phpretard Posted November 2, 2009 Share Posted November 2, 2009 I need to convert numbers to text. The instructions are to "call the convertNumber function" I would like to be able to post to this, so I thought convertNumber($_POST['theNumber']); I can't seem to get it to work. Can anyone read this and help out? <? /* Copyright 2007-2008 Brenton Fletcher. http://bloople[dot]net/num2text You can use this freely and modify it however you want. */ function convertNumber($num) { list($num, $dec) = explode(".", $num); $output = ""; if($num{0} == "-") { $output = "negative "; $num = ltrim($num, "-"); } else if($num{0} == "+") { $output = "positive "; $num = ltrim($num, "+"); } if($num{0} == "0") { $output .= "zero"; } else { $num = str_pad($num, 36, "0", STR_PAD_LEFT); $group = rtrim(chunk_split($num, 3, " "), " "); $groups = explode(" ", $group); $groups2 = array(); foreach($groups as $g) $groups2[] = convertThreeDigit($g{0}, $g{1}, $g{2}); for($z = 0; $z < count($groups2); $z++) { if($groups2[$z] != "") { $output .= $groups2[$z].convertGroup(11 - $z).($z < 11 && !array_search('', array_slice($groups2, $z + 1, -1)) && $groups2[11] != '' && $groups[11]{0} == '0' ? " and " : ", "); } } $output = rtrim($output, ", "); } if($dec > 0) { $output .= " point"; for($i = 0; $i < strlen($dec); $i++) $output .= " ".convertDigit($dec{$i}); } return $output; } function convertGroup($index) { switch($index) { case 11: return " decillion"; case 10: return " nonillion"; case 9: return " octillion"; case 8: return " septillion"; case 7: return " sextillion"; case 6: return " quintrillion"; case 5: return " quadrillion"; case 4: return " trillion"; case 3: return " billion"; case 2: return " million"; case 1: return " thousand"; case 0: return ""; } } function convertThreeDigit($dig1, $dig2, $dig3) { $output = ""; if($dig1 == "0" && $dig2 == "0" && $dig3 == "0") return ""; if($dig1 != "0") { $output .= convertDigit($dig1)." hundred"; if($dig2 != "0" || $dig3 != "0") $output .= " and "; } if($dig2 != "0") $output .= convertTwoDigit($dig2, $dig3); else if($dig3 != "0") $output .= convertDigit($dig3); return $output; } function convertTwoDigit($dig1, $dig2) { if($dig2 == "0") { switch($dig1) { case "1": return "ten"; case "2": return "twenty"; case "3": return "thirty"; case "4": return "forty"; case "5": return "fifty"; case "6": return "sixty"; case "7": return "seventy"; case "8": return "eighty"; case "9": return "ninety"; } } else if($dig1 == "1") { switch($dig2) { case "1": return "eleven"; case "2": return "twelve"; case "3": return "thirteen"; case "4": return "fourteen"; case "5": return "fifteen"; case "6": return "sixteen"; case "7": return "seventeen"; case "8": return "eighteen"; case "9": return "nineteen"; } } else { $temp = convertDigit($dig2); switch($dig1) { case "2": return "twenty-$temp"; case "3": return "thirty-$temp"; case "4": return "forty-$temp"; case "5": return "fifty-$temp"; case "6": return "sixty-$temp"; case "7": return "seventy-$temp"; case "8": return "eighty-$temp"; case "9": return "ninety-$temp"; } } } function convertDigit($digit) { switch($digit) { case "0": return "zero"; case "1": return "one"; case "2": return "two"; case "3": return "three"; case "4": return "four"; case "5": return "five"; case "6": return "six"; case "7": return "seven"; case "8": return "eight"; case "9": return "nine"; } } ?> Thank you Quote Link to comment Share on other sites More sharing options...
seanlim Posted November 2, 2009 Share Posted November 2, 2009 Hi, If you used the code without modifying anything, I'm sure it should work as long as a proper parameter is passed? Try to output $_POST['theNumber'] to ensure that you are, indeed, passing a proper parameter.. Quote Link to comment Share on other sites More sharing options...
phpretard Posted November 2, 2009 Author Share Posted November 2, 2009 I added this to the bottom of the page... <? if (isset($_POST['submit'])){ echo $_POST['theNumber']."<br />"; convertNumber($_POST['theNumber']); } ?> <form action="" method="post"> <input type="text" name="theNumber" /><input type="submit" name="submit" value="go" /> </form> It output the number but still not the text...? Quote Link to comment Share on other sites More sharing options...
seanlim Posted November 2, 2009 Share Posted November 2, 2009 it is probably because you didn't echo the output? the function convertNumber only returns the textual value Quote Link to comment Share on other sites More sharing options...
phpretard Posted November 2, 2009 Author Share Posted November 2, 2009 What the heck am I doing wrong??? <? if (isset($_POST['submit'])){ echo $_POST['theNumber']."<br />"; convertNumber($_POST['theNumber'])."<br />"; echo $output; // <-- ??? } ?> <form action="" method="post"> <input type="text" name="theNumber" /><input type="submit" name="submit" value="go" /> </form> Still notta. Quote Link to comment Share on other sites More sharing options...
seanlim Posted November 2, 2009 Share Posted November 2, 2009 haha.. $output does not exist! it is only exists locally within the convertNumber function. try: echo $_POST['theNumber']."<br />"; echo convertNumber($_POST['theNumber'])."<br />"; Quote Link to comment Share on other sites More sharing options...
phpretard Posted November 2, 2009 Author Share Posted November 2, 2009 Thank you!!!!!! Quote Link to comment Share on other sites More sharing options...
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