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Bash The shell script function gives "find: missing argument to -exec"

I wrote a function in a Bash script shell to search for a Linux tree for file names matching a pattern containing a regular expression with color highlighting:

function ggrep {
 LS_="ls --color {}|sed s~./~~"
 [ -n "$1" -a "$1" != "*" ] && NAME_="-iname $1" || NAME_=
 [ -n "$2" ] && EXEC_="egrep -q \"$2\" \"{}\" && $LS_ && egrep -n \"$2\" --color=always \"{}\"|sed s~^B~\ B~" || EXEC_=$LS_
 FIND_="find . -type f $NAME_ -exec sh -c \"$EXEC_\" \\;"
 echo -e \\e[7m $FIND_ \\e[0m
 $FIND_
}

eg. ggrep a*lists all files starting with aunder the current directory tree,

and ggrep a* xlists of files starting with aand containingx

When I run it, I get:

find: missing argument to `-exec'

although I get the correct output when I copy and paste the output of a string using "echo" to the terminal. Can someone please tell me what I did wrong?

-, , ggrep * x , x, * , \* '*'. ? !

+4
2
eval $FIND_

.

BASH , , , "eval", , . , .

, , . *, -, . , . , , . , , . ( , .)

+1

find \; \\;.

find . -type f $NAME_ -exec sh -c \"$EXEC_\" \;
+2

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