Bash變量在複合評估中不起作用

Question

預先感謝您對這個問題的幫助，這個問題看起來很簡單，但我一直無法解決。下面的代碼是的情況自明：

#INPUTS 
    PID=8255 
    echo PID is $PID 

    #WORKING REFERENCE (THIS WORSKS!) 
    GPID=$(ps fxao pid,ppid,pgid,comm |awk '$1 == "8255"'| awk '{print $3}') 
    echo GPID is $GPID 

    #TAKE 1 (DOES NOT WORK) 
    GPID1=$(ps fxao pid,ppid,pgid,comm |awk '$1 == "$PID"'| awk '{print $3}') 
    echo GPID1 is $GPID1 

    #TAKE 2 (DOES NOT WORK) 
    GPID2=$(ps fxao pid,ppid,pgid,comm |awk '$1 == "$($PID)"'| awk '{print $3}') 
    echo GPID2 is $GPID2 

執行的輸出：

PID is 8255 
GPID is 8251 
GPID1 is 
GPID2 is 

Answer 1

你必須給它每文檔頁面的值PARAM：http://www.gnu.org/software/gawk/manual/gawk.html#Using-Shell-Variables

PID=8255 
echo PID is $PID 
GPID1=$(ps fxao pid,ppid,pgid,comm | awk -v d1="$PID" '$1 == d1'| awk '{print $3}') 
echo GPID1 is $GPID1