我想使用shell脚本将所有出现的$[a-z|A-Z|0-9]$
替换为&[a-z|A-Z|0-9]
。例如,$HELLO$
应该变为&HELLO
。
I/P - $HELLO$
o/P - &HELLO
我尝试使用sed替换字符串。
sed -i -e 's/$*$/&/g filename
I/P - $HELLO$
O/P - &HELLO&
但是我无法确保最后一个$被替换为空格而不是&
答案 0 :(得分:1)
使用分组和反向引用::
import flash.events.Event;
import flash.geom.Point;
var i: Number;
var size: int = 80;
var down: Boolean = false;
var up: Boolean = true;
var inside: Boolean = true;
var outside: Boolean = true;
var circle: Shape = new Shape();
stage.addEventListener(Event.ENTER_FRAME, loop);
stage.addEventListener(MouseEvent.MOUSE_UP, mouseup);
char.addEventListener(MouseEvent.MOUSE_DOWN, mousedown);
function loop(e: Event) {
if (down == true) {
// Checks if mouse pointer is on path i.e 'S' alphabet
if (s.hitTestPoint(stage.mouseX, stage.mouseY, true)) {
inside = true;
outside = true;
var point: Point = maskobj.globalToLocal(new Point(stage.mouseX, stage.mouseY));
var point2: Point = new Point();
//Checks if mouse pointer is completely outside of drawn area
for (i = 0; i < 2 * Math.PI; i += (2 * Math.PI) / 10) {
point2.x = stage.mouseX + (size / 3) * Math.cos(i);
point2.y = stage.mouseY + (size / 3) * Math.sin(i);
if ((maskobj.hitTestPoint(point2.x, point2.y, true))) {
outside = false;
break;
}
}
//Checks if mouse pointer is completely inside drawn area
for (i = 0; i < 2 * Math.PI; i += (2 * Math.PI) / 10) {
point2.x = stage.mouseX + (size / 3) * Math.cos(i);
point2.y = stage.mouseY + (size / 3) * Math.sin(i);
if (!(maskobj.hitTestPoint(point2.x, point2.y, true))) {
inside = false;
break;
}
}
//Character will be moved only if mouse position not to far from current position
if (outside == false) {
if (inside == false) {
//Increases drawn area by drawing a circle shape in 'maskobj' MovieClip
circle.graphics.beginFill(0x0000ff);
circle.graphics.drawCircle(point.x, point.y, size);
circle.graphics.endFill();
maskobj.addChild(circle);
//Moves character to new position
char.x = stage.mouseX;
char.y = stage.mouseY;
}
}
}
}
}
function mouseup(e: MouseEvent): void {
up = true;
down = false;
}
function mousedown(e: MouseEvent): void {
down = true;
up = false;
}
答案 1 :(得分:1)
使用awk
的解决方案:这使用awk的内置函数sub
来搜索和替换$
符号,然后sub
再次用于替换{{1}到$
。
&