Topics by old_apache

<div class="form-group"> <label for='image' class="col-sm-2 control-label">Leader Board</label> <div class="col-sm-10"> <input id= "txt_lead" type = "text" name="image_lead" class="form-control input-sm" onclick ="javascript:document.getElementById('file_lead').click();"> <input id = "file_lead" type= "file" accept="image/*" name="file_ads_lead" style='visibility: hidden;' onchange="ChangeText(this, 'txt_lead');"/> <?php echo form_error('image_lead', '<small class="pull-right req">', '</small>');?> </div> </div> <script> function ChangeText(oFileInput, sTargetID) { document.getElementById(sTargetID).value = oFileInput.value.replace("C:\\fakepath\\", "");; …

I have Skype installed on my PC, as i knew using port 80 as default. Then, i changed httpd.config to 8080, but my apache not starting After did a little googling, i have to uncheck 443 on Skype setting. My question: is the apache uses 2 ports? And why?

hello fellas i'm stuck in this function: function getDataForPagination() { $sql = "select * from t_book limit 1, 15; $result = mysql_query($sql); if (!$result) return false; $num_rows = mysql_num_rows($result); if ($num_rows == 0) return false; $result = dbResultToArray2($result); return $result; } i want to make the above function more flexible …

i developed a system in xampp 1.7.2, and now i run it in xampp 1.8.2 (lastest version) i got this error Parse error: syntax error, unexpected end of file in C:\xampp\htdocs\ hint: in xampp 1.7.2 no error anybody can tell me why?

i have 2 files, one file created on Windows and another was made on Ubuntu, i want to use this command: comm -12 file1 file2 how do i do that?

i got a string var=96.33% from 120 how do i get 96.33 from $var?