compare array data with single column in database

Question

I need to check if the array stored data matches with the data present in the database. Meaning that, user will select check box data 1,4,5 which will be stored as $data[]. In database there is a table with a column ID which will have 1, 2, 3, 4, 5. Every time user selects different check box, which will be stored in $data. I want to check those values with the database ID and i want to get only those values which is stored in-front of that ID in other column called 'value'

Example: if $data has value 3, 4, then it should get the value from the database only for 4 and 5.

I am not getting how to do this comparison and get the value.

$query = "select values from customer where bname = '".$_POST['bname']."' ";

$update = DB_query($query,$db);
while($row = DB_fetch_array($update))
{
    $array=explode(',',$row["values"]);
}

@sentilbp: thanks for your time. i used in_array() as suggested by Avin Varghese , it worked fine!. — user2727979
– user2727979, Commented Aug 29, 2013 at 7:14

Avin Varghese · Accepted Answer · 2013-08-29 06:36:11Z

1

Use php array_diff(); OR in_array(); to compare array with your MySQL column.

  $query = "select values from customer where bname = '".$_POST['bname']."' ";

$update = DB_query($query,$db);
while($row = DB_fetch_array($update))
{
    $array=explode(',',$row["values"]);
}

    if (in_array("one", $array)) {
        echo "one Found";
    }

edited Aug 29, 2013 at 6:36

answered Aug 29, 2013 at 6:26

Avin Varghese

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1 Comment

user2727979 Over a year ago

Thank you very much Avin Varghese , It worked as i wanted!. Thanks a ton

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