How to check an array in a sql query?

Question

So I need to check an array in my query:

$query = "SELECT * FROM post WHERE ".$select." LIKE '%".$search."%'  AND  ID NOT IN '" .$Lastvar."'";

$Lastvar is my array. I have no idea where to go from here an any help would be appreciated.

EDIT: Here's my full query:

$Lastvar = array();
mysql_select_db('submisions', $dbconn);
$query = "SELECT * FROM post WHERE ".$select." LIKE '%".$search."%' AND ID NOT IN (" . join(", ", $Lastvar) . ")";

$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
if ($num_rows == 0) {
    echo 'No results were found';
    exit;
}

What do you mean by 'an array to make in my query'? If you just want to retrieve the result as array, use mysql_fetch_array(mysql_query($query)); — Dennis Hunink
– Dennis Hunink, Commented Aug 20, 2012 at 8:45
Right, that makes more sense. Don't worry about the English, can happy to all of use :) I've nothing to add to the answers below, should do the trick for you. Good luck! — Dennis Hunink
– Dennis Hunink, Commented Aug 20, 2012 at 9:02

Andrius Naruševičius · Accepted Answer · 2012-08-20 08:45:50Z

5

If the ID is of integer type:

$query = "SELECT * FROM post WHERE ".$select." LIKE '%".$search."%'  
AND ID NOT IN (" . join(", ", $Lastvar) . ")";

answered Aug 20, 2012 at 8:45

Andrius Naruševičius

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3 Comments

Follett Over a year ago

Im not getting any results any more. Even if what I search is exactly word for word what I need. Any chance you could check and make sure I set up my array right? I posted the full query in OP.

Andrius Naruševičius Over a year ago

Well you do set your array up, but you don't add any data to it! Thus your query gets an error because of AND ID NOT IN ()

Follett Over a year ago

Well that's actually encased in a for loop. At the end of the for loop I put the ID I got in the array and then the next time I go through it wont grab that ID.

Dan Barzilay · Accepted Answer · 2012-08-20 08:46:33Z

0

Do it like this:

$query = "SELECT * FROM post WHERE ".$select." LIKE '%".$search."%'  AND  ID NOT IN ('" .join("', '", $Lastvar).")";

answered Aug 20, 2012 at 8:46

Dan Barzilay

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