PHP exec $PATH 变量缺少元素

Question

当我在命令行中回显 $PATH 时，它返回

/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/Applications/MAMP/Library/bin:/usr/local/git/bin:/usr/X11/bin

当我执行这段 php 代码时

exec('echo $PATH; whoami; less /etc/paths; 2>&1')

我明白了

string 'echo $PATH; whoami; less /etc/paths; 2>&1' (length=56)
array
  0 => string '/usr/bin:/bin:/usr/sbin:/sbin' (length=29)
  1 => string 'eric' (length=4)
  2 => string '/usr/bin' (length=8)
  3 => string '/bin' (length=4)
  4 => string '/usr/sbin' (length=9)
  5 => string '/sbin' (length=5)
  6 => string '/usr/local/bin' (length=14)
  7 => string '/Applications/MAMP/Library/bin' (length=30)
  8 => string '/usr/bin:/bin:/usr/sbin:/sbin' (length=29)

这是在 Mac OS X 上。谁能告诉我为什么最后两个路径元素丢失了？

Answer 1

在调用 exec 之前尝试执行此操作:

putenv("PATH=" .$_ENV["PATH"]. ':/usr/local/git/bin:/usr/X11/bin');