What's Wrong with this?

Anon

print ("<form action=\"record_ed_sql.php\">
<SELECT name=\"dropdown\">
<OPTION>phpa</OPTION>
<OPTION>phpb</OPTION>
</SELECT>
<input type=\"submit\">");
if (!$dbtable || !in_array($dbtable, 'phpa', 'phpb')) {
$dbtable = 'phpa';
}

$sql = "select * from $dbtable";

Anon

What is it supposed to do?

What do you think this line is doing?

if (!$dbtable || !in_array($dbtable, 'phpa', 'phpb')) {

As I read it you are saying if not $dbtable ($dbtable == 0) OR not in_array($dbtable, 'phpa', 'phpb') (neither phpa or phpb exist in the array $dbtable) then set $dbtable = 'phpa'.

Now I believe that what you are doing is displaying a form and letting the user select a database to work from. If this is the case and your form is calling itself where do you populate the $dbtable variable, since your select is labeled as dropdown.

Anon

I believe the syntax you are using for in_array() is incorrect. Here is the manual definition of in_array.

bool in_array (mixed needle, array haystack [, bool strict])

I believe you can only search an array for one value.

If this is your problem, you can try using the if statement below:

if(!$dbtable || (!in_array('phpa',$dbtable) && !in_array('phpb',$dbtable))) {
// yada
}

Hope this helps.

Anon

The form is for editing records in a table specified by the dropdown. However, when I populate the $dbtable with $dropdown it does not give me a default table and I get a "Warning: Supplied..." error for $dbtable = $dropdown. Techek provided the variable to give me a default table but when I use the default table it locks the form onto the default table and bypasses the dropdown box. When I use the dropdown box it changes the display to the table indicated in the dropdown but when the data is entered it goes to the default table. ??? The script can be viewed entirely at: http://www.codesx.com/phpstuff/record_editor.txt

Thanks for any input

Anon

I tried it the way you suggested also and that does not resolve the problem either. I don't need to call two tables. I just need one default to populate the $dbtable = until I use the dropdown to select a table I want to view the records for. You can view the script at http://www.codesx.com/phpstuff/record_editor.txt

Thanks Josh

Anon

Your problem is that you are relying on someone else to tell you how this works.

First when you come to this page you don't have any value in $dropdown, so your line $dbtable = $dropdown will cause an error when you have it uncommented.

Second in your code you have an if..else..elseif structure. This is completely wrong. You can have as many elseif's as you want, but you can only have one else and it has to come at the end of your block (if..elseif..elseif...else).

I would suggest that you change your if (!$dbtable) line to be something alot more readable like.

if (!isset($dropdown)) {
$dbtable = "phpa";
}
else
{
$dbtable = $dropdown;
}

That will set your $dbtable to be "phpa" unless $dropdown has something in it. If you are paranoid (and you probably should be) you can add data validation to your $dropdown variable in the else section.