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About Max45

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  1. Thaaaaaaaaaaaaaaaaaaaankyou , You really save my time for this advice. It worked Thanks
  2. You mean like this? th, td { text-align: left display: flexbox; flex-order: 1; } th, td { text-align: left display: flexbox; flex-order: 1; }
  3. The following code is showing my result horizontally and I want to show them vertically $id = $_GET['id']; $sql = "SELECT * FROM users WHERE id = $id"; $result = $conn->query($sql); if ($result->num_rows > 0) { echo "<table align=\"center\">; <tr> <th>ID</th> <th>First Name</th> <th>Last Name</th> <th>Email</th> <th>Join Date</th> </tr>"; // output data of each row while($row = $result->fetch_assoc()) { echo "<tr> <td>".$ro
  4. Thanks a lot buddy. And sorry if my post is not understandable. See ya
  5. How to user insert... select... with my two tables based on id? Hi all, I have two tables Table1_Name= users ->Which has all users Tabel2_Name=addfriend -> which I to want add users to it from table1(users) based on their id Here my code that didn't work $id = $_GET['id']; $sql = "INSERT INTO addfriend(fname) SELECT fname, FROM users WHERE id = $id";
  6. I am Trying to put my php echo into bootstrap div like this $result = $conn->query($sql); <div class="alert-success"> echo " <tr> <th>New Users</th> </tr> "; echo "<br />" </div>;
  7. I am trying to add a bootstrap class to php echo in mysql query but it doesn't work Here the code that I using $result = $conn->query($sql); echo ""; echo " New Users "; echo " "; echo ""; Any ides ?
  8. Hi I have made this simple login page by setcookies function and I want now to make a logout button. Here the code <?php /* PHP Form Login Remember Functionality with Cookies */ if(!empty($_POST["remember"])) { setcookie ("username",$_POST["username"],time()+ 3600); setcookie ("password",$_POST["password"],time()+ 3600); setcookie ("color",$_POST["color"],time()+ 3600); //3600 = 1 hour //86400 = 1 day //(8640*30) = 1 month echo "Cookies Set Successfuly"; } else { setcookie("username",""); setcookie("password",""); setcookie("color",""); echo "Cookies
  9. I have created a simple mysql database with php code and I put it on server using on server phpmyadmin . I added an insert button and everything was ok.But then the code stop working without no reason. Here the URL of my database site: mathcalc.appleschat.com Here the insertion code <?php include "connectToDB.php"; function mksafe($data){ $data=trim($data); $data=strip_tags($data); $data=htmlspecialchars($data); $data=addslashes( $data); return $data } $value1 = mksafe($_POST['fname']); $value2 = mksafe($_POST['lname']); $value3 = mksafe($_PO
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