skill：solidworks-assembly 实测笔记与脚本、easyeda-sch-geometry 补充
Co-Authored-By: Claude Opus 5.5 <[email protected]>
这个提交包含在：
zhou yang
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# -*- coding: utf-8 -*-
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u"""决定性判据：底板上的 PCIE 插槽 3D 模型，是没建腔的实心块，还是真连接器？
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结论口径：
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实心块 → 实体体积/包围盒 ≈ 1，面数个位~十几 ⇒ 3D **量不出**插入深度 d，
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也判不了"舌头是不是比腔长"
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真连接器 → 体积比明显 <1，面数几百 ⇒ 干涉结论可信，可以继续追
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为什么独立成一个脚本：它**不需要装配体、不需要刀片、更不碰 IInterferenceDetectionMgr**。
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重度用 IDM 之后 SolidWorks 会丢掉运行对象表注册（NOTES §22/§24），
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所以把这个判据做成"一个进程内从头到尾跑完"的形式，不依赖既有连接。
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"""
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import io
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import os
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import re
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import subprocess
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import sys
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import time
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sys.stdout = io.TextIOWrapper(sys.stdout.buffer, encoding='utf-8')
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import pythoncom
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import win32com.client as wc
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BASE_STEP = r'D:\DCDC\新建文件夹\3D_BASE-PCB_2026-09-12.step'
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def sw_pids():
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out = subprocess.run(['tasklist', '/FI', 'IMAGENAME eq SLDWORKS.EXE', '/FO', 'CSV', '/NH'],
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capture_output=True, text=True, encoding='gbk', errors='ignore').stdout
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return sorted(re.findall(r'"SLDWORKS\.exe","(\d+)"', out, re.I))
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def rot_count():
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try:
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rot = pythoncom.GetRunningObjectTable()
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return sum(1 for _ in rot.EnumRunning())
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except Exception as e:
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return u'枚举失败 %s' % str(e)[:40]
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print(u'=== 诊断 ===')
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print(u' SLDWORKS 进程: %s' % sw_pids())
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print(u' 运行对象表条数: %s' % rot_count())
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app = None
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try:
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app = wc.GetActiveObject('SldWorks.Application')
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print(u' ✅ 现有实例可连，不重启')
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except Exception as e:
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print(u' ✗ 连不上 (%s)' % str(e)[:50])
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if app is None:
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if sw_pids():
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print(u'\n=== 结束现有实例（已失联，救不回来）===')
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subprocess.run(['taskkill', '/F', '/IM', 'SLDWORKS.exe'],
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capture_output=True, text=True)
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time.sleep(4)
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print(u' 剩余: %s' % sw_pids())
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print(u'\n=== 启动新实例 ===')
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for k in range(20):
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try:
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app = wc.Dispatch('SldWorks.Application')
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app.Visible = True
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_ = app.RevisionNumber
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break
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except Exception as e:
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if k == 19:
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raise
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print(u' 重试 %d/20 (%s)' % (k + 1, str(e)[:50]))
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time.sleep(5)
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print(u' Revision %s 文档数 %s' % (app.RevisionNumber, app.GetDocumentCount))
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if app.GetDocumentCount == 0:
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print(u'\n=== 导入底板 STEP（≈50 s）===')
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t0 = time.time()
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data = app.GetImportFileData(BASE_STEP)
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err = wc.VARIANT(pythoncom.VT_BYREF | pythoncom.VT_I4, 0)
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app.LoadFile4(BASE_STEP, 'r', data, err)
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print(u' err=%s 用时 %.1fs 文档数 %s' % (err.value, time.time() - t0, app.GetDocumentCount))
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# ── 直接在打开的文档里按标题找插槽**零件**，不经装配树 ──────────────────
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print(u'\n=== 找插槽零件 ===')
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cand = []
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for d in app.GetDocuments:
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try:
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t, ty = d.GetTitle, d.GetType
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except Exception:
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continue
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if ty == 1 and 'PCIE' in t.upper() and '64P11L' in t.upper():
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cand.append(d)
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print(u' 命中 %d 个零件文档' % len(cand))
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if not cand:
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print(u' ⛔ 没找到；打开的零件标题样本：')
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for d in list(app.GetDocuments)[:12]:
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try:
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if d.GetType == 1:
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print(u' %s' % d.GetTitle[:80])
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except Exception:
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pass
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sys.exit(1)
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part = cand[0]
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print(u' 用: %s' % part.GetTitle)
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bb = [v * 1000 for v in part.GetPartBox(True)]
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bbv = (bb[3] - bb[0]) * (bb[4] - bb[1]) * (bb[5] - bb[2])
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print(u' 包围盒 %.2f x %.2f x %.2f = %.1f mm^3'
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% (bb[3] - bb[0], bb[4] - bb[1], bb[5] - bb[2], bbv))
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bodies = None
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for args in ((0, True), (-1, True), (0, False)):
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try:
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bodies = part.GetBodies2(*args)
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if bodies:
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print(u' GetBodies2%s → %d 个实体' % (str(args), len(bodies)))
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break
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except Exception as e:
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print(u' GetBodies2%s 失败 %s' % (str(args), str(e)[:60]))
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if not bodies:
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print(u' ⛔ 取不到实体'); sys.exit(1)
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tot, faces = 0.0, 0
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for i, bd in enumerate(bodies):
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o = bd if hasattr(bd, 'GetMassProperties') else wc.Dispatch(bd)
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try:
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tot += o.GetMassProperties(1.0)[3] * 1e9
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except Exception as e:
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print(u' 实体%d 体积失败 %s' % (i, str(e)[:60]))
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try:
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nf = o.GetFaceCount
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faces += nf
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if i < 8:
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print(u' 实体%-2d 面数 %s' % (i, nf))
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except Exception:
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pass
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r = tot / bbv if bbv else 0
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print(u'\n 实体体积 %.1f mm^3 / 包围盒 %.1f = **%.3f**' % (tot, bbv, r))
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print(u' 总面数 **%d**%s' % (faces, u' ⚠ 0 是假值（实体不可能 0 面），忽略它' if faces == 0 else u''))
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print()
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# ⛔⛔ 教训：原来这里写的是 `if r > 0.80 or faces < 30`，
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# 结果 `GetFaceCount` 返回假值 0 → `faces < 30` 成立 → 一票推翻了
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# 体积比 0.492 给出的相反证据，打印出错误的"实心块"结论。
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# ⇒ 不要把强弱不同的信号用 `or` 并列；面数只在**读到非 0** 时才参与判断。
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faces_ok = faces > 0
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if r > 0.80 or (faces_ok and faces < 30):
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print(u' ⇒ **实心块，没有槽腔**。')
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print(u' 3D 量不出插入深度 d；"舌头 36.0 比腔长"也**无法由 3D 判定** ——')
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print(u' 刀片停在塑料体顶面纯粹是因为模型没开口。只能靠卡尺实测。')
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else:
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print(u' ⇒ **模型建了腔**，几何冲突是真的，干涉结论可信，可以继续追。')
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在新工单中引用
屏蔽一个用户