php 无法打开流：无效的参数
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failed to open stream: Invalid argument
提问by Martin Wickman
In this code :
在这段代码中：
$path = "C:\NucServ\www\vv\static\arrays\news.php";
$fp = fopen($path, "w");
if(fwrite($fp=fopen($path,"w"),$text))
{
echo "ok";
}
fclose($fp);
I have this error message:
我有这个错误信息：
failed to open stream: Invalid argument
What is wrong in my code?
我的代码有什么问题？
回答by Martin Wickman
Your backslashes is converted into special chars by PHP. For instance, ...arrays\news.phpgets turned into
PHP 将您的反斜杠转换为特殊字符。例如，...arrays\news.php变成
...arrays
ews.php
You should escape them like this:
你应该像这样逃避他们：
$path = "C:\NucServ\www\vv\static\arrays\news.php";
Or use singles, like this:
或者使用单打，像这样：
$path = 'C:\NucServ\www\vv\static\arrays\news.php';
Also, your ifis messed up. You shouldn't fopenthe file again. Just use your $fpwhich you already have.
另外，你的情况if很糟糕。你不应该fopen再次文件。只需使用您$fp已经拥有的。
回答by user7609550
path error:
$path = 'C:/NucServ/www/vv/static/arrays/news.php';file lock:
user file_get_contents replace fopen
路径错误：
$path = 'C:/NucServ/www/vv/static/arrays/news.php';文件锁：
user file_get_contents replace fopen
