stevie.whalen

I think you have entered .js script in .css file. If you want to add .css code then you have to write like this: body { background-image: url('../images/indonusa3.jpg'); } or else you have to write your code in .js file <script> $('body').css('background-image', 'url("<?php echo base_url('assets/images/indonusa3.jpg'); ?>")'); </script> Hope this will …

Both have really bright futures according to their own advantages and features. PHP is most speedy and easy to use for web development. Java won't be going anywhere either. Both are good choices for their respective domains.

Here are the simplest code for addfile.php <?php $target = 'files/'.$_FILES['upfile']['name']; if (file_exists("files/" . $_FILES['upfile']['name'])) { echo "<script type='text/javascript'> var answer = confirm('Overwrite? Y/N'); if (answer == true) { alert('You pressed OK!') var overwrite = ".move_uploaded_file($_FILES['upfile']['tmp_name'],$target)." if(overwrite == 1) { alert('File Overwrite successfully'); window.location = 'search.php'; }else alert('File Overwrite Failed'); …

I couldn't get the value of `$offerpricepl` echo "<tr><td>" . $row1['selection']."<td>".$offerpricepl. "</tr>"; // There is missing the value of $offerpricepl variable and it will not define anywhere.. I think It should be `$offerprice1=$row1['fsomefieldname'];` then we can get the iterate value. One more suggestion I would like to add here: echo …