我正在尝试编写“身份流包装器”。第一个目的是让PHP通过这个包装器进行传输,这将与PHP本身完全相同。第二种是修改这个包装器以便做出决策。
所以第一项工作是将stream_open()映射到fopen(),将stream_write()映射到fwrite(),依此类推。 fopen()映射似乎工作正常,但是当我尝试调用fwrite()时,由于某种原因,包装器的内部资源变为布尔值,并且无法写入文件。 fread()会发生同样的事情。
任何人都可以解释为什么以及如何避免/解决?感谢...
以下是我的脚本输出:
36: resource(6) of type (stream)
60: resource(7) of type (stream)
49: bool(true)
PHP Warning: fwrite() expects parameter 1 to be resource, boolean given in /var/www/projects/stream/test.php on line 51
PHP Stack trace:
PHP 1. {main}() /var/www/projects/stream/test.php:0
PHP 2. fwrite() /var/www/projects/stream/test.php:62
PHP 3. IdentityStreamWrapper->stream_write() /var/www/projects/stream/test.php:62
PHP 4. fwrite() /var/www/projects/stream/test.php:51
63: resource(7) of type (stream)
这是它的代码(并不足以重现):
<?php
class IdentityStreamWrapper {
var $fileHd;
var $fileName;
function __construct() {
static::unwrap();
}
function __destruct() {
static::wrap();
}
static function unwrap() {
stream_wrapper_restore("file");
}
static function wrap() {
stream_wrapper_unregister("file");
stream_wrapper_register("file", "IdentityStreamWrapper");
}
function stream_open($path, $mode, $options, &$opened_path)
{
if ($options >0) {
$tOptions = $options;
$tOptions = ($tOptions > STREAM_USE_PATH) ? ($tOptions - STREAM_REPORT_ERRORS) : $tOptions;
if ($tOptions === STREAM_USE_PATH) {
$opened_path = realpath($path);
}
}
$this->fileName = $path;
$this->fileHd = fopen($path,$mode);
$this->position = 0;
echo __LINE__.": ";
var_dump($this->fileHd);
return $this->fileHd;
}
function stream_write($data)
{
echo __LINE__.": ";
var_dump($this->fileHd);
$ret = fwrite($this->fileHd,$data);
return $ret;
}
}
IdentityStreamWrapper::wrap();
$fp = fopen("test.txt", "w+");
echo __LINE__.": ";
var_dump($fp);
fwrite($fp, "line1\n");
echo __LINE__.": ";
var_dump($fp);
IdentityStreamWrapper::unwrap();
?>
答案 0 :(得分:0)
方法stream_open必须返回一个布尔值,而不是fopen()。这就是修复课程的全部内容。