MySQL数据库xampp

Question

我试图建立一个论坛，但为此，我需要一个数据库与数据。这是我的剧本：

(连接和打印数据脚本)

    <!DOCTYPE html>
<html>
<head>
    <title>Login Page</title>
</head>
<body>
    <div id ="frm">
        
            <?php $con = connect(); ?>

<div>
    <?php renderContent($con); ?>
</div>
        
    </div>

</body>
</html>



<?php
    function connect() {
        $con=mysqli_connect("localhost", "root", "", "forum");
        var_dump($con);
        // Check connection
        if (mysqli_connect_errno()) {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        } else {
            return $con;
        }
    }
    
    function renderContent($con) {
        $sql = "SELECT * FROM textforum";
        $result = $con->query($sql);
        
        
        
            // output data of each row
            
                echo $row["text"]. "<br>";
            
        
        
    }

输出错误是：

对象(Mysqli)#1 (19) { "affected_rows"=> int(0) "client_info"=> string(79) "mysqlnd 5.0.11-dev - 20120503 -$Id“：76b08b24596e12d4553bd41fc93cccd5bac2fe7a $”client_version"=> int(50011) "connect_errno"=> int(0) "connect_error"=> NULL“errno "errno"=> int”(0)“错误”=> string(0) "error_list"=> (0)数组{} "field_“=>(0){}。计数“=> int(0) "host_info"=> string”(20) "localhost 30 TCP/IP“info"=> NULL "insert_id"=> int(0)”=>int(0) "server_info"=> string(21) "5.5.5-10.1.21-MariaDB“"server_version"=> int(50505) "stat"=> string(134) "Uptime: 7020 Thread:1个问题:1674个慢速查询:0打开:38个平行表:一个打开表:每秒30个查询：0.238“sqlstate"=>字符串(5) "00000”"protocol_version"=> int(10) "thread_id"=> int(177) "warning_count"=> int(0) } 注意:未定义变量:第42行C:\xampp\htdocs\LoginTest\Login\process.php中的行

这是我的输出: object(mysqli)#1 (19) { "affected_rows"=> int(0) "client_info"=> string(79) "mysqlnd 5.0.11-dev - 20120503 - $Id: 76b08b24596e12d4553bd41fc93cccd5b7a $“”client_version"=> int(50011) "connect_errno"=> int(0) "connect_error"=> "errno"=> int(0)“错误”=> string(0) "error_list"=> (0)数组(0){} "field_count"=> int(0) "host_info"=> string(20) "localhost 20 TCP/IP“info "info"=> NULL "insert_id"=> int(0) "server_info"=> string(21) "5.5.5-10.1.21-MariaDB”"server_version"=> int(50505) "stat"=> string(134) "Uptime: 8825 Thread:1个问题: 1683个慢查询:0 Opens: 38个平行表:1个打开表“：每秒30次查询: 0.190“sqlstate"=> string(5) "00000”"protocol_version"=> int(10) "thread_id"=> int(182) "warning_count"=> int(0) } Text blalblalblbText blalblalblb

blalblalblbText是正确的输出，但其他输出不是我如何摆脱它。

Answer 1

mysqli_connect()接受第四个参数作为databasename

1. 主机名
2. 用户名
3. 密码
4. 数据库名

您错过了在连接中传递数据库。

试试这个：

$con=mysqli_connect("localhost", "root", "","database_name_here");

更新您的函数renderContent

function renderContent($con) {
    $sql = "SELECT * FROM textforum";
    $result = mysqli_query($con,$sql);
    if (mysqli_num_rows($result) > 0) {
      // output data of each row
      while($row = mysqli_fetch_assoc($result)) {
        echo $row["text"]. "<br>";
      }
    }else{
      echo "0 results";
    }
}