CLI ( command-line ) PHP displaying source code in-line when programis executed

[Double Echo]

Hi all,

I'm using PHP 4.4.2, and use PHP on both the command-line and the web.

I am running PHP on SuSE 10 Linux , in a VMware 5.5 workstation, using
Apache 2.0.55 , on my Dell laptop. Everything has been running flawlessly
without problems. Very amazing to use VMware, it has worked beautifully.

uname -a

Linux xxxxxxx 2.6.13-15.8-default #1 Tue Feb 7 11:07:24 UTC 2006 i686 i686 i386
GNU/Linux

php -v
PHP 4.4.2 (cli) (built: Feb 7 2006 20:13:29)
Copyright (c) 1997-2006 The PHP Group
Zend Engine v1.3.0, Copyright (c) 1998-2004 Zend Technologies

php -m

[PHP Modules]
bcmath
bz2
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ctype
curl
dbx
dio
domxml
exif
ftp
gd
gettext
gmp
iconv
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mssql
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ncurses
openssl
overload
pcre
posix
session
shmop
sockets
standard
sysvsem
sysvshm
tokenizer
wddx
xml
yp
zlib

[Zend Modules]


When I run the following program from the command line it prints out the
source code as well as the output from the program. I run other PHP programs
on the command-line and this doesn't happen. It's really peculiar. The
error output is below.

I copied this off of php.net, and have modified it only slightly to allow
command-line input. I was trying to find a CSV solution, which I ended up
finding elsewhere, but am curious why this program would error out the
way it does--have I misconfigured something? My php.ini is basically
unchanged, if at all. I think I might have set the path to mysql but that's
about it. When I run this program from the web it works without error.

<?php

if ( $argv[0] )
{
$file = $argv[0] ;
}
else {
$file = $_GET['file'] ;
print "<font face=arial>\n" ;
}

$row = 1;

$handle = fopen("$file", "r");

while ( ( $data = fgetcsv($handle , 1000, ",") ) !== FALSE )
{

$num = count($data);

print "$num fields in line $row:<br> \n";

$row++;

for ($c = 0; $c < $num; $c++)
{
print "$data[$c]<br> \n" ;
}

}

fclose($handle) ;

?>


=== COMMAND-LINE OUTPUT ===


1 fields in line 1:<br>
<?php<br>
1 fields in line 2:<br>
<br>
1 fields in line 3:<br>
if ( $argv[0] )<br>
1 fields in line 4:<br>
{<br>
1 fields in line 5:<br>
$file = $argv[0] ;<br>
1 fields in line 6:<br>
}<br>
1 fields in line 7:<br>
else {<br>
1 fields in line 8:<br>
$file = $_GET['file'] ;<br>
1 fields in line 9:<br>
print "<font face=arial>\n" ;<br>
1 fields in line 10:<br>
}<br>
1 fields in line 11:<br>
<br>
1 fields in line 12:<br>
$row = 1;<br>
1 fields in line 13:<br>
<br>
2 fields in line 14:<br>
$handle = fopen("$file"<b r>
r);<br>
1 fields in line 15:<br>
<br>
3 fields in line 16:<br>
while ( ( $data = fgetcsv($handle <br>
1000<br>
,) ) !== FALSE ) <br>
1 fields in line 17:<br>
{<br>
1 fields in line 18:<br>
<br>
1 fields in line 19:<br>
$num = count($data);<b r>
1 fields in line 20:<br>
<br>
1 fields in line 21:<br>
print "$num fields in line $row:<br> \n";<br>
1 fields in line 22:<br>
<br>
1 fields in line 23:<br>
$row++;<br>
1 fields in line 24:<br>
<br>
1 fields in line 25:<br>
for ($c = 0; $c < $num; $c++) <br>
1 fields in line 26:<br>
{<br>
1 fields in line 27:<br>
print "$data[$c]<br> \n" ;<br>
1 fields in line 28:<br>
}<br>
1 fields in line 29:<br>
<br>
1 fields in line 30:<br>
}<br>
1 fields in line 31:<br>
<br>
1 fields in line 32:<br>
fclose($handle) ;<br>
1 fields in line 33:<br>
<br>
1 fields in line 34:<br>
?> <br>

[rlee0001]

Re: CLI ( command-line ) PHP displaying source code in-line when program is executed

Echo Echo,

At the beginning output $file. For some reason the file is reading
itself as input. If you are using:

php myprog.php infile.csv

....on the command line then I think you should be trying to get
"infile.csv " (or whatever) using $argv[1] not $argv[0]. Otherwise you
are passing the files own source to itself and the line

print "$data[$c]<br> \n" ;

Is causing the source to print.

-Robert

[Double Echo]

Re: CLI ( command-line ) PHP displaying source code in-line whenprogram is executed

You're right!!

I'm smacking myself--too much coding!!

I changed it on my other progs but forgot to on this one.

Thanks!


rlee0001 wrote:[color=blue]
> Echo Echo,
>
> At the beginning output $file. For some reason the file is reading
> itself as input. If you are using:
>
> php myprog.php infile.csv
>
> ...on the command line then I think you should be trying to get
> "infile.csv " (or whatever) using $argv[1] not $argv[0]. Otherwise you
> are passing the files own source to itself and the line
>
> print "$data[$c]<br> \n" ;
>
> Is causing the source to print.
>
> -Robert
>[/color]

[Jasen Betts]

Re: CLI ( command-line ) PHP displaying source code in-line when program is executed

On 2006-02-12, Double Echo <doubleecho@you r.com> wrote:[color=blue]
> Hi all,
>
> I'm using PHP 4.4.2, and use PHP on both the command-line and the web.[/color]
[color=blue]
> When I run the following program from the command line it prints out the
> source code as well as the output from the program. I run other PHP programs
> on the command-line and this doesn't happen. It's really peculiar. The
> error output is below.
>
>
> 1 fields in line 1:<br>
><?php<br>
> 1 fields in line 2:<br>
><br>
> 1 fields in line 3:<br>
> if ( $argv[0] )<br>
> 1 fields in line 4:<br>
> {<br>
> 1 fields in line 5:<br>
> $file = $argv[0] ;<br>[/color]

.....



[color=blue]
><?php
>
> if ( $argv[0] )
> {
> $file = $argv[0] ;
> }
> else {
> $file = $_GET['file'] ;
> print "<font face=arial>\n" ;
> }[/color]

argv[0] is the name of the PHP script.

your program is reading itself and printing itself out.
use argv[1];



--

Bye.
Jasen