Batch file error handle

Question

I have batch file that execute my php script, but I dont know how to handle the empty argument, can you help me?

this is my batch script

echo off
set phppath=%1
set resultfolder=%2
set jsonfile=%3
shift
shift
shift

set PHP_HOME=%phppath%
set PATH=%PATH%;%PHP_HOME%
echo Set PHP_HOME in "%phppath%"
echo Locate result in folder "%resultfolder%"
echo JSON file is "%jsonfile%"


pause;
php -f cxense_json_to_xml.php %resultfolder% %jsonfile%

[read me](http://stackoverflow.com/questions/4953170/how-to-check-command-line-parameter-in-bat-file) — Shawn Mehan, Nov 02 '15 at 00:39
@ShawnMehan - wrong shell environment, that won't help at all. — SomethingDark, Nov 02 '15 at 00:41
Which argument could be empty and what exactly do you want to do if one of them is missing? Is there a kind of default value or do you want to output an error message? — MichaelS, Nov 02 '15 at 08:05

somebadhat · Answer 1 · 2019-04-03T00:35:30.463

@echo off
set phppath=%1
set resultfolder=%2
set jsonfile=%3
set PHP_HOME=%phppath%
set PATH=%PATH%;%PHP_HOME%
echo Set PHP_HOME in "%phppath%"
echo Locate result in folder "%resultfolder%"
echo JSON file is "%jsonfile%"
php -f cxense_json_to_xml.php %resultfolder% %jsonfile%
Pause

And the batch would have to be started like this:

Batch.bat "path to PHP" "path to result folder" "name of jsonfile"

there is no semicolon after pause.

Batch file error handle

1 Answers1