我有以下代码:// db connection info set up earlier $sql= "SELECT `TABLE_1.ID`, `TABLE_2.ID`, `POTATO` FROM `TABLE_1.ID` LEFT JOIN `TABLE_2` ON `TABLE_1`.`ID` = `TABLE_2`.`ID_OF_OTHER_TABLE`;"; $rows = mysqli_query($connection, $sql); foreach ($rows as $row){$potato = $row["POTATO"];$id = $row["TABLE_2.ID"]; }我无法获得TABLE_2.ID.我试图做一个print_r以获得正确的格式,但是它说这是一个mysqli对象,我没有...
我很困惑为什么以下代码成功地向我的数据库表添加了一个新行,而mysqli_affected_rows($dbc)返回“-1”,因此在signup.php中出现错误: dbc.inc.php:DEFINE ('DB_USER', 'root'); DEFINE ('DB_PASSWORD', ''); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'v');$dbc = mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die('Error connecting to MySQL server.');mysqli_set_charset($dbc, 'utf8');signup.ph...
直升机,我使用“bind_result”,“LIKE CONCAT”…通过两个查询字符串达到全文搜索和分页. 但是如何将bind_result方法更改为fetch_assoc?<?php $mysqli = new mysqli("localhost", "", "", "test");$query_string="hello"; //keywords$sqltxt="SELECT * FROM `table` WHERE text1 LIKE CONCAT('%', ?, '%')";//first query : for get the total number of data $stmt = $mysqli->prepare($sqltxt); $stmt->bind_param("s",$query_s...
我以为我可以使用MYSQLI_STMT_NUM_ROWS和MYSQLI_STMT_STORE_RESULT来检查否.返回的行数. (参见注释行/// 1 ///,/// 2 ///,/// 3 ///) 但它似乎不在下面的上下文中. 这段代码确实有效(没有注释行),但我试图添加额外的检查,以确认返回的记录不超过1条. (尽管情况应该总是这样,因为表中的电子邮件字段是唯一的,但无论如何都不会受到影响). 有人能说清楚我做错了吗? 这是我得到的错误(第84行,如果WHILE …行):An error occurred in s...
这是我在php中的简单查询,使用mysqli面向对象的样式:$query = "SELECT name FROM usertable WHERE id = ?"; $stmt = $mysqli->prepare($query); $stmt->bind_param('i', $id); $id= $_GET['id']; $stmt->execute(); $stmt->bind_result($name);while($stmt->fetch()){echo $name." "; }$stmt->free_result(); $stmt->close();这很好用.我获取了从select语句中检索的名称列表. 现在,我希望在内部使用$name变量作为另一个查询的参数...
参见英文答案 > Can I mix MySQL APIs in PHP? 4个当我尝试调用我的函数时,我正在调用非对象上的成员函数query(). 我的代码看起来像这样:function add_profile() {$hostname = "localhost";$dbusername = "username";$dbname = "dbname";$dbpassword = "password";$link = mysqli_connect($hostname, $dbusername, $dbpassword, $dbname); if (!$link) { die('Connect Error (' . mysqli_conn...
参见英文答案 > Single Value Mysqli 8个我只需要从mysql数据库中获取用户名为X的成员的id.这只能通过while循环完成,还是有其他方法吗? 我在想的是:$id = mysqli_query($con,'SELECT id FROM membrs WHERE username = '$username' LIMIT 1)谢谢,解决方法:您可以使用:mysqli_fetch_array();// For Instance$id_get = mysqli_query($con, "SELECT id FROM membrs WHERE username='$username' ...
我有一个mysql连接,它包含在一个单独的文件中:require 'settings.php';我有一个包含所有功能的文件,还包括:require 'functions.php';在那里的设置看起来像这样:$db = mysqli_connect("host", "username", "passwort", "database");if(!$db) {exit("Error: ".mysqli_connect_error());}并且函数使用此连接,如下所示:function includehomepage() {$data = array();$query = "SELECT pagecontent FROM `pages` WHERE `id` = `0`";...
我有以下代码用于注册表单.我想检查电子邮件,看它是否已经登录数据库,如果是,则返回错误.$name = $_POST['name']; $email = $_POST['email'];$result = mysqli_query($connection,"SELECT * FROM users WHERE email='$email'");if($result=="") { echo("email was not found"); } else {echo("email was found"); }如果结果变回空洞,我会遇到麻烦.我已经尝试过使用几件事,如果找不到任何东西,就无法弄清楚如何让它正常工作.解决方法...
我有办法获取表的列的名称.它工作正常,但现在我想更新到新的mysqli? (我尝试了mysqli_fetch_field,但我不知道如何应用于这种情况,我不确定它是否是wright选项) 如何用mysqli做同样的事情? :$sql = "SELECT * from myTable"; $result = mysql_query($sql,$con); $id = mysql_field_name($result, 0); $a = mysql_field_name($result, 1);echo $id; echo $a;解决方法:这是实现这个缺失函数的方法:function mysqli_field_name($re...
我希望能够创建扩展MySQLi类的类来执行所有SQL查询.$mysql = new mysqli('localhost', 'root', 'password', 'database') or die('error connecting to the database');我不知道如何在不全局化$mysql对象的情况下执行此操作以在我的其他方法或类中使用.class Blog {public function comment() {global $mysql;//rest here }}任何帮助,将不胜感激. 谢谢.解决方法:我的建议是创建一个Singleton DataAccess类,在全局配置文件中实例化该...
我想将下面的MySQL查询更改为MySQLi(预处理语句),但我不知道如何操作,因为它有多行要选择.任何人都可以指出正确的方式.$check_added_files = mysql_query("select * from `vpb_uploads` where `username` = '".mysql_real_escape_string($username)."' and `firstname` = '' and `image_one` != '' and `image_two` != '' and `image_three` != '' and `image_four` != '' and `image_five` != ''");if(mysql_num_rows($check_adde...
我是一名初学者,也是一名文凭学生…我使用localhost创建了数据库…我在查看我的数据库时遇到问题…请帮助我…我希望你能用完整的代码帮我…这是错误…Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\SLR\View S110 PC01.php on line 10 cannot select DB这是我的代码……<?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password ...
我正在编写一个简单的代码来从我的数据库中名为Plasma的表中获取2个元素.似乎连接正常,但由于未知原因我无法选择任何数据库. 代码:<?php $db = "Plasma"; $dbH = "localhost"; $dbU = "plasma"; $dbP = "plasma";$dbCon = mysqli_connect($dbH,$dbU,$dbP,$db);if(!dbCon){echo "Conenction Fail";}mysqli_select_db($dbCon,$db);$qry = "select Mid,Mname from ya_movies order by DOA limit 5;";$Response = mysqli_query($dbC...
我遇到了一些麻烦.所以我试图使用MySQLi连接到我的数据库,但我收到此错误:Warning: mysqli::mysqli(): (28000/1045): Access denied for user 'root'@'localhost' (using password: NO) in /home/venge/public_html/library/classes/database.class.php on line 16Warning: Missing argument 1 for Users::__construct(), called in /home/venge/public_html/routing.php on line 4 and defined in /home/venge/public_html/libra...