kristo5747 Posted December 22, 2011 Share Posted December 22, 2011 I wrote a very simple web form that allows my user to view text files from within their Internet browser. Occasionally, the search criteria entered returns more than one file. So I want to implement a feature whereby the text files returned by the search are compressed into a ZIP. I got a prototype working but it only compresses the first file. The second or third files are ignored. Here's my code <HTML><body><form name="myform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset><label for="DBRIDs">RIDs</label><input type="text" id="DBRIDs" name="DBRIDs" > </fieldset></form></body></HTML> <?php function check_search() { if (isset($_POST['submit'])) {if (!empty($_POST['DBRIDs'])) { $results = getFiles(); } } else $errors = "Please enter something before you hit SUBMIT."; return Array($results, $errors); } function getFiles() { $result = null; $ZIPresult = null; if (empty($_POST['DBRIDs'])) { return null; } $mydir = MYDIR; $dir = opendir($mydir); $DBRIDs = $_POST['DBRIDs']; $getfilename = mysql_query("select filename from search_table where rid in (" . $DBRIDs . ")") or die(mysql_error()); while ($row = mysql_fetch_array($getfilename)) { $filename = $row['filename']; $result .= '<tr><td><a href="' . basename($mydir) . '/' . $filename . '" target="_blank">' . $filename . '</a></td></tr>'; $ZIPresult .= basename($mydir) . '/' . $filename; } if ($result) { $result = "<table><tbody><tr><td>Search Results.</td></tr> $result</table>"; shell_exec("zip -9 SearchResult.zip ". $ZIPresult ." > /dev/null "); } return $result; } The hyperlinks pointing to the file(s) are generated just fine. The ZIP file however only contains the first file listed in the results. How can I get the ZIP file to capture ALL the files returned?? Thanks for your input. **PS: The new ZipArchive() library/class is not available on our production environment so I chose to use the Unix utility ZIP instead.** Quote Link to comment Share on other sites More sharing options...
kicken Posted December 22, 2011 Share Posted December 22, 2011 You probably just need to put a space between the filenames. Right now your $ZIPResult variable is going to hold all the filenames squished up together in one long string without any spaces separating them. $ZIPresult .= basename($mydir) . '/' . $filename.' '; Quote Link to comment Share on other sites More sharing options...
kristo5747 Posted December 22, 2011 Author Share Posted December 22, 2011 You probably just need to put a space between the filenames. Right now your $ZIPResult variable is going to hold all the filenames squished up together in one long string without any spaces separating them. $ZIPresult .= basename($mydir) . '/' . $filename.' '; That was it . I can now ZIP multiple files. Thanks a million! Quote Link to comment Share on other sites More sharing options...
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