PHP array from checkbox values returns only Array instead of values

Question

According to suggestion here to use $_POST method to declare array from checkbox values I made it. But it isn't working.

In mysql table below, it stores only text that says Array. It doesn't store any of the values checked.

Even if I put echo like in the code below, it prints out "Array".

HTML PAGE WITH QUESTIONNAIRE

<input type="checkbox" name="cb1[]" value="Mike">
<input type="checkbox" name="cb1[]" value="Irena">
<input type="checkbox" name="cb1[]" value="Sonya">

<input type="checkbox" name="cb2[]" value="Samsung">
<input type="checkbox" name="cb2[]" value="Apple">

PHP SCRIPT

for($i=1;$i<101;$i++) {
    if(isset($_POST['cb'.$i])) {
        $row[$i] = $_POST['cb'.$i];
    }}
echo $row[1]; //it gives Array insted of values selected in the cb1

<?php
$db =& JFactory::getDBO(); 
$query = "INSERT INTO storeresults(V1, V2) VALUES ('$row[1]','$row[2]')"; 
$db->setQuery($query); 
$db->query();
?>

Use var_dump($row[1]); instead, because the $row[1] is an array of the selected values. If you want to get the value you need to put it in another for loop — Javad
– Javad, Commented Apr 22, 2014 at 18:51
Do you want to insert each selected value as a record, or I think you want to join them with something like , then insert the joined value in DB? — Javad
– Javad, Commented Apr 22, 2014 at 18:53

Marc B · Accepted Answer · 2014-04-22 18:52:02Z

6

Given your name="cb1[]", your $_POST value will simply be $_POST['cb1'], which happens to be an array:

foreach($_POST['cb1'] as $key => $value) {
   echo $value; // spit out a name
   echo $_POST['cb1'][$key]; // same thing, in a different way
}

Plus, as written, your query is vulnerable to SQL injection attacks.

answered Apr 22, 2014 at 18:52

Marc B

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