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基于问题Menu: wrap every X item，如果我不知道菜单元素的确切数量，如何包装每个x元素？
如果我使用the answer of chrisBerlin，则问题在于它仅适用于3个元素。
但是对于每个孩子的动态计数，它会输出一个无效的代码 我怎么解决这个问题？我的代码显示了HMENU的3个级别。我需要第2级的x-wrap。我每行使用7个元素。
在我的页面上，第一个元素有例如5个孩子。所以<ul class="level-2">没有关闭。
tmp.nav_main = COA
tmp.nav_main{
10 = COA
10{
10 = HMENU
10{
wrap = <ul class="level-1">|</ul>
entryLevel = {$t3d_navigation.nav_main_entryLevel}
excludeUidList = {$t3d_navigation.nav_main_excludeUidList}
1 = TMENU
1 {
IProcFunc = user_tmenulinebreak->main
expAll = 1
NO{
wrapItemAndSub = <li class="level-1 norm uid_{field:uid}">|</li>
wrapItemAndSub.insertData = 1
ATagParams = class="level-1 norm"
stdWrap.wrap = <span>|</span>
}
CUR < .NO
CUR.wrapItemAndSub = <li class="level-1 click uid_{field:uid}">|</li>
CUR.wrapItemAndSub.insertData = 1
CUR.ATagParams = class="level-1 click"
CUR = 1
ACT < .CUR
ACT = 1
}
2 < .1
2 {
wrap = <div class="level-2-wrapper-outer"><div class="level-2-wrapper"><div class="level-2-wrapper-inner">|</div></div></div>
expAll = 1
NO{
wrapItemAndSub = |*|<ul class="level-2"><li class="level-2 norm">|</li> || <li class="level-2 norm">|</li> || <li class="level-2 norm">|</li> || <li class="level-2 norm">|</li> || <li class="level-2 norm">|</li> || <li class="level-2 norm">|</li> || <li class="level-2 norm">|</li></ul>|*|
ATagParams = class="level-2 norm"
}
CUR.wrapItemAndSub = |*|<ul class="level-2"><li class="level-2 click">|</li> || <li class="level-2 click">|</li> || <li class="level-2 click">|</li> || <li class="level-2 click">|</li> || <li class="level-2 click">|</li> || <li class="level-2 click">|</li> || <li class="level-2 click">|</li></ul>|*|
CUR.ATagParams = class="level-2 click"
ACT < .CUR
ACT = 1
}
3 < .1
3 {
wrap = <ul class="level-3">|</ul>
expAll = 1
NO{
wrapItemAndSub = <li class="level-3 norm">|</li>
doNotShowLink = 1
after{
cObject = TEXT
cObject{
typolink{
parameter = {field:pid}#{field:uid}
parameter.insertData = 1
ATagParams = class="level-3 norm"
}
data = field:title
}
}
}
CUR{
wrapItemAndSub = <li class="level-3 click">|</li>
doNotShowLink = 1
after{
cObject = TEXT
cObject {
typolink{
parameter = {field:pid}#{field:uid}
parameter.insertData = 1
ATagParams = class="level-3 click"
}
data = field:title
}
}
}
ACT < .CUR
ACT = 1
}
}
}
}
在我的特殊情况下，我需要<ul class="level-2"></ul>中的1-7个项目如果有超过7项，则新行必须以下一个8-14元素开头。如果新行超过14个。
答案 0 :(得分：2)
Read the docs它提供了很多线索，optionSplit的有效语法是：
first-element(s) |*| middle-element(s) |*| last-element(s)
last，first，middle，这意味着如果您有一个项目，则只会使用last段，first, last first, middle(s), last 每个细分可以包含编号的子细分除以||。
您只在中间部分使用带编号的子分段，您需要修复它。
另外一点是，选项Split必须在所有情况下都有效，当只有1,2或3个项目可用时，你使用它的方式会导致有ul未封闭的情况，换句话说就是你在 optionSplit的部分或子部分需要正确的声明...
无论如何，optionSplit是一个不错但技巧娴熟的技术，它是多年前发明的，当没有任何可靠的JS库时，通过某种因素增加了样式操作的可能性，目前我会使用jQuery和/或新的CSS功能来执行此操作，只需生成普通菜单，然后将所需的类添加到每个x元素。即：