所以这个语法确实连接并获取发布到数据库的信息,但我不知道如何获取图像的 url 以发布到数据库。无论我将 test1 更改为什么,它都不会发布。
protected function trim_file_name($name, $type, $index) {
// Remove path information and dots around the filename, to prevent uploading
// into different directories or replacing hidden system files.
// Also remove control characters and spaces (\x00..\x20) around the filename:
$file_name = trim(basename(stripslashes($name)), ".\x00..\x20");
// Add missing file extension for known image types:
if (strpos($file_name, '.') === false &&
preg_match('/^image\/(gif|jpe?g|png)/', $type, $matches)) {
$file_name .= '.'.$matches[1];
}
if ($this->options['discard_aborted_uploads']) {
while(is_file($this->options['upload_dir'].$file_name)) {
$file_name = $this->upcount_name($file_name);
}
}
$con = mysql_connect("localhost","-----","-----");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("-----", $con);
mysql_query("INSERT INTO posts (postid, post_content)
VALUES ('', 'test1')");
mysql_close($con);
{
header("Location: http://snarb.com/index.php");
}
return $file_name;
}