pretty new to php and i think i might be missing some fundamental limitation of isset. i have two php scripts below that are executed by crond, one using --host X and one that does not. and below that are three different attempts at generating a command line that will be executed. the first example using strlen works as expected. please see bottom of thread for the command line that is generated for each php script respectively. when i came across isset it seemed to make more sense to use that instead, especially the third example. HOWEVER, both isset examples fail when $server is the value for -host. as you might guess -host is $host, --server is $server and --path is $path. is this an issue of having $server twice on the same line when using isset? i've been tinkering with this off and on for a few days and i'm stumped. i need not be spoon fed but some plain speak on what i might be misunderstanding would be great, thanks.
How do you set $host? Do you assume $host is automatically set by appending --host=$host to your php-program-call? If that is so: That's not happening - you have to parse command line args. Or did you just leave the command line parsing code out? If so, please show the code, how you parse the command line args.
First of all. There are functions and libs for everything. Command-Line-Option Parsing may be done with getopt: Look here: PHP: getopt - Manual
(But you may ignore that, if your code works).
The following does not work:
Code:
<?php
#!/usr/bin/php -q
If you decide to use #!, it has to be at the very beginning of the Script-File. Only then the kernel will read the following binary as interpreter for the actual file.
Further without testing the code myself, I suggest you test, if your Arg-parsing produces correct output. I suspect it may not fully work as expected and the values are not what you assume them to be. I suggest you insert some debug statements before+after Arg parsing like:
A logging function would help you to debug your code independently of the running environment(webserver or command line).
Code:
#!/usr/bin/php
<?PHP
# global log-file handle
$log = NULL;
function mylog($msg=NULL) {
$logfile="/tmp/myapp.log"; # There may be better locations than /tmp for the logfile
global $log;
if(!$log) {
# only open logfile once per script run. After the first
# mylog() invocation the open filehandle will be used.
# Also be lazy and let it automatically be closed at
# end of programm by php itself
$log = fopen($logfile,"a");
if(! $log) die("Cannot open Logfile: " . error_get_last());
}
if($msg) {
fwrite($log,date("Y/m/d h:i:s",mktime())." : ".getmypid()." : ".(string)$msg."\n");
}
}
The above mentioned debug statements are easily usable with such a log function(print_r needs to be told, that it returns the formated dump data and so it does not print it to stdout itself).
PHP.net is a great and fabulous source of documentation for PHP. Nearly all functions documentation pages have directly attached real-life coding examples down at the bottom of each page.
before i posted my query i commented all the parsing, processing and db statements and simply put echoes at the bottom so it would print the ssh command and all of the variables i expected to get set. the variables were all set as expected and so was the ssh command when i use strlen to determine if the $host variable was set. however, when i use either of the isset chunks the $server after the -host argument is null but the $server after the ConnectTimeout=60 is okay. i just don't get why my isset chunks won't put $server in both places i expect it to be when $host is not set. so let me ask this, do the two isset pieces look okay? thanks.
finally figured this one out. if i delete or comment the line in the php script:
Code:
$host = "";
the code works as expected now. it finally sunk in after reading the isset documentation untold times that a NULL variable is still a set variable. so although i had not supplied the:
Code:
--host=denasv004
as a command line arg to the script it would not put the $server value there because i had set $host as NULL with the:
Hello,
This function was copied into my code, which was compiled without error/warning, but when executed there is always Segmentation fault at the end after the output (which seems correct!):
void get_hashes(unsigned int hash, unsigned char *in)
{
unsigned char *str = in;
int pos =... (7 Replies)
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The line numbers and the statements of the program are separated by a space..
1 #include<stdio.h>
2 char a,b,c;
3 float x,y,z;
4 int main()
5 {
6 int d,e,f;
7 // further declarations
8 // further declarations
9 /* body*/
10 }
11 void fun1()
12... (1 Reply)
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hello,
i have a problem with strlen. I have written this:
for(y=13,z=0; cInBuf!=' ';y++)
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cBuf=cInBuf;
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}
len = strlen(cBuf);
out=len/2;
fprintf(outfile,"F%i",out);
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