PHP MySQL check if items in array exist in Table, THEN SHOW - php

Question

PHP 5.4 with MySQL
Building off of this:
<?php
$arr = ...
$sql = 'SELECT COUNT(*) FROM table WHERE field IN ( ' . implode( ',', $arr ) . ' );';
$result = $db->query( $sql );
?>
How would I display which items of the array ARE NOT in the table?

Answer 1

Instead of returning the count, return the field itself:
$sql = 'SELECT field WHERE field IN ' . implode( ',', $arr ) . ' );';
$result = $db->query($sql);
$found = array();
while ($row = $result->fetch_assoc()) {
$found[] = $row['field'];
}
Then you can compare the two arrays:
$not_found = array_diff($arr, $found);

Answer 2

If your $arr variable is sanitized and safe to build a query with, you can use union operators to build a table containing all the words in the array, then you can left join that table to table.field and only keep array items where there was no match.
$arr = array('one','two');
$sql = "select item from (
select '" . implode("' item union select '",$arr). "' item
) t1 left join table t2 on t2.field = t1.item
where t2.field is null";
this will produce the following sql
select item from (select 'one' item union select 'two' item) t1
left join table t2 on t2.field = t1.item
where t2.field is null

Answer 3

Just to give an idea (this probably won't work)
$arr = ...
$sql = 'SELECT field FROM table WHERE field IN ( ' . implode( ',', $arr ) . ' );';
$result = $db->query( $sql );
while($r = $result->fetch())
$arr2[] = $r['field'];
print_r(array_diff($arr,$arr2));

Answer 4

after you've done imloud, array ceases to be an array, this is a string, so you do not have to select count(*). You need to select id after do the while and compare arrays

Answer 5

Just use "NOT IN", if your items is up to 10 elements:
<?php
$arr = array('one','two');
$filterExisting = "'" . implode("', '", $arr ) . "'";
$sql = "SELECT COUNT(*) FROM table WHERE field NOT IN (". $filterExisting .")";
$result = $db->query($sql);
?>