From: Stephen G. B. <sg_...@mc...> - 2009-02-26 16:13:16
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Hi All, I am trying to display the file name and line number that is currently being executed in a bash file. Test_Bash.sh contains --------------------------------------------------------------------------------------------------------- #! /bin/sh # src_build() { echo "Entering src_build" echo "BASH = " $BASH echo "LINENO = " $LINENO echo "COMP_LINE = " $COMP_LINE echo "BASH_SOURCE = " $BASH_SOURCE echo "BASH_COMMAND = " $BASH_COMMAND echo ${BASH_SOURCE[$i]} echo '${BASH_LINENO[$i]} = ' ${BASH_LINENO[$i]} echo "$i = " $i pwd ls } fail() { echo echo "=============" echo "fail " echo "Build failed." echo "=============" exit 1 } succeed() { echo echo "================" echo "Build succeeded." echo "================" } if src_build; then succeed else echo "src_build fail" fail fi ------------------------------------------------------------------------------------------------------------------ Part of the msys output is GNUCash@ELSHADAI ~ $ Test_Bash.sh Entering src_build BASH = /usr/bin/sh LINENO = 3 COMP_LINE = BASH_SOURCE = BASH_COMMAND = ${BASH_LINENO[$i]} = = /home/GNUCash Packaging Test_Bash.sh src ================ Build succeeded. ================ GNUCash@ELSHADAI ~ How do I display the line number and filename of the currently executing bash file? Thanks in advance for your help Yours Sincerely Stephen Grant Brown |