我试图在sql数据库中获取文件路径和文件名。
我在'/ music / single /'中有一个名为'run.mp3'的文件,为此我需要填写表格:
id artist filename filepath
ai single run /music/single/run.mp3
我搜索谷歌有关此问题的任何解释,因为我对php和mysql很新,我真的需要帮助。
我尝试过:
$files = scandir(../music/);
foreach($files as $file) {
$paths = explode("/", $file);
$artist = $paths[1];
$song = $paths[2];
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"INSERT INTO playlist (artist, song,filepath) VALUES ('$artist', '$song', '$file')";
mysqli_close($con);
$print_r($files);
的编辑输出:
Array ( [0] => . [1] => .. [2] => C418 - Alpha [3] => C418 - Beta ),
Array ( [0] => . [1] => .. [2] => C418 - Alpha [3] => C418 - Beta ),
Array ( [0] => . [1] => .. [2] => C418 - Alpha [3] => C418 - Beta ),
Array ( [0] => . [1] => .. [2] => C418 - Alpha [3] => C418 - Beta )
答案 0 :(得分:0)
是SQL还是MySQL数据库?对于MySQL,您可以使用它。首先打开你的连接。所以填写你的数据库,用户名和密码。
$con=mysqli_connect("localhost","username","password","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"INSERT INTO your_table (artists, filename, filepath)
VALUES ('single', 'run', ' /music/single/run.mp3' )");
mysqli_close($con);
请点击此处:http://www.w3schools.com/php/php_mysql_insert.asp了解更多信息。
答案 1 :(得分:0)