mysqli

问题 im trying to write a sign up code in php mysqli and phpmyadmin. im having struggle in that. i have written the code as shown below but nothing happens when its run ie when a user tries to signs up. all the major coding is done I can not figure out where the problem/s is/are. i have considered all that steps that should be taken in the process ie validate user data, checking the db for current usernames. I would be grateful if anyone could assist me with this and point in the right direction!

问题 I have updated my original post based on what I learned from your comments below. It is a much simpler process than I originally thought. require '../database.php'; $pdo = Database::connect(); $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $sql = "SELECT * FROM Orders WHERE id = 430"; $q = $pdo->prepare($sql); $q->execute(array($id)); $data = $q->fetch(PDO::FETCH_ASSOC); echo 'Order Num: ' . $data['id'] . '<br>'; $sql = "SELECT * FROM Order_items JOIN Parts ON Parts.id = Order

问题 I have updated my original post based on what I learned from your comments below. It is a much simpler process than I originally thought. require '../database.php'; $pdo = Database::connect(); $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $sql = "SELECT * FROM Orders WHERE id = 430"; $q = $pdo->prepare($sql); $q->execute(array($id)); $data = $q->fetch(PDO::FETCH_ASSOC); echo 'Order Num: ' . $data['id'] . '<br>'; $sql = "SELECT * FROM Order_items JOIN Parts ON Parts.id = Order

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 6 years ago . I was impressed by the pricing for microsoft's azure cloud platform and want to switch. However, I have 976 php scripts that are written in mysqli. There was not a clear answer if it was or was not supported. Is

问题 This question already has answers here : How can I prevent SQL injection in PHP? (28 answers) mysqli_fetch_assoc() expects parameter / Call to a member function bind_param() errors. How to get the actual mysql error and fix it? (1 answer) Reference - What does this error mean in PHP? (35 answers) Closed 2 days ago . I'm new to PHP and I'm trying to authenticate the users by their username and password.(basically login form) If their password matches then they are redirected to welcome.php . I

问题 I have asked another question for my problem with collation setting, which seems to be ignored/overriden by some unwanted defaults. To workaround this, I want to use COLLATE in the queries returning "illegal mix of collations" error. However, when I try to, I get "COLLATION is not valid" error instead. The query in question (I didn't face this problem with other queries yet), this time with COLLATE on the parameter ( ? translates to a string of comma-separated numbers): SELECT k.url FROM kml

问题 Is it correct to use one view file with say 2 functions inside it. For example I have project table in my database and I have 2 functions inside the view file for this specific MySQL query. Each query returns a specific view so say function 1 = num_rows function 2 = data inside the rows Would this be correct or should I do something different ? 回答1: Your View Files should not have functions in them. The idea of MVC is to separate the Model, e.g. everything not related to presentation from the

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 5 years ago . I have a mysql column with the following attributes listDate TIMESTAMP DEFAULT = CURRENT_TIME and I am displaying it in my table like this echo "<td>" . $row['listDate'] . "</td>"; How would one go about converting the output to time since posted instead of the default: 2014-12-17 00:32:50 回答1: In

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . $check="SELECT * FROM users WHERE name = '$_POST[name]'"; $rs = mysqli_query($conn,$check); $data = mysqli_fetch_array($rs, MYSQLI_NUM); if($data[0] > 1) { echo "User Already in Exists<br/>"; }else echo "User does Not exist"; Is this the correct way? It says User does Not exist no matter what; and I know there

问题 either a issue with the code or csv is causing trouble for the import. Anything that could be a problem with the code, or with the csv? When i run the script i get a default messeage of : CSV File not readable This works for 5 out of the 15 files and cannot see an obvious issue or solution. http://www.dodgejeffgen.org/gs/issue1.csv <?php ini_set('auto_detect_line_endings',TRUE); ini_set('post_max_size', '128M'); ini_set('upload_max_filesize', '128M'); function clean($link, $str, $default ='')