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13994853095 avatar 13994853095 commented on June 24, 2024 1

好的明白了,我去尝试一下,谢谢

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hizzgdev avatar hizzgdev commented on June 24, 2024

leading-line-color 设置的是这个node指向其父节点的线条的颜色,仅作用于设置的那个节点(线条)

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13994853095 avatar 13994853095 commented on June 24, 2024

好的谢谢,所以按照这个逻辑,假设我的节点id是1,2,3,4,5,6,其下的子节点是1-1,1-2,2-1,2-1-1这种形式的话,我需要对每一个节点进行处理,给id以1开头的节点线条一个颜色,以2开头的一个颜色,然后以此类推,给这6类及其子节点分别设置6种线条颜色对吗

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hizzgdev avatar hizzgdev commented on June 24, 2024

https://hizzgdev.github.io/jsmind/docs/zh/1.usage.html

在这个文档的1.4 样式里提到了用法,感觉用这个还挺符合你的需求的。你可以在设置节点颜色的同时设置引导线的颜色。

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hizzgdev avatar hizzgdev commented on June 24, 2024

好的谢谢,所以按照这个逻辑,假设我的节点id是1,2,3,4,5,6,其下的子节点是1-1,1-2,2-1,2-1-1这种形式的话,我需要对每一个节点进行处理,给id以1开头的节点线条一个颜色,以2开头的一个颜色,然后以此类推,给这6类及其子节点分别设置6种线条颜色对吗

逻辑上符合你的需求。至于如何标识是六类中的哪一类,你可以从数据上看怎么方便。如果数据在db里的话,不用id来标识也问题不大。还是看你方便。

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13994853095 avatar 13994853095 commented on June 24, 2024

哈喽哈喽,再叨扰一句,如果我的6类分支想实现:前三个默认展开至3级,后三个默认展开至2级,这个只能通过遍历设置每个节点的expanded来实现嘛

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hizzgdev avatar hizzgdev commented on June 24, 2024

应该不需要遍历所有节点,最好是在数据层面把前三个的所有二级节点id找出来并记下来,这样的话就可以用 jm.expand_to_depth(depth) 展开到2级,然后再遍历上面记下来的节点,给它展开就好了。

不过如果你能在 show 之前对脑图数据进行修改的话,其实就会更容易实现。

node 的数据结构大概是这样的:

{
    "id": "xxxx",
    "topic": "topic-xxx",
    "expanded": true
}

你只要把上面的 expanded 改成 false,它就默认会把子节点给收起来了。

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