php - fwrite()：提供的参数不是有效的流资源

Question

使用此代码时，我不断收到 3 个错误：

Warning: fopen() [function.fopen]: Filename cannot be empty
Warning: fwrite(): supplied argument is not a valid stream resource
Warning: fclose(): supplied argument is not a valid stream resource

我不知道该怎么办。我是一个php菜鸟。

<?php

$random = rand(1, 9999999999);
$location = "saves/".$random;


while (file_exists($location)) {
$random = rand(1, 999999999999);
$location = "saves/".$random;
}

$content = "some text here";
$fp = fopen($location,"wb");
fwrite($fp,$content);
fclose($fp);
?>

Answer 1

根据您在编辑之前的原始问题：

由于该文件尚不存在，因此您的while条件将不起作用，这就是您收到这些错误消息的原因。

而且由于您对文件使用随机数，因此您永远不会知道首先要打开哪个文件。只需删除while循环。

尝试这个：

<?php
$random = rand(1, 999999999999);
$location = "saves/".$random;

$content = "some text here";
$fp = fopen($location,"wb");
fwrite($fp,$content);
fclose($fp);
?>

Answer 2

从您拥有的代码看来，$location 似乎只存在于 while 循环的范围内。尝试

<?php

$location = "";
while (file_exists($location)) {
    $random = rand(1, 999999999999);
    $location = "saves/".$random;
}

$content = "some text here";
$fp = fopen($location,"wb");
fwrite($fp,$content);
fclose($fp);
?>

Answer 3

首先，您必须为$location变量设置值，或者由于尚未创建文件，请尝试以下操作：

$random = rand(1, 999999999999);
$location = "saves/".$random;


$content = "some text here";

//if(file_exists($location)) $fp = fopen($location,"wb");
$fp = fopen($location, 'wb') or die('Cannot open file:  '.$location); //implicitly creates file

fwrite($fp,$content);
fclose($fp);