Populating an HTML select list from a PHP MySQL array

Question

I'm fetching a list of 'actors' from my MySQL database table actors & trying to populate the result into a HTML Select box:

    <?php
$query = "SELECT actor_name FROM actors";
$result = mysql_query($query) or die("<h1>Error - the query could not be executed</h1>\n");
$num_rows = mysql_num_rows($result);
$row = mysql_fetch_array($result);

print("<h3>Actors</h3>\n");
print($num_rows);
if($num_rows == 0){
    print("<h3>No items are currently recorded in table Actors</h3>\n");
}
else{
    print("<select id=\"actors\" name=\"actors\">\n");
    for($i = 0; $i < $num_rows; $i++){
        print("<option>$row[$i]</option>");
        $row = mysql_fetch_array($result);
    }
    print("</select>");
}
?>

Error:

Notice: Undefined offset: 1 in C:\xampp\htdocs\actors.php on line 16

I'm getting a Undefined offset notice from the second record in the array onwards. When I add an isset check the select box is only populated with the first record. This would indicate that there is something wrong my query? I've checked my table, and there are 113 records.

Any help would be greatly appreciated.

Jason Heo · Accepted Answer · 2013-12-05 11:59:56Z

1

I think below code looks better and works well.

$query = "SELECT actor_name FROM actors";
$result = mysql_query($query) or die("<h1>Error - the query could not be executed</h1>\n");
$num_rows = mysql_num_rows($result);

print("<h3>Actors</h3>\n");
print($num_rows);

if($num_rows == 0)
{
    print("<h3>No items are currently recorded in table Actors</h3>\n");
}
else
{
    print("<select id=\"actors\" name=\"actors\">\n");
    while ($row = mysql_fetch_array($result))
    {
            print("<option>$row[0]</option>");
    )
    print("</select>");
    mysql_free_result($result);
}

answered Dec 5, 2013 at 11:59

Jason Heo

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4 Comments

Peon Over a year ago

I starter writing something like that. You beat me to it + I suggest you change that to mysqli better too.

Jason Heo Over a year ago

someone gave me -1, could you explain in detail?

R R Over a year ago

i am not the one who downvoted you but if someone ask a question try to tell him where he/she is doing wrong rather than giving him a new code.

Jason Heo Over a year ago

@RishabhRaj honestly, my English is bad. so there is no time to explain in detail. in later, I'll try to give more information. Many thanks.

R R · Accepted Answer · 2013-12-05 12:07:05Z

1

 $query = "SELECT actor_name FROM actors";
    $result = mysql_query($query) or die("<h1>Error -
    the query could not be    executed</h1>\n");
    $num_rows = mysql_num_rows($result);
    $row = mysql_fetch_array($result);
-`--------^^^^^^ //**you are already fetching the array once`**

you are using $row = mysql_fetch_array($result); twice

for($i = 0; $i < $num_rows; $i++){
        print("<option>$row[$i]</option>");
        $row = mysql_fetch_array($result);
----------------^^^^//**remove this**

    }

the correct loop will be:

for($i = 0; $i < $num_rows; $i++){
        print("<option>$row[$i]</option>");

    }

answered Dec 5, 2013 at 12:07

R R

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