fopen不允许为文件名创建变量

Question

我正在尝试创建一个文件并逐行写入。当尝试使用变量中的文件名创建文件时，会失败。但如果我对文件名进行硬编码，它就会起作用。

回声为：

无法打开/创建文件：CleanStatements\Clean_IRS01HTAX.TEXT

这是有效的：

if ($handle) {
$cleanFileHandle = fopen( "CleanStatements\Clean_IRS01HHTAX.TXT", "w") or die("Unable to open/create file: ".$this->CleanFilePath);
while (($line = fgets($handle)) !== false) {
fwrite($cleanFileHandle, $line);
}
fclose($cleanFileHandle);
fclose($handle);
} else {
// error opening the file.
}

这不是

if ($handle) {
$cleanFileHandle = fopen( $this->CleanFilePath, "w") or die("Unable to open/create file: ".$this->CleanFilePath);
while (($line = fgets($handle)) !== false) {
fwrite($cleanFileHandle, $line);
}
fclose($cleanFileHandle);
fclose($handle);
} else {
// error opening the file.
}

这是完整的课程：

/**
* Class StatementFile
*/
class StatementFile
{
var $Name;
var $FilePath;
var $Type;
var $CleanFileName;
var $CleanFilePath;
function __construct($filePath){
$this->Name = basename($filePath).PHP_EOL;
$this->FilePath = 'Statements\'.$filePath;
$this->Type = null;
$this->CleanFileName = "Clean_".$this->Name;
$this->CleanFilePath = "CleanStatements\" . $this->CleanFileName;
}
function cleanStatement(){
$handle = fopen($this->FilePath, "r");
echo "Opening file: ".$this->FilePath ."<br/>";
if ($handle) {
$cleanFileHandle = fopen( $this->CleanFilePath, "w") or die("Unable to open/create file: ".$this->CleanFilePath);
while (($line = fgets($handle)) !== false) {
// process the line read.
// clean line here
fwrite($cleanFileHandle, $line);
}
fclose($cleanFileHandle);
fclose($handle);
} else {
// error opening the file.
}
}

Answer 1

tl；dr

从$this->Name中移除PHP_EOL

---

这只是猜测，但我敢打赌PHP_EOL不包含有效的文件名字符：

$this->Name = basename($filePath).PHP_EOL;
// which ultimate ends up concatenated in
$this->CleanFilePath

如果你真的，真的，出于任何原因，真的需要将其保持在$this->name中，那么应用trim()

$cleanFileHandle = fopen( trim( $this->CleanFilePath ), "w") or die("Unable to open/create file: ".trim( $this->CleanFilePath) );