Please help me on..passing a variable from a file using unix to a php file..

[shara]

Hello there,

I have to pass a variable from unix as an input to a php file.
The syntax i used is

var1="SUCCESS"
report_errors.php var1
 

Now report_errors is not being executed and the php start symbols "<?" etc all are being thrown as syntax errors.I dont have any idea why this is happening with php scripts on UNIX platform.Can anybody please help me with this and come up with a solution?

[prn]

I have to admit that I'm not particularly familiar with php, but I have a pretty good guess as to what the problem is.

You moved this from windows, right? Under windows, the filename extension tells the system how to interpret the file contents. Under unix/linux, the filename extension is just to remind you of what's in it. the system pays it no attention.

If you are going to run a script (any language) on a unix/linux system, invoked by just using its name, then the first line of the script needs to tell the OS/shell what interpreter to use, so a php script needs:

#! /usr/bin/php

as its first line.

Alternatively, you could invoke the script with

php report_errors.php

BTW, if your script refers to an environmental variable for its data, you don't need to both set the variable and pass it on the command line. If it refers to "VAR1" then you should be able to omit it from the command line. If it refers to whatever was passed to it, then you can equally well invoke the script with

report_errors.php "SUCCESS"

HTH,
Paul