad, Commented Jan 14, 2014 at 19:10

user3141031 · Accepted Answer · 2014-01-14 19:07:19Z

2

The NOT IN clause requires an array -- you're providing a string.

Try this:

$array_following_array = implode(", ", $array_following_array);

$suggestions_query = "SELECT * FROM users WHERE user_id NOT IN   ('".$already_following_array."') AND user_id != '".$user_id."'";

answered Jan 14, 2014 at 19:07

user3141031

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