Question

这是第一次在Stack Overflow上访问，但肯定在这里找到了很多有用的信息！

当前，我正在尝试根据当前选择来确定如何选择表格中的下一项或上一项。

我当前的表格如下：

maleSkins = { 7,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,32,33,34,35,
        36,37,43,44,45,46,47,48,49,51,52,57,58,59,60,61,66,67,72,73,80,82,83,
        84,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,
        112,113,114,115,116,117,118,120,121,122,123,124,125,126,127,128,132,
        133,134,135,136,137,142,143,144,146,147,153,154,156,159,160,161,162,
        168,170,173,174,175,176,177,179,180,181,182,183,184,185,186,187,188,
        189,200,202,203,204,206,210,212,213,217,220,221,222,223,227,228,229,
        230,234,235,236,239,240,241,242,247,248,249,250,252,254,258,259,260,261,262 }

femaleSkins = { 9,10,11,12,13,31,38,39,40,41,53,54,55,56,69,76,77,88,89,90,91,92,
        93,129,130,131,138,139,140,141,145,148,151,152,157,190,191,192,193,195,
        196,197,198,199,201,207,211,214,215,216,218,219,224,225,226,232,233,237,238,243,244,245,246,251,256,257 }

maleSkins表中的默认选择为“ 7”，当他们为性别选择“ female”时，我将使用femaleSkins表。

当前我的函数调用如下所示

function selSkin(button,state)
if button ~= "left" and state ~= "up" then
    return
end
if source == createChar.maleButt then
    femaleSkin = false
    maleSkin = true
elseif source == createChar.femaleButt then
    maleSkin = false
    femaleSkin = true
end
if source == createChar.nextSkin then
    if maleSkin == true then
        newModel = table.concat(maleSkins,)
    elseif femaleSkin == true then
    end
elseif source == createChar.prevSkin then
    if maleSkin == true then
    elseif femaleSkin == true then
    end
end
end

因此，我试图在“ createChar.nextSkin”和“ createChar.prevSkin”内部基于当前皮肤对表进行排序，但是我不确定如何进行操作。

如果有人能给我做这些的基础，我会很乐意，剩下的我会自己做！

（注意：我将使用预定义变量来实现此目的）

maleSkin = true
femaleSkin = true
curSkin = 7
newModel = nil

Answer 1

首先欢迎堆栈溢出：）

我注意到的一件小事：如果角色是男性或女性，则有两个变量要存储。这允许4种组合，其中2种没有意义。您也可以只使用一个变量，例如<template> <div id="cyto" ref="cyto"></div> </template> <script> import cytoscape from "cytoscape"; export default { name: "HelloWorld", data: function() { return { cy: null }; }, props: { msg: String }, mounted() { let cy = cytoscape({ container: this.$refs.cyto, elements: [ // list of graph elements to start with { // node a data: { id: "a" } }, { // node b data: { id: "b" } }, { // edge ab data: { id: "ab", source: "a", target: "b" } } ], style: [ // the stylesheet for the graph { selector: "node", style: { "background-color": "#666", label: "data(id)" } }, { selector: "edge", style: { width: 3, "line-color": "#ccc", "target-arrow-color": "#ccc", "target-arrow-shape": "triangle" } } ], layout: { name: "grid", rows: 1 } }); //this line below breaks the browser this.cy = cy; } }; </script>  <style scoped> #cyto { height: 100%; display: block; border: 1px solid blue; } </style>，如果该变量为假，则使用女性皮肤。另外，您可以只做maleSkin和skinType = 'male'（Lua实习生字符串，所以这和比较整数一样快）

现在，除非您打算拥有几百万个或更多的皮肤，否则您可以遍历表以查找当前皮肤，然后使用上一个。

skinType = 'female'

现在，要获得下一个皮肤，您可以function skinOffset(skin, skinList, offset) for i,current_skin in ipairs(skinList) do if current_skin == skin then return skinList[i + offset] else end end，对上一个皮肤skinOffset(curSkin, maleSkins, 1)。

Answer 2

将表索引另外保存在全局变量中。有几种方法，例如skinIndex = 1，您的代码可能如下所示：

if source == createChar.nextSkin then
  newModel = maleSkin and maleSkins[skinIndex + 1] or femaleSkins[skinIndex + 1]
elseif source == createChar.prevSkin then
  newModel = maleSkin and maleSkins[skinIndex - 1] or femaleSkins[skinIndex - 1]
end

但是您必须提防skinIndex <1或> #maleSkins / #femaleSkins

通过表前进和后退

2 个答案: