asai
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Everything posted by asai
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Heres were I am at right now: <html> <link href="css/style.css" rel="stylesheet" type="text/css" media="screen" /> <?php include 'config.php'; include 'opendb.php'; $result = mysql_query("SELECT DISTINCT DATE_FORMAT(FileDate, '%b %Y') as adate, DATE_FORMAT(FileDate, '%d.%m.%Y') as ndate, FileName as filename FROM DayMovie ORDER BY FileDate DESC") or die(mysql_error()); echo "<table border='0'>"; while($row = mysql_fetch_array( $result )) { echo "<tr><td>"; echo $row['adate']; echo "</td><td>"; echo date($row['ndate']); echo "</td><td>"; echo "<a href='alldaymovies/{$row['filename']}'>Se film</a>"; echo "</td></tr>"; } echo "</table>"; include 'closedb.php'; ?> </html> This generates a table with 3 columns. In the first column it is the adate value (Month and year). What I would like is to select this value from a combobox and then only show the content that has this value in this column. The result can be viewed here: http://81.166.2.19/movies.php
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Now I think we are getting closer. However when I put this code together with the rest of my code I did something wrong. With the result that the the webpage is empty. I would like that only the combobox shows on the page and when I select a value, the page is updated with the result in the table.
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Is there any suggestions to my last question?
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I have tried this: <?php include 'config.php'; include 'opendb.php'; $query_disp="SELECT * FROM DayMovie ORDER BY FileDate"; $result_disp = mysql_query($query_disp, $conn); $options = array(); while ($query_data = mysql_fetch_array($result_disp)) { $options[$query_data["Id"]] = $query_data["FileDate"]; } ?> <select name="Id" onClick="submitCboSemester();"> <?php foreach ($options as $key => $value) : ?> <?php $selected = ($key == $_POST['Id']) ? 'selected="selected"' : ''; ?> <option value="<?php echo $key ?>" <?php echo $selected ?>> <?php echo $value ?> </option> <?php endforeach; ?> </select> Can be viewed here: http://81.166.2.19/combo.php However, I would like it to look a little different. Instead of all the dates, I would like it to only show december 2014 and january 2015 After that the result of this choice displays the content of this table accordingly: http://81.166.2.19/movies.php Is that possible?
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Yes something like that. Is it possible to do this choice from the webpage? i.e. from a dropdown box? The dropdown should only have available choices from the available data. If there is no data from february 2015 then it not an option in the dropdown. Is this possible?
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Sorry for my bad explaining. I am norwegian, så my english isn't too good. Heres my php: <html> <link href="css/style.css" rel="stylesheet" type="text/css" media="screen" /> <?php include 'config.php'; include 'opendb.php'; $result = mysql_query("SELECT * FROM DayMovie WHERE YEAR(FileDate) = 2015 ORDER BY FileDate DESC") or die(mysql_error()); echo "<table border='0'>"; while($row = mysql_fetch_array( $result )) { echo "<tr><td>"; echo date('d.m.Y', strtotime($row['FileDate'])); echo "</td><td>"; echo "<a href='alldaymovies/{$row['FileName']}'>Watch movie</a>"; echo "</td></tr>"; } echo "</table>"; include 'closedb.php'; ?> </html> This lists the content of the table very nice. However this list is going to be very long. What I could use is a way to choose which year and month i would like to view.
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I have one more question: In the table theres a row with date. Is there a way to use this date and create a row heading with the months? ie: All the rows with date in january, have a heading with January, and so on.
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Perfect!! Thank you. One small detail: How can I adjust the colum alignment?
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Fantastic! Work perfect! Thanks for good help. One last question: The date field has sql date format: yyyy-mm-dd Is there a way to display this in another format, i.e. dd.mm.yyyy instead?
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Thanks for the reply! Almost there: echo "<a href='alldaymovies/'FileName''>View</a>"; As you can see I would like to put in the filename from one of the rows in the table. The path works great, but I didn't get the filename with this code. How can i do that?
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Hi, A little new to php programming... I have a question: I have written this code: <html> <head> <title>Test</title> </head> <FONT FACE="Verdana, sans-serif"> <H3>Movies</h3> <?php include 'config.php'; include 'opendb.php'; $result = mysql_query("SELECT * FROM DayMovie ORDER BY FileDate") or die(mysql_error()); echo "<table border='1'>"; echo "<tr> <th>Filename</th><th>Path</th><th>Date</th><th>Filetype</th> <th>Size</th></tr>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo "<tr><td>"; echo $row['FileName']; echo "</td><td>"; echo $row['FilePath']; echo "</td><td>"; echo $row['FileDate']; echo "</td><td>"; echo $row['FileType']; echo "</td><td>"; echo $row['FileSize']; echo "</td></tr>"; } echo "</table>"; include 'closedb.php'; ?> I would like to have a link to these files on all the rows to open the files. How can I do that? I would also like that the link from the mysql row is not shown in the table. Only a text with the text "link" or "view".