11Tami Posted September 22, 2006 Share Posted September 22, 2006 Hello, I don't know whats wrong with this code. Its supposed to include a new web site url daily. Since its a web site url it won't be in the same directory, it will just be calling the url address online. Thank you!!! Tami<?php $content = array(); $content["01"] = "day1.htm"; $content["02"] = "day2.htm"; $content["03"] = "day3.htm"; $content["04"] = "day4.htm"; $content["05"] = "day5.htm"; $content["06"] = "day6.htm"; $content["07"] = "day7.htm"; $content["08"] = "day8.htm"; $content["09"] = "day9.htm"; $content["10"] = "day10.htm"; $content["11"] = "day11.htm"; $content["12"] = "day12.htm"; $content["13"] = "day13.htm"; $content["14"] = "day14.htm"; $content["15"] = "day15.htm"; $content["16"] = "day16.htm"; $content["17"] = "day17.htm"; $content["18"] = "day18.htm"; $content["19"] = "day19.htm"; $content["20"] = "http://www.website.com"; $content["21"] = "day20.htm";$content["22"] = "day21.htm"; $content["23"] = "day23.htm"; $content["22"] = "day23.htm"; $content["24"] = "day24.htm"; $content["25"] = "day25.htm"; $content["26"] = "day26.htm"; $content["27"] = "day27.htm"; $content["28"] = "day28.htm"; $content["29"] = "day29.htm"; $content["30"] = "day30.htm"; $content["31"] = "day31.htm"; $date = date('d'); if (isset($content[$date]) include($content[$date]);} ?> Quote Link to comment Share on other sites More sharing options...
freakus_maximus Posted September 22, 2006 Share Posted September 22, 2006 Looks like your missing your first curly brace:[code]if (isset($content[$date]) {include($content[$date]);} [/code] Quote Link to comment Share on other sites More sharing options...
SharkBait Posted September 22, 2006 Share Posted September 22, 2006 You could shorten up the array code a bit:[code]<?php$content = array(1 => "day1.htm", "day2.htm", "day3.html".....etc);$date = date('d');include($content[$date]);?>[/code]Don't think you need the if(isset( part though Quote Link to comment Share on other sites More sharing options...
11Tami Posted September 22, 2006 Author Share Posted September 22, 2006 The error message I am getting says it can't open the url that I put on line 22 for today's date. I try using real url's not the ones below. It gives me the correct address for it, but it won't open it. It says suitable wrapper can't be found even though my include command works fine, otherwise it wouldn't bring up slot 22's web address.I've tackled php include commands and now know how to do it really well, so there is still something wrong with the script. PS. I changed it to this in my php. $date = date('d');include($content[$date]);Any ideas? I notice this has no brackets surrounding any commands, does it need any? ThanksHere is the one I am trying now before I try trimming it down.<?php $content = array(); $content["01"] = "day1.htm"; $content["02"] = "day2.htm"; $content["03"] = "day3.htm"; $content["04"] = "day4.htm"; $content["05"] = "day5.htm"; $content["06"] = "day6.htm"; $content["07"] = "day7.htm"; $content["08"] = "day8.htm"; $content["09"] = "day9.htm"; $content["10"] = "day10.htm"; $content["11"] = "day11.htm"; $content["12"] = "day12.htm"; $content["13"] = "day13.htm"; $content["14"] = "day14.htm"; $content["15"] = "day15.htm"; $content["16"] = "day16.htm"; $content["17"] = "day17.htm"; $content["18"] = "day18.htm"; $content["19"] = "day19.htm"; $content["20"] = "http://www.website.com"; $content["21"] = "http://www.website.com"; $content["22"] = "http://www.website.com"; $content["23"] = "http://www.website.com"; $content["24"] = "http://www.website.com"; $content["25"] = "day25.htm"; $content["26"] = "day26.htm"; $content["27"] = "day27.htm"; $content["28"] = "day28.htm"; $content["29"] = "day29.htm"; $content["30"] = "day30.htm"; $content["31"] = "day31.htm"; $date = date('d');include($content[$date]);?> Quote Link to comment Share on other sites More sharing options...
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