Re: Problem with each()

well, that didn't work either. Same error. Show me where my thinking is wrong, if I've got an array ($HTTP_POST_VARS) in this case. and $HTTP_POST_VARS[var_list] is an array and the first entry in the array. if I did $x = each($HTTP_POST_VARS) in a while loop, the first combination that is returned is $x[0] with value of 'var_list' and $x[1] with value of 'Array'. In the first instance, I passed the name of the array ($HTTP_POST_VARS) to each(), so in this instance I would again pass the name of the array to each() to pull the value pairs out of it. This would mean that $each_array2 = each($each_array[0]), not $each_array[1]. Am I way off? > In article <B6F2A4D6.626C%dave@dminor.com>, dave@dminor.com says... > > }print("key: $each_array[0] value: $each_array[1]"); > } > }When I get to an array within HTTP_POST_VARS, it prints: > }key: var_name value: Array > } > }ok, but when I try to do an each on that array: (and maybe this is where I > }am wrong) > }$each_array2 = each($each_array[0]) > } > > There's your problem.. Your trying to do each(..) on, what in your > example above, is set to "var_name" ... > > Try changing that to: > > $each_array2 = each($each_array[1]); > > It's the second variable in your example that's an Array, (or it may > just be a string set to the word "Array".. Who knows?.. =) > > -- > Jeff Carnahan - jcarnahan@networq.com