mysqli

问题 I have an issue where I wrote a stored procedure that inserts data into a bunch of tables and then finally returns and id by selecting it. when I execute the stored procedure in PHPMyAdmin, the data is inserted and the ID returned. When executing the procedure from PHP using MySQLi, the data is inserted into the tables however when I check the results to retrieve the new id, the result seems to be blank. Any help here would be appreciated. I am not sure if the procedure returns empty results

问题 I am writing a simple test script to get my head around MySQLi prepared statements. This script will become the basis for a function that needs to be slightly dynamic in its handling of results. I need the fetch method to return as an array similar to how mysql_fetch_array() behaves. I am aware some of these methods require mysqlnd. I have checked phpinfo. Mysqlnd is installed and present. See screenshot. Please note, PDO is available on my server, but I need to get this script working with

问题 I am writing a simple test script to get my head around MySQLi prepared statements. This script will become the basis for a function that needs to be slightly dynamic in its handling of results. I need the fetch method to return as an array similar to how mysql_fetch_array() behaves. I am aware some of these methods require mysqlnd. I have checked phpinfo. Mysqlnd is installed and present. See screenshot. Please note, PDO is available on my server, but I need to get this script working with

问题 I have several query strings which I want to execute at once using "mysqli_multi_query". This works. When I insert a query again to check each item in joined tables using "mysqli_query" it doesn't return any result nor any error from PHP. When I run the query string manually in phpmyadmin, everything works fine as it should. Here's my code: <?php $connect = mysqli_connect('localhost','root','','database'); $strquery = ""; $strquery .= "1st Query"; $strquyer .= "2nd Query"; if($multi = mysqli

问题 This question already has answers here : How to make mysqli connect function? (2 answers) Closed 4 years ago . I have to select data from MySQL Database. I have been looking for the answer, but still haven't found any. I am learning from W3School My MySQL doesn't have any username or password, so my code looks like this: <?php $servername = "localhost"; $dbName = "db_Test"; //Create Connection $conn = mysqli_connect($servername, $dbName); //Check Connection if(!$conn){ die("Connection Failed.

问题 I had a script for this that worked great in PHP 5.3. New host only supports 5.5 so I've done some modifying. So far, I can only get one row to be returned. I've been agonising for hours and I just can't see why. Is this possibly because I'm not using array_push() after the while loop? In my original script I had it but found out with this one I got a 500 Internal Server Error with it which disappeared if I took it out. <?php $mysqli = mysqli_connect("localhost", "user", "password", "database

问题 This seems to always return 1 for $item_result->num_rows; even though there are 0 rows in the DB. However, if an item exists it updates the row correctly. I'm sure something is wrong with my syntax but I'm having a hard time wrapping my head around this mysqli. $item_query = "SELECT COUNT(*) FROM `rel` WHERE `cart_id` = '".$cartId."' && `id_item` = '".$item_id."'"; $item_result = $mysqli->query($item_query) or die($mysqli->error.__LINE__); if($item_result->num_rows==1) { $item_query2 =

问题 Code: // get value of id that sent from address bar $id=filter_input(INPUT_GET, 'id', FILTER_VALIDATE_INT); if ($id === false) { //filter failed die('id not a number'); } if ($id === null) { //variable was not set die('id not set'); } //white list table $safe_tbl_name = ''; switch($tbl_name){ case 'Table1': $safe_tbl_name = 'MyTable1'; break; case 'Table2': $safe_tbl_name = 'MyTable2'; break; default: $safe_tbl_name = 'forum_questions'; }; $sql="SELECT * FROM `$safe_tbl_name` WHERE id=?"; if

问题 I'm kinda new to mysqli and I'm trying to display data on a table. Here's the code: <tbody> <?php $query = $mysqli->query("SELECT username,password,FName,LName,userAddress FROM tbl_users"); $no = 1; while($row = $query->fetch_assoc()){ ?> <tr> <td><?php echo $no++ ?></td> <td><?php echo $row['username'] ?></td> <td><?php echo $row['password'] ?></td> <td><?php echo $row['FName'] .' '. $row['LName'] ?></td> <td><?php echo $row['userAddress'] ?></td> <td> <a href="update.php?id=<?php echo $row[

问题 I'm kinda new to mysqli and I'm trying to display data on a table. Here's the code: <tbody> <?php $query = $mysqli->query("SELECT username,password,FName,LName,userAddress FROM tbl_users"); $no = 1; while($row = $query->fetch_assoc()){ ?> <tr> <td><?php echo $no++ ?></td> <td><?php echo $row['username'] ?></td> <td><?php echo $row['password'] ?></td> <td><?php echo $row['FName'] .' '. $row['LName'] ?></td> <td><?php echo $row['userAddress'] ?></td> <td> <a href="update.php?id=<?php echo $row[