我正在尝试使用compose:
通过单个函数创建2个面板make-panel: func [sentchar][
probe compose/deep [
text "N1:"
(to-set-word rejoin["fld1" sentchar ":"]) field ; TO BE NAMED fld1A and fld1B for 2 panels
text "N2: "
(to-set-word rejoin["fld1" sentchar ":"]) field ; TO BE NAMED fld2A and fld2B for 2 panels
text "Product: "
(to-set-word rejoin ["txt_out" sentchar ":"]) text ; TO BE NAMED txt_outA and txt_outB for 2 panels
button "Get product" [
x: to-path to-word (rejoin ["face/parent/pane/fld1" sentchar "/text"])
y: to-path to-word (rejoin ["face/parent/pane/fld2" sentchar "/text"])
(to-set-path (to-path rejoin ["face/parent/pane/txt_out" sentchar "text"] ))
form multiply get x get y ] ] ]
view compose [
(make-panel "A") return
(make-panel "B") return ]
然而,即使我尝试过不同的组合,我也会遇到关于to-word和to-path的错误。问题在哪里?
答案 0 :(得分:5)
您的错误是尝试使用" /"创建一个单词。字符。
>> to-word "foo/bar"
*** Syntax Error: invalid character in: "foo/bar"
*** Where: to
*** Stack: to-word
我的第二个倾向是你不应该使用字符串来组成值引用 - 如果没有别的东西你会失去约束力。可以尝试以下方法:
to path! compose [face parent pane (to word! rejoin ["fld2" sentchar]) text]
我的第一个倾向是你过度复杂,但这超出了你的问题的范围。
我将尝试解决此代码中的一些其他问题:
关于make-panel
的说明 - 由于您没有制作panel
,这只是一个误称,只是将一些元素规格组合在一起。出于本答案的目的,我将使用名称make-row
。此外,我永远不会喜欢fld1
或tout
这个名字(这是一个真实的词!)但会坚持不懈。
正如我上面提到的,你总是更好地从单词和字符串开始,如在Rebol / Red中,单词在评估期间获取上下文 - 从字符串加载的单词不会。例如:
make object! [
foo: "bar"
probe get first [foo] ; "bar"
probe get first load "[foo]" ; error
]
在您创建三个新单词时,让我们明确地这样做:
make-row: function [row-id [string!]][
fld1: to word! rejoin ["fld1-" row-id]
fld2: to word! rejoin ["fld2-" row-id]
tout: to word! rejoin ["tout-" row-id] ; note that 'tout is an English word
...
]
从这里开始,我们可以在规范中开始构建独特的参考资料。
make-row: func [row-id [string!] /local fld1 fld2 tout][
fld1: to word! rejoin ["fld1-" row-id]
fld2: to word! rejoin ["fld2-" row-id]
tout: to word! rejoin ["tout-" row-id]
compose/deep [
text "N1:"
(to set-word! fld1) field
text "N2:"
(to set-word! fld2) field
text "Product:"
(to set-word! tout) text
button "Get Product" [
...
]
return
]
]
现在我们通过此按钮操作进入粘性区域:
x: to-path to-word (rejoin ["face/parent/pane/fld1" sentchar "/text"])
y: to-path to-word (rejoin ["face/parent/pane/fld2" sentchar "/text"])
(to-set-path (to-path rejoin ["face/parent/pane/tout" sentchar "text"] ))
form multiply get x get y ] ] ]
我认为可以用伪代码表达你想要做的事情:
Product = text of product of N1 for this row * N2 for this row
此处代码中的主要错误是您将邻近引用与命名引用混合在一起。如果您检查face/parent/pane
,那里没有fld1*
,fld2*
或tout*
引用,它只是一个面对象块。在您努力创造独特名称的同时,让我们暂时继续这样做。请记住,我们仍处于compose/deep
操作的深处:
x: get in (fld1) 'data
y: get in (fld2) 'data
set in (tout) 'text form x * y
我们现在更加简洁,一切都应该有效(请注意'data
为您提供'text
的加载值。
我关注的是,我们有很多新的词汇,我们需要x
和y
。因此,让我们回到接近的想法。
当您查看撰写的View规范时:
view probe compose [
(make-row "A")
(make-row "B")
]
您会看到您的主视图面将包含很多孩子。要查找您要点击的按钮附近的面孔,我们首先需要找到面部内的按钮。让我们这样做:
button "Get Product" [
this: find face/parent/pane face
]
由于与该按钮相关联的前六个面孔,让我们转到这一组的开头:
button "Get Product" [
this: skip find face/parent/pane face -6
]
现在我们可以根据接近程度进行计算:
button "Get Product" [
here: find face/parent/pane face
here/6/text: form here/2/data * here/4/data
]
轰!我们的产品只有一个here
而不是rows-count * 3 + x + y
。真棒!
由于我们没有生成任何其他字词,我们甚至不需要一个函数来生成我们的行,归结为以下几点:
row: [
text "N1:" field
text "N2: " field
text "Product: " text 100
button "Get product" [
; go back six faces from current face
here: skip find face/parent/pane face -6
here/6/text: form here/2/data * here/4/data
]
return
]
view compose [
(row)
(row)
]
由于您似乎有复杂的需求并且无法始终枚举您需要的字段,因此您可以使用extra
字段将字段组合在一起。我们可以使用块来包含row-id
和field-id
:
make-row: func [row-id][
compose/deep [
text "N1:" field extra [(row-id) "N1"]
text "N2: " field extra [(row-id) "N2"]
text "Product: " text 100 extra [(row-id) "Output"]
button "Get product" extra (row-id) [
...
]
return
]
]
view compose [
(make-row "A")
(make-row "B")
]
在按钮操作中,我们可以收集与该行相关联的所有面:
faces: make map! collect [
foreach kid face/parent/pane [
if all [
block? kid/extra
face/extra = kid/extra/1
][
keep kid/extra/2
keep kid
]
]
]
这给你一张漂亮的地图!与所有相关的面和一个简单的计算:
faces/("Output")/text: form faces/("N1")/data * faces/("N2")/data
如果您只是将其用于产品,那么您甚至不需要收集:
product: 0
foreach kid face/parent/pane [
if all [
block? kid/extra
face/extra = kid/extra/1
][
product: product + kid/value
]
]
答案 1 :(得分:1)
真正的挑战
make-panel: func [sentchar][
compose/deep [
text "N1:" (to-set-word rejoin['fld1 sentchar ]) field ; TO BE NAMED fld1A and fld1B for 2 panels
text "N2:" (to-set-word rejoin['fld2 sentchar ]) field ; TO BE NAMED fld2A and fld2B for 2 panels
text "Product: " (to-set-word rejoin ['tout sentchar]) text ; TO BE NAMED toutA and toutB for 2 panels
button "Get product" [
x: ( to-path reduce [ to-word rejoin ["fld1" sentchar] 'text ])
y: (to-path reduce [ to-word rejoin ["fld2" sentchar] 'text ])
(to-set-path reduce [to-word rejoin ["tout" sentchar] 'text]) form multiply load x load y
]
]
]
view v: compose [
(make-panel "A") return
(make-panel "B") return
]
当然,您不需要中间词 x 和 y 。但是你可以自己做。