- approval_control 重写:带 base_id 按主键绝对判定;无 id 才名称100%一致 AND 核心规格(_code_of)一致兜底,杜绝同名不同规误杀 - 借/出库服务命中需审批但未选审批人时,报错列出需审批物料名/规格
89 lines
3.0 KiB
Python
89 lines
3.0 KiB
Python
from app.models.base import MaterialBase
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def _code_of(spec):
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"""取规格的首段编码('S0013/S0013' -> 'S0013';'S0013' -> 'S0013')"""
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s = (spec or '').strip()
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if not s:
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return ''
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return s.split('/')[0].strip().upper()
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def _materials_flagged():
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"""一次性取出所有“启用且需审批”的物料(量小,逐项判定前先全量拉一次)"""
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rows = MaterialBase.query.filter(
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MaterialBase.is_enabled == True,
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MaterialBase.is_approval_required == True
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).all()
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return rows
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def _is_int(v):
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try:
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int(v)
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return True
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except (TypeError, ValueError):
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return False
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def resolve_approval_control(items):
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"""
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判断一批出库/借库申请明细是否需要审批。
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判定优先级:
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1) 方案1(最高):明细带有效 base_id → 直接按物料主键查 is_approval_required,
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且只要有有效 ID,无论命中与否都不再做任何字符串兜底(杜绝同名/同规格误杀)。
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2) 方案2(无 ID 兜底):名称必须 100% 精确一致 AND 核心规格一致(斜杠前段,
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或前端未传规格视为防呆拦截),两条件同时成立才判定命中。
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"""
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flagged = []
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seen = set()
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def _record(m_obj, req_name):
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if m_obj.id in seen:
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return
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seen.add(m_obj.id)
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flagged.append({
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'name': m_obj.name or req_name,
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'spec_model': m_obj.spec_model or '',
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'base_id': m_obj.id,
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})
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# 1. 查出所有“启用且需审批”的危险物料库
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flagged_materials = _materials_flagged()
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for item in items or []:
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name = str(item.get('name') or '').strip()
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spec = str(item.get('spec_model') or '').strip()
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base_id = item.get('base_id')
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# --- 方案 1:最高优先级,依据 base_id 绝对判定 ---
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if base_id is not None:
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bid = int(base_id) if _is_int(base_id) else 0
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if bid > 0:
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m = MaterialBase.query.get(bid)
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if m and m.is_enabled and m.is_approval_required:
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_record(m, name)
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# ★ 关键:只要有有效 ID,无论是否需审批,绝不允许往下走字符串兜底
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continue
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# --- 方案 2:无 ID 时的降级兜底 (名称精确相等 AND 核心规格相等) ---
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if not name:
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continue
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# 提取当前申请行的核心规格
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req_core_spec = _code_of(spec)
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for fm in flagged_materials:
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# 条件 A:名称必须 100% 绝对一致
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if fm.name != name:
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continue
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# 条件 B:核心规格一致 (或前端未传规格视为防呆拦截)
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fm_core_spec = _code_of(fm.spec_model)
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if not req_core_spec or req_core_spec == fm_core_spec:
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_record(fm, name)
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break # 命中即判定本申请行需审批,跳出内层循环
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return bool(flagged), flagged
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