implode output does not gives proper output - php

Question

This shows all the elements from an array in a string but without a separator. I used ',' as a separator in this code. And it does not work. How can i separate them with separator.
$con = mysql_connect("localhost","root","");
if(!$con)
{
die("Database Connection failed.".mysql_error());
}
$db = mysql_select_db("test1",$con);
if(!$db)
{
die("Database selection failed.".mysql_error());
}
$query = "select * from menu limit 10";
$result = mysql_query($query,$con);
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
//$menu_name = $row['m_name'];
//$menu_image = $row['m_image'];
$menu_name = array($row['m_name']);
$menu_image = array($row['m_image']);
echo implode(',',$menu_name);
//echo "<img src='images/$menu_image' style='height:200px;width:200px;'>";
}
if(!$result)
{
echo mysql_error;
}
?>

Answer 1

you are storing array into variable . so if you will use imploade on an variable it will not show the result so best way is to store your result into an array and then use implode .
Don't use depricated mysql_* function use mysqli_* or pdo
instead
$menu_image=array(); // start array to store
$menu_name=array();
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$menu_name[] = $row['m_name']; // store menu name
$menu_image[] = $row['m_image']; // store menu image
}
echo implode(',',$menu_name); // finaly use implode

Answer 2

You can try this:
$menu_name[];
$menu_image[];
while($row = mysql_fetch_array($result)) {
$id = $row['id'];
$menu_name[] = $row['m_name']; // I change your $menu_name into an array so that when you fetch $row['m_name'] it returns an array
$menu_image[] = $row['m_image']; //same thing
}
echo implode(',',$menu_name);
//echo "<img src='images/$menu_image' style='height:200px;width:200px;'>";

Answer 3

try this in place of your while loop
$menu_name = array();
$menu_image = array();
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$menu_name[] = $row['m_name'];
$menu_image[] = $row['m_image'];
}
echo implode(',',$menu_name);