在另一个脚本中运行bash脚本时，父脚本中定义的所有变量是否都继承给孩子？

Question

For example, say I have script1.sh and I need to call script2.sh. Will script2.sh be able to use any variables defined script1.sh? If yes, if I went a step deeper and called script3.sh within script2.sh, would that have access to script1.sh variables?

If not, what would have to be done to achieve that?

Answer 1

Only the exported variables are accessible in the called script:

1.sh

#! /bin/bash
export exported=1
not_exported=2
2.sh

2.sh

#! /bin/bash
echo $exported
echo $not_exported

Running 1.sh only outputs 1 .

You can export a variable only for a dingle call by assigning to it before the call:

not_exported=3 2.sh

Answer 2

There are three ways a variable may be made available for use by a script launched by another script :

• Marking the variable for export (such as with the export keyword used at the time of declaration or afterwards, or by using the - x option of declare , local or typeset ) ;

• Putting an assignment for the variable as a prefix to the command being executed (eg varname=1 command and args ), separated by spaces only ;

• Calling the child script with source or . , which causes the script to be read and interpreted by the current shell instead of being launched as a separate process, and therefore makes all variables of the current context (including local variables) available to the child script.

Note that marking a variable for export will cause this variable to be copied to the memory space of the child process, and this copy then becomes independent from the variable of the parent shell : modifying it in the child will not change the value in the parent.

Using source or . is the only way you can cause a child script to modify variables in the parent script, as a child process never has access to the memory space of its parent.