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rubyarat2010

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  1. Thank you so much!
  2. It is a little "program" I am making to help me with somthign else. I know it is not that safe of a method but only I will be using it. I tried what you put but it kind of gives me the same error you said ( ! ) Parse error: syntax error, unexpected '(', expecting T_STRING or T_VARIABLE or '{' or '$' in C:\wamp\www\NotesWebsite\ViewNote.php on line 4 Thank you for the quick reply!
  3. This code is meant to retrieve a row in a form but it returns nothing, I have tried it many different times even with HTML in it it returns nothing at all. <?php $conn = new mysqli("****", "***", "****"); $conn->select_db('mynotesdatabase'); $result = mysqli_query($conn, "SELECT * FROM" . $_GET['Folder'] . "WHERE Name='" . $_GET['Name'] . "'") or die(mysql_error()); while($row = mysqli_fetch_array($result) or die(mysql_error())) { echo "<br>"; echo $row['Main']; echo "<br>"; } echo "<br/> <a href='javascript:history.back(-1);'>Back</a>"; ?>
  4. Thank you so much man! Yea I was mxixing em up >.<
  5. Also I realized that the variable database is not database but tables. Just an FYI. Here is the full code if you want to see it, sorry to double post $con=mysql_connect("*", "*", "*", "mynotesdatabase") or die(mysql_error()); $query = mysql_query("SHOW TABLES FROM mynotesdatabase") or die(mysql_error()); while($row = mysql_fetch_array($query)) { $table = $query; $result = mysql_query("SELECT * FROM $table", $con) or die(mysql_error()); echo "<b>+" . $table . "</b><br/>"; while($row = mysql_fetch_array(mysql_query($result, $con))) { $type = ""; if($row['Type'] == 0){ $type = "Link"; } else{ $type="Note"; } echo "&nbsp -" . $row['Name'] . " (" . $type . ")"; echo "<br>"; } } mysql_close($con);
  6. Hey man thank you so much! I did not see those, so I changed them all to mysql. Now I have this but all it says is "No database selected": $query = mysql_query("SHOW TABLES FROM mynotesdatabase") or die(mysql_error()); while($row = mysql_fetch_array($query)) { $database = $query; $result = mysql_query("SELECT * FROM $database", $con) or die(mysql_error()); echo "<b>+" . $database . "</b><br/>"; while($row = mysql_fetch_array(mysql_query($result, $con))) { $type = ""; if($row['Type'] == 0){ $type = "Link"; } else{ $type="Note"; } echo "&nbsp -" . $row['Name'] . " (" . $type . ")"; echo "<br>"; } } mysql_close($con);
  7. So I want to get each table name in a MYSQL database and put it in the variable database. This is what I have so far but I cant seam to get it working, thanks in advance! It just says "Access denied for user ''@'localhost' to database 'mynotesdatabase'" But when I print out the actual rows of the tables it works but not when I want to print out the table names. if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } else{ echo "Connected", "<br/>"; } $query = mysql_query("SHOW TABLES FROM mynotesdatabase") or die(mysql_error()); while($row = mysql_fetch_array($query)) { $database = $query; $result = mysqli_query($con,"SELECT * FROM $database"); echo "<b>+" . $database . "</b><br/>"; while($row = mysqli_fetch_array($result)) { $type = ""; if($row['Type'] == 0){ $type = "Link"; } else{ $type="Note"; } echo "&nbsp -" . $row['Name'] . " (" . $type . ")"; echo "<br>"; } } mysqli_close($con);
  8. Ok i got it one other thing i assighn the header at the end but it gives me this error Warning: Cannot modify header information - headers already sent by (output started at /home/zendavis/public_html/insert.php:2) in /home/zendavis/public_html/insert.php on line 76?
  9. I did even tried puting periods like he showed Author:'$author; '<br/> Date: '$date;' <br/> ' $body;'
  10. Still Gives me erorrs:(((((( <?php $author=$_POST["author"]; $date=$_POST["date"]; $title=$_POST["title"]; $body=$_POST["body"]; $imagelink=$_POST["imgl"]; $link = $title ; $con = mysql_connect("localhost","zendavis_tim","sherlock11<>"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("zendavis_posts", $con); mysql_query("INSERT INTO posts1 (author, imagelink,date,body,title,link) VALUES ('$author', '$imagelink','$date','$body','$title','$link')"); $title = str_replace(" ", "_", $title); $conatent = ' <!doctype html> <html> <head> <title>Zen Davis.com</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link href="style.css" type="text/css" rel="stylesheet" /> </head> <body link="#5e6959" h3="#5e6959" vlink="#5e6959" > <div class="content"> <div class="top_block NavMenu"> <div class="content"> <!--Navigation line --> <h1><div style="text-align:center"><p><a href="Index.php" style="text-decoration: none" >Home</a>&nbsp <a href="About.html" style="text-decoration: none" >About</a> &nbsp <a href="Contact.php" style="text-decoration: none" >Contact</a> &nbsp <a href="Games.php" style="text-decoration: none" size="25">Games</a></p></div></h1> </div> </div> <div class="top_block MiddlePhoto"> <div class="content"> Author:'$author; '<br/> Date: '$date;' <br/> ' $body;' </div> </div> <div class="top_block BottomTables"> <div class="content"> </div> </div> </div> <!-- * Layout generated with http://layzilla.com * Layout generator is free of use. * We appreciate if you leave this comment block in commercial use of generator. * All comment and ideas can be submitted to us using contacts below. * * site: www.jmholding.com * email: [email protected] * twitter: @jmholding --> </body> </html> '; $file = "$title.html"; $open = fopen($file, "w"); fwrite($open, $conatent); fclose($open); header("Location: http://zendavis.com/$link.html"); ?>
  11. <?php echo $author ?> this didint work either.
  12. sorry but this didint work Author: "$author" <br/> Date: " $date" <br/> "$body"
  13. mikosiko works now but now the php variables in the html dont work?
  14. mikosiko works now but now the php variables in the html dont work?
  15. but thats not in the html?
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