我想列出2目录中的文件,但我收到了以下错误:
参数#1不是C:''wamp''www''new.php中第62行的数组:
$directory = '/openssl/bin/';
$extension = '.pem';
$directory2 = '/openssl/try/bin/';
$extension2 = '*.*';
if ( file_exists($directory) ) {
foreach(glob($directory.'*'.$extension) as $file){
foreach(glob($directory2.'*'.$extension2) as $file2){
$result = array_merge($file, $file2); // line 76 error 1
?>
<tr >
我希望输出是两个目录中的文件名列表
<td> <?php echo basename($result); ?></td> // this should be list all the filename inside the 2 directory in table
file_get_contents():文件名在C:''wamp''www''new.php中不能为空,位于下方的第76行
$data = openssl_x509_parse(file_get_contents($result));
Hi glob需要文件的物理路径。所以试着运行这个代码
$dir = __DIR__;
$directory = $dir.'/openssl/bin/';
$extension = '.pem';
$directory2 = $dir.'/openssl/try/bin/';
$extension2 = '*.*';
if ( file_exists($directory) ) {
$myArr = array_merge(glob($directory.'*'.$extension), glob($directory2.'*'.$extension2));
foreach($myArr as $file){
echo $file;
echo "<br>";
}
}
<?php
$directory = '/openssl/bin/';
$extension = '.pem';
$directory2 = '/openssl/try/bin/';
$extension2 = '*.*';
if ( file_exists($directory) ) {
$myArr = array_merge(glob($directory.'*'.$extension), glob($directory2.'*'.$extension2));
foreach($myArr as $file){
echo $file;
echo "<br>";
}
}
?>
感谢volkinc和其他人的帮助,分享他们的意见和建议。我在这里得到了答案。我只是对glob执行代码有点困惑。感谢你为我安排了volkinc干杯