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Jedijon

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Everything posted by Jedijon

  1. How to add the ability to login with username or email for login? <?php ob_start(); include('../header.php'); include_once("../db_connect.php"); session_start(); if(isset($_SESSION['user_id'])!="") { header("Location: ../dashboard"); } if (isset($_POST['login'])) { $email = mysqli_real_escape_string($conn, $_POST['email']); $password = mysqli_real_escape_string($conn, $_POST['password']); $result = mysqli_query($conn, "SELECT * FROM users WHERE email = '" . $email. "' and pass = '" . md5($password). "'"); if ($row = mysqli_fetch_array($result)) { $_SESSION['user_id'] = $row['uid']; $_SESSION['user_name'] = $row['user']; $_SESSION['user_email'] = $row['email']; header("Location: ../dashboard"); } else { $error_message = "Incorrect Email or Password!!!"; } } ?>
  2. .dropbtn { background-color: #4CAF50; color: white; padding: 10px; font-size: 16px; border: none; } .dropdown { position: relative; display: inline-block; } .dropdown-content { display: none; position: absolute; background-color: #f1f1f1; min-width: 800px; box-shadow: 0px 8px 16px 0px rgba(0,0,0,0.2); z-index: 1; } .dropdown-content img { width: 32px; height: 32px; margin: 3px; position: relative; float: left; } .dropdown-content a { color: black; padding: 12px 16px; text-decoration: none; display: block; } .dropdown-content a:hover {background-color: #ddd;} .dropdown:hover .dropdown-content {display: block;} .dropdown:hover .dropbtn {background-color: #3e8e41;} /* HIDE RADIO */ [type=radio] { position: absolute; opacity: 0; width: 0; height: 0; } /* IMAGE STYLES */ [type=radio] + img { cursor: pointer; } /* CHECKED STYLES */ [type=radio]:checked + img { outline: 2px solid #f00; } <div class="dropdown"> <button class="dropbtn">Dropdown</button> <div class="dropdown-content"> <form name="form" action="icons-submit/submit_icon-default.php" method="post"> <input type="radio" id="icon_default" name="icon_default" value="icons/default.jpg" onchange='if(this.value != 0) { this.form.submit(); }'> <img src="icons/default.jpg"> <input type="text" id="id" name="id" value=" . $row['id'] . " style="background: transparent;border: none;font-size: 0;"> </form> <form name="form" action="icons-submit/submit_icon-default.php" method="post"> <input type="radio" id="icon_default" name="icon_default" value="icons/default.jpg" onchange='if(this.value != 0) { this.form.submit(); }'> <img src="icons/default.jpg"> <input type="text" id="id" name="id" value=" . $row['id'] . " style="background: transparent;border: none;font-size: 0;"> </form> <form name="form" action="icons-submit/submit_icon-Grass_Block.php" method="post"> <label> <input type="radio" id="icon_grass-block" name="icon_grass-block" value="icons/Grass_Block.png"onchange='if(this.value != 0) { this.form.submit(); }'> <img src="icons/Grass_Block.png"> </label> <input type="text" id="id" name="id" value=" . $row['id'] . " style="background: transparent;border: none;font-size: 0;"> </form> <form name="form" action="icons-submit/submit_icon-gear.php" method="post"> <label> <input type="radio" id="icon_gear" name="icon_gear" value="icons/Gear.png"onchange='if(this.value != 0) { this.form.submit(); }'> <img src="icons/Gear.png"> </label> <input type="text" id="id" name="id" value=" . $row['id'] . " style="background: transparent;border: none;font-size: 0;"> </form> </div> </div>
  3. Query looks like: SELECT * FROM datakeyWarning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in /srv/disk1/2648096/www/site/test/index.php on line 35 It shows the session data
  4. Error message Parse error: syntax error, unexpected ';', expecting ',' or ')' in /srv/disk1/2648096/www/site/test/index.php on line 20
  5. I have that <?php // Initialize the session session_start(); // Check if the user is logged in, if not then redirect him to login page if(!isset($_SESSION["loggedin"]) || $_SESSION["loggedin"] !== true){ header("location: ../login"); exit; } ?>
  6. <?php $con=mysqli_connect("IP","Username","Password","Database name"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT * FROM {$_SESSION['datakey']}"; echo "<table border='1'> <tr> <th></th> <th></th> <th></th> <th></th> <th></th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['column1'] . "</td>"; echo "<td>" . $row['column2'] . "</td>"; echo "<td>" . $row['column3'] . "</td>"; echo "<td>" . $row['column4'] . "</td>"; echo "<td>" . $row['column5'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?>
  7. $_SESSION['datakey'] contains the name of the table
  8. It will grave the info from the login SESSION
  9. Didn't work $result = mysqli_query($con,"SELECT * FROM {$_SESSION['datakey']}";
  10. $result = mysqli_query($con,"SELECT * FROM table name"); but where the table name goes I wan't it use $_SESSION['datakey'] to get the table name
  11. I don't wan't it to be inserted into the database but use the session info as the name of the table to grave data from for the html table
  12. Like this $result = mysqli_query($con,"INSERT * FROM $_SESSION['datakey']";
  13. $result = mysqli_query($con,"SELECT * FROM $_SESSION['datakey']";
  14. <?php $con=mysqli_connect("IP","Username","Password","Database"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT * FROM support"); echo "<table border='1'> <tr> <th>Name</th> <th>Email</th> <th>Priority</th> <th>Type</th> <th>Subject</th> <th>Message</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['email'] . "</td>"; echo "<td>" . $row['priority'] . "</td>"; echo "<td>" . $row['type'] . "</td>"; echo "<td>" . $row['subject'] . "</td>"; echo "<td>" . $row['message'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?>
  15. <?php $connection = mysql_connect("ip","user","password"); $db = mysql_select_db("database"); $query = mysql_query("select * from users"); echo '<table>'; echo '<tr><td>Username</td> <td>Email</td></tr>'; while ($record = mysql_fetch_array($query)) { echo '<tr> <td>' . $record['username'] . '</td> <td>' . $record['email'] . '</td> </tr>'; } echo '</table>'; ?>
  16. How to get specific data from MySQL and echo it inside <p> <?php echo "$_POST['id']; ?> </p>
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