doumi9661 2014-10-27 23:39
浏览 7
已采纳

使用php [关闭]从网页上的数据库显示信息

Trying to get information from my database displayed in a webpage. Here is my code:

<body>
<?php
$con=mysqli_connect("localhost", "root", "");
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

/*
$db_selected = mysqli_select_db("music" , $con);
*/

$strSQL = "SELECT * FROM music";

$rs = mysqli_query($strSQL, $con);

while($row = mysqli_fetch_array($rs, MYSQLI_BOTH)) {

   // Write the value of the column FirstName (which is now in the array $row)
  echo $row['artist'] . "<br />";
  echo $row['title'] . "<br />";
  echo $row['format'] . "<br />";
  echo $row['notes'] . "<br />";

  }
?> 
  • 写回答

2条回答 默认 最新

  • duanping2005 2014-10-27 23:56
    关注

    Using mysqli_, the connection comes first for everything. Whereas the connection comes as the last parameter with mysql_ functions and you may have confused those by having used mysql_ functions in the past and have decided to start using mysqli_ now.

    Let's start with where you've commented out:

    /*
    $db_selected = mysqli_select_db("music" , $con);
    */
    

    Either you change that to:

    $db_selected = mysqli_select_db($con, "music");
    

    Or change:

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

    to:

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

    as I've made it that way below, using all four parameters in one line.

    Important sidenote:

    Make sure that both database and table are indeed called music.

    If your database is a different name than music, then that's what you need to use for the fourth parameter and not the table name.

    <body>
    <?php
    $con=mysqli_connect("localhost", "root", "", "music");
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    
    $strSQL = "SELECT * FROM music";
    
    $rs = mysqli_query($con, $strSQL);
    
    while($row = mysqli_fetch_array($rs, MYSQLI_BOTH)) {
    
       // Write the value of the column FirstName (which is now in the array $row)
      echo $row['artist'] . "<br />";
      echo $row['title'] . "<br />";
      echo $row['format'] . "<br />";
      echo $row['notes'] . "<br />";
    
      }
    ?> 
    

    Having used or die(mysqli_error($con)) to mysqli_query() would have signaled the error.

    More precisely:

    $rs = mysqli_query($con, $strSQL) or die(mysqli_error($con));
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥15 BP神经网络控制倒立摆
  • ¥20 要这个数学建模编程的代码 并且能完整允许出来结果 完整的过程和数据的结果
  • ¥15 html5+css和javascript有人可以帮吗?图片要怎么插入代码里面啊
  • ¥30 Unity接入微信SDK 无法开启摄像头
  • ¥20 有偿 写代码 要用特定的软件anaconda 里的jvpyter 用python3写
  • ¥20 cad图纸,chx-3六轴码垛机器人
  • ¥15 移动摄像头专网需要解vlan
  • ¥20 access多表提取相同字段数据并合并
  • ¥20 基于MSP430f5529的MPU6050驱动,求出欧拉角
  • ¥20 Java-Oj-桌布的计算