megglz Posted February 21, 2011 Share Posted February 21, 2011 I have the following array which builds a concatenation of the items in tree-like form. $arrayT = array(); $arrayT[0] = "a"; $arrayT[1] = "b"; $arrayT[2] = "c"; $arrayT[3] = "d"; $arrayT[4] = "e"; $arrayT[5] = "f"; $arrayT[6] = "g"; $arrayT[7] = "h"; $arrayT = buildTree($arrayT); print_r($arrayT); function buildTree($array) { $arrayTree = $array; $start = 0; $end = count($arrayTree); $i = 0; while ($start != $end - 1) { if ($i % 2 == 1) { $arrayTree[count($arrayTree) - 1].=$arrayTree[$start + $i]; } else $arrayTree[] .= $arrayTree[$start + $i]; $i++; if (($start + $i) == $end) { $i = 0; $start = $end; $end = count($arrayTree); } } return $arrayTree; } Example output is Array ( [0] => a [1] => b [2] => c [3] => d [4] => e [5] => f [6] => g [7] => h [8] => ab [9] => cd [10] => ef [11] => gh [12] => abcd [13] => efgh [14] => abcdefgh ) I would like the values to be hashed using sha1. e.g value at [8] ab would be the hash of [0]a and [1] b, value [12]abcd would be the hash of the values at [8]ab and [9]cd. I modified it myself in the next code snippet but I still don't feel it's doing what it's meant function buildTree($array) { $arrayTree = $array; $start = 0; $end = count($arrayTree); $i = 0; while ($start != $end - 1) { if ($i % 2 == 1) { $arrayTree[count($arrayTree) - 1].=$arrayTree[$start + $i]; sha1($arrayTree[count($arrayTree)-1].=$arrayTree[$start + $i]); } else $arrayTree[] .= sha1($arrayTree[$start + $i]); $i++; if (($start + $i) == $end) { $i = 0; $start = $end; $end = count($arrayTree); } } return $arrayTree; } $val1 = sha1($arrayT[0]); $val2 = sha1($arrayT[1]); $val3 = sha1($val1.$val2); $val4 = sha1("ab"); echo "VAL1 ".$val1; echo "<br/>"; echo "VAL2 ".$val2; echo "<br/>"; echo "VAL3 ".$val3; echo "<br/>"; echo "VAL4 ".$val4; Anyone any ideas? Thankyou. Link to comment https://forums.phpfreaks.com/topic/228405-using-arrays-and-sha1/ Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.