[Resolved] Why does my array return as "Array"?

Jaxeed

$enum_field = mysql_field_name($field_list, $i);
$select_command = mysql_query("SELECT $enum_field FROM $selected_table", $conn) or die (mysql_error());
$fetch_enum_array = mysql_fetch_row($select_command);
$enumarray = explode("','",preg_replace("/(enum|set)\('(.+?)'\)/","\\2",$fetch_enum_array));

This command should eledgedly return the enum values in an array in $enumarray. I need to break appart this array so I can put the different values into section with radiobuttons in a from.

I've tried doing a foreach command:

foreach ($enumarray as $enumvalue)
{
$render_insert .= "<input name=".mysql_field_name($field_list, $i)." type=\"radio\" value=\"$enumvalue\"> $enumvalue";
}

Which I thought would return one radiobutton for each value. However this returns only as one radiobutton saying Array. I obviously have not broken this array down correctly, so what is the correct way of doing that? Or does it return as Array when the array is empty, so I must have done something wrong earlier?

Appreciate any info on thisone, thanks 🙂

dnast

Looks like you didn't pull the string out of the result array.

Try this:

$enumarray = explode("','",preg_replace("/(enum|set)('(.+?)')/","\2",$fetch_enum_array[$enum_field]));

Also to see the structure of an array. Use something like:

echo "<pre>";
print_r($array_var);
echo "</pre>";

It's very useful. I end up doing this quite a bit.

Jaxeed

Thanks for your input dnast!

Seems then my array is empty as print_r($enumarray); returns Array ( [0] => Array ). That the enum field is correctly configured I'm sure of, and the alternatives should be 'yes' or 'no'. The field length is said to be 3, so I assume thats 'yes', 'no' and 'NULL'.

hmmmm... Any ideas why my array turns out empty?

dnast

Try using:

$enum_field = mysql_field_name($field_list, $i);
$query = "SELECT $enum_field FROM $selected_table";
echo $query;

$select_command = mysql_query($query, $conn) or die (mysql_error());
$fetch_enum_array = mysql_fetch_row($select_command);

echo "<pre>";
print_r($fetch_enum_array);
echo "<pre>";

My guess is that one of your variables isn't set to what you think it is. This should help narrow down the problem.

BuzzLY

Originally posted by Jaxeed
Thanks for your input dnast!

Seems then my array is empty as print_r($enumarray); returns Array ( [0] => Array ). That the enum field is correctly configured I'm sure of, and the alternatives should be 'yes' or 'no'. The field length is said to be 3, so I assume thats 'yes', 'no' and 'NULL'.

hmmmm... Any ideas why my array turns out empty?

Actually, it's not empty. What that tells you is that your array $enumarray contains a single element, which is also an array. You need to delve a little deeper... try:

print_r($enumarray[0]);

Jaxeed

for ($i = 0; $i < $columns; $i++)
{
$fieldtypetest = mysql_field_type($field_list, $i);
$fieldlengthtest = mysql_field_len($field_list, $i);
$fieldflagtest = explode (" ", mysql_field_flags($field_list, $i));

if(in_array("enum", $fieldflagtest))
{
$render_insert .= "<tr><td bgcolor=\"#B0BADA\"><div align=right class=\"text\">" .mysql_field_name($field_list, $i). ":</div></td>";
$render_insert .= "<td bgcolor=\"#B0BADA\" class=\"text\">";

$enum_length = $fieldlengthtest - 1;
$enum_field = mysql_field_name($field_list, $i);
$enum_show = "SHOW COLUMNS FROM $selected_table LIKE '$enum_field'";
$select_command = mysql_query($enum_show, $conn) or die (mysql_error());
$fetch_enum_array = mysql_fetch_row($select_command);
$enumarray = explode("','",preg_replace("/(enum|set)\('(.+?)'\)/","\\2",$fetch_enum_array[1]));

foreach($enumarray as $enumvalue)
{
$render_insert .= "<input name=".mysql_field_name($field_list, $i)." type=\"radio\" value=\"$enumvalue\"> $enumvalue";
}
$render_insert .= "<input name=".mysql_field_name($field_list, $i)." type=\"radio\" value=\"NULL\"> NULL </td></tr>";
}
}

I had chosen the wrong MySQL query, and did also not realise that it is different to type the MySQL query as a own variable first, insteady of directly into the mysql_query() tag.

THANK YOU dnast! You've taught me some good problem solving skills. Thanks for your input BuzzLY too.

BuzzLY

Excellent 🙂 It's always nice to see someone actually learn how to solve problems, rather than just look for the answer to their question(s). Debugging is a valuable skill you need to become a good developer. Pretty soon, you'll be answering others' questions! 🙂

dnast

Glad you figured it out! And pretty much on your own. 🙂

And I completely agree with BuzzLY. Debugging skills really reduce the downtime of trying to sovle problems that are sometimes as simple as a typo or variable mixup.