0

对不起,这可能是一个愚蠢的问题,但我是新手,我在谷歌上找不到答案。这段代码给了我两个错误:

Warning: mysql_query() expects parameter 2 to be resource, boolean given in C:\xampp\htdocs\Music Collection\submitmusic.php on line 27

Warning: mysql_error() expects parameter 1 to be resource, string given in C:\xampp\htdocs\Music Collection\submitmusic.php on line 29

不知道如何解决这个问题请帮助我。

<html>
    <head>
        <title> Music Collection </title>
    </head>

    <body>
    <?php
    $con = "mysql_connect ('localhost','root','','music')";
    // Check Connection
    if (mysql_errno())
    {
        echo "Failed to connect: " . mysql_error();
    }
    else
    {
        $title = $_POST['title'];
        $artist = $_POST['artist'];
        $album = $_POST['album'];
        $location = $_POST['location'];
        $media = $_POST['media'];

        $sql = mysql_query("INSERT INTO entries (Title, Artist, Album, Location, Media) VALUES ('$title','$artist','$album','$location','$media')");

        if (!mysql_query($con,$sql))
        {
            die ('Error: ' . mysql_error($con));
        }
        else
        {
            echo "record added!";
        }
    }
    mysql_close($con);
    ?>

    </body>
</html>
4

5 回答 5

2

删除周围的双引号:

$con = "mysql_connect ('localhost','root','','music')";
于 2013-11-12T13:50:56.323 回答
0
  1. 只使用 mysqli_。mysql_ 已弃用,可能很快会从 php 中删除
  2. 你的连接是错误的。应该是这样的:

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

于 2013-11-12T13:50:44.543 回答
0

首先删除双引号:

$con = "mysql_connect ('localhost','root','','music')";

IE

$con = mysql_connect ('localhost','root','','music');

然后更改以下行

if (!mysql_query($con,$sql))

if (!mysql_query($sql,$con))

因为 mysql_query 的第一个参数需要是 sql 查询和第二个数据库标识符。

其余代码没问题

于 2013-11-12T14:10:40.770 回答
0

您还调用了 mysql_query 两次,除非您想要两次插入,否则您可能想要执行以下操作:

$sql = mysql_query("INSERT INTO entries (Title, Artist, Album, Location, Media) VALUES ('$title','$artist','$album','$location','$media')");
if (!sql)
...
于 2013-11-12T13:52:32.417 回答
0

将您的声明更改为

$sql = mysql_query("INSERT INTO entries (Title, Artist, Album, Location, Media) VALUES ('$title','$artist','$album','$location','$media')",$con);
于 2013-11-12T13:52:53.000 回答