Notice: Array to string conversion in line 6! Problem about SQL - php

Question

I have a simple code (very simple one) that I was using to try something out for a work and I was trying a function to work with the variables of a form radio in post method, to update my SQL table with the output of the form. But when I'm going to try it, it doesn't update and gives me a notice.
It has something to do with the query (because the error says is in that line of the code) but I still don't know what it is.
I tried to change the syntax of the SQL sentence in different ways. I changed the user I was going to use to change the "image_value" column. I even checked the syntax of the query in phpmyadmin, and it worked.
Here is the php code:
<?php
mysql_connect("localhost","root","");
function user_image($value){
print_r($value);
//This is the problem
$query = "UPDATE users SET image_value = '$value' WHERE (ID) = '6'";
mysql_query($query);
}
?>
And here i have the code of the form and how I'm using the function (if there is any mistake that I haven't seen)
<form method="post" action="">
<input type="radio" name="1" value="1">imagen1
<br>
<input type="radio" name="2" value="2">imagen2
<br>
<input type="radio" name="3" value="3">imagen3
<br>
<input type="radio" name="4" value="4">imagen4
<br>
<button type="submit"><span>Submit</span></button>
</form>
<?php
user_image($_POST);
?>

Answer 1

Your problem is you are passing the complete object $_POST.
You don't specify if your radio name is correct (The name of your radio as a number from 1 to 4).
In the case, you are trying to set the value of image_value from a radio button should be.
<form method="post" action="">
<input type="radio" name="image_value" value="1">imagen1
<br>
<input type="radio" name="image_value" value="2">imagen2
<br>
<input type="radio" name="image_value" value="3">imagen3
<br>
<input type="radio" name="image_value" value="4">imagen4
<br>
<button type="submit"><span>Submit</span></button>
</form>
<?php
if (isset($_POST['image_value'])) {
user_image($_POST['image_value']);
}
?>
and your function
function user_image($value) {
mysql_connect("localhost","root","");
print_r($value);
//This is the problem
$query = "UPDATE users SET image_value = '$value' WHERE (ID) = '6'"; //ID should be dynamic base on the user I guess
mysql_query($query);
}
?>

Answer 2

$_POST pass an object value to the server, you need to specify the property you want to use, var_dump $value to understand all what it contains.