[PHP]
<?php
class FileRemoval {
var $fileNameArray, $isRemoved, $errorMsg = '';
function FileRemoval() {
$this->fileNameArra y = array();
$this->isRemoved = 0;
}
function getErrorMsg() {
return $this->errorMsg;
}
function getIsRemoved() {
return $this->isRemoved;
}
function setFileNameArra y() {
global $argv;
$junk = array_shift($ar gv); // LOB OFF FIRST ARRAY ELEMENT YOU
DON'T NEED THE SCRIPT NAME
$this->fileNameArra y = $argv;
}
function remove() {
$this->setFileNameArr ay();
if (sizeof($this->fileNameArra y) == 0) {
$this->isRemoved = 0;
$this->errorMsg = "No files provided for removal\n";
}
if (!$this->errorMsg) {
foreach ($this->fileNameArra y as $val) {
if (!file_exists($ val) || strlen($val) == 0) {
$this->isRemoved = 0;
$this->errorMsg = "File: $val does not exist\n";
echo $this->errorMsg;
break;
}
}
if (!$this->errorMsg) @unlink($file);
}
if (!$this->errorMsg) $this->isRemoved = 1;
}
function writeErr() {
if ($this->errorMsg) {
$fileID = fopen('./cmds.err', 'r') or die("Could not find error
log");
$myErr = '[' . date("%d/%b/%Y:%H:%I:%S"). ']' . $this->errorMsg .
"\n";
fputs($fileID, $myErr);
fflush($fileID) ;
fclose($fileID) ;
fwrite(STDOUT, $myErr);
}
}
}
$fileRemoval =& new FileRemoval();
$fileRemoval->remove();
echo "**" . $fileRemoval->getIsRemoved() ;
if (!$fileRemoval->getIsRemoved() ) $fileRemoval->writeErr();
$fileRemoval = null;
?>
[/PHP]
I am writing a very simple PHP script that will be called from a
front-end bash script (and a TCL script - long story) that will remove
a list of files passed into the PHP script as $argv array. Problem is
right now that I am unable to get it to write to the error log (see
method writeErr) nor am I able to produce the error in stdout (again
see method writeErr), I get the error message:
Supplied argument is not a valid File-Resource handle in line 53:
line 53: fwrite(STDOUT, $myErr);
I am not sure now how to do this so I can use some help.
Thanx
Phil 3 3569
If you open your file for reading, of course you can't write to it. Pass
"a+" as the second argument to fopen() for appending to a file.
And there is not STDOUT keyword in PHP. Do an echo if you want to output
something to stdout.
Uzytkownik "Phil Powell" <so*****@erols. com> napisal w wiadomosci
news:1c******** *************** ***@posting.goo gle.com... [PHP] <?php
class FileRemoval {
var $fileNameArray, $isRemoved, $errorMsg = '';
function FileRemoval() { $this->fileNameArra y = array(); $this->isRemoved = 0; }
function getErrorMsg() { return $this->errorMsg; }
function getIsRemoved() { return $this->isRemoved; }
function setFileNameArra y() { global $argv; $junk = array_shift($ar gv); // LOB OFF FIRST ARRAY ELEMENT YOU DON'T NEED THE SCRIPT NAME $this->fileNameArra y = $argv; }
function remove() { $this->setFileNameArr ay(); if (sizeof($this->fileNameArra y) == 0) { $this->isRemoved = 0; $this->errorMsg = "No files provided for removal\n"; } if (!$this->errorMsg) { foreach ($this->fileNameArra y as $val) { if (!file_exists($ val) || strlen($val) == 0) { $this->isRemoved = 0; $this->errorMsg = "File: $val does not exist\n"; echo $this->errorMsg; break; } } if (!$this->errorMsg) @unlink($file); } if (!$this->errorMsg) $this->isRemoved = 1; }
function writeErr() { if ($this->errorMsg) { $fileID = fopen('./cmds.err', 'r') or die("Could not find error log"); $myErr = '[' . date("%d/%b/%Y:%H:%I:%S"). ']' . $this->errorMsg . "\n"; fputs($fileID, $myErr); fflush($fileID) ; fclose($fileID) ; fwrite(STDOUT, $myErr); } }
}
$fileRemoval =& new FileRemoval(); $fileRemoval->remove(); echo "**" . $fileRemoval->getIsRemoved() ; if (!$fileRemoval->getIsRemoved() ) $fileRemoval->writeErr(); $fileRemoval = null;
?> [/PHP]
I am writing a very simple PHP script that will be called from a front-end bash script (and a TCL script - long story) that will remove a list of files passed into the PHP script as $argv array. Problem is right now that I am unable to get it to write to the error log (see method writeErr) nor am I able to produce the error in stdout (again see method writeErr), I get the error message:
Supplied argument is not a valid File-Resource handle in line 53:
line 53: fwrite(STDOUT, $myErr);
I am not sure now how to do this so I can use some help.
Thanx Phil
"Phil Powell" <so*****@erols. com> wrote in message
news:1c******** *************** ***@posting.goo gle.com...
<SNIP CODE> I am writing a very simple PHP script that will be called from a front-end bash script (and a TCL script - long story) that will remove a list of files passed into the PHP script as $argv array. Problem is right now that I am unable to get it to write to the error log (see method writeErr) nor am I able to produce the error in stdout (again see method writeErr), I get the error message:
Supplied argument is not a valid File-Resource handle in line 53:
line 53: fwrite(STDOUT, $myErr);
I am not sure now how to do this so I can use some help.
Sounds like your're using a PHP version older than 4.3.0 where STDOUT et al,
are not automatically opened. You could do this yourself, using something
like:
define('STDOUT' , fopen("php://stdout", "r"));
define('STDERR' , fopen("php://stderr", "r"));
...
See: http://www.php.net/manual/en/features.commandline.php
for more details.
I hope this helps.
Anthony Borla
"Anthony Borla" <aj*****@bigpon d.com> wrote in message
news:KV******** **********@news-server.bigpond. net.au... "Phil Powell" <so*****@erols. com> wrote in message news:1c******** *************** ***@posting.goo gle.com... <SNIP CODE> I am writing a very simple PHP script that will be called from a front-end bash script (and a TCL script - long story) that will remove a list of files passed into the PHP script as $argv array. Problem is right now that I am unable to get it to write to the error log (see method writeErr) nor am I able to produce the error in stdout (again see method writeErr), I get the error message:
Supplied argument is not a valid File-Resource handle in line 53:
line 53: fwrite(STDOUT, $myErr);
I am not sure now how to do this so I can use some help.
Sounds like your're using a PHP version older than 4.3.0 where STDOUT et
al, are not automatically opened. You could do this yourself, using something like:
define('STDOUT' , fopen("php://stdout", "r")); define('STDERR' , fopen("php://stderr", "r")); ...
See:
http://www.php.net/manual/en/features.commandline.php
for more details.
I hope this helps.
Anthony Borla
Actually none of that worked for me, including define(). The solution was a
bit esoteric and I don't have a technical explanation for it, but in the TCL
script that calls the PHP code I did this
set blah [eval "exec php -q /./fileremoval.php $fileList]
puts $blah
Apparently the TCL script that calls the PHP script might have control of
STDOUT so there was nothing that PHP could do to output its results, so I
simply allowed for any results to go to the TCL script variable $blah that
would be set to whatever the PHP script sends it.
I can't explain it any better than that.
Phil This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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