PHP 读取文件作为字节数组
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PHP read file as an array of bytes
提问by BlackRoot
I have a file written using JAVAprogram which has an array of integers written as binary I made an for loop over this array and write it using this method
我有一个使用JAVA程序编写的文件，它有一个整数数组写成二进制我在这个数组上做了一个 for 循环并使用这个方法编写它
public static void writeInt(OutputStream out, int v) throws IOException {
out.write((v >>> 24) & 0xFF);
out.write((v >>> 16) & 0xFF);
out.write((v >>> 8) & 0xFF);
out.write((v >>> 0) & 0xFF);
}
I'm ask about how to read this file in PHP.
我问的是如何在PHP.
回答by Orangepill
I believe the code you are looking for is:
我相信您正在寻找的代码是：
$byteArray = unpack("N*",file_get_contents($filename));
UPDATE:Working code supplied by OP
更新：OP 提供的工作代码
$filename = "myFile.sav";
$handle = fopen($filename, "rb");
$fsize = filesize($filename);
$contents = fread($handle, $fsize);
$byteArray = unpack("N*",$contents);
print_r($byteArray);
for($n = 0; $n < 16; $n++)
{
echo $byteArray [$n].'<br/>';
}
回答by Joni
You should read the file as a string and then use unpack to parse it. In this case you have stored a 32-bit integer in big endian byte order, so use:
您应该将文件作为字符串读取，然后使用 unpack 来解析它。在这种情况下，您以大端字节顺序存储了一个 32 位整数，因此请使用：
$out = unpack("N", $in);
The format codes are documented here: http://www.php.net/manual/en/function.pack.php
格式代码记录在此处：http: //www.php.net/manual/en/function.pack.php
