将参数传递到bat文件，然后将其传递到php脚本

Question

I having trouble passing command line parameters, that were passed to a bat file, onto a php script

This is what I have:

@echo off
echo %1%
set foo=%1%
php %~dp0%myscript.php %1% %2% %3%

The first three lines work. The 4th line works if it is just:

php %~dp0%myscript.php

also works if I hard code the parameters:

php %~dp0%myscript.php a b c

but if it is:

php %~dp0%myscript.php %1% %2% %3%

I get the following error:

Could not open input file: c:\dev\123

I have tried all of the following syntax for the parameters:

%1
%1%
"%1"
"%1%"
%*
%*%
"%*"
"%*%"

What am I doing wrong?

Answer 1

Never mind figured it out:

set pth=%~dp0%myscript.php
php %pth% %1 %2 %3

seems I have to do it in two steps - looks as if too much concatenation is confusing it.

Answer 2

There are evil superabundant % percent signs in your line:

php %~dp0%myscript.php %1% %2% %3%
         ^               ^   ^   ^

This line is then parsed as follows:

• php :taken literally;
• %~dp0 :results to full path of your bat or cmd script;
• %myscript.php % : results to empty string;
• 1% %2% %3% : results to 123 as every % % results to empty string and trailing % as well.

As per Command Line arguments (Parameters) , use

php %~dp0myscript.php %1 %2 %3