可以将两个条件进行合并，这样在提交时可以根据不同的情况执行相应的逻辑。下面是合并后的代码示例：
handleFilterSubmit(isSave) {
if (isSave || this.state.activeKey === 6) {
const values = {
...this.props.productData,
status: 'prodPass',
note: this.basicRef.formVales.note,
// seq: '',
fpmsBaseInfo: this.basicIfo,
fpmsControlInfo: this.caculateRef.formVales,
fpmsCostAccountingInfo: this.controllRef.formVales,
fpmsPeriodPublicationInfo: this.periodRef.terms,
fpmsFileInfo: this.fileRef.files,
limitCode: this.limitCodeRef.limitCode
};
const { type } = this.props.params;
console.log(values, type, '903420943295u238y5782352390502359');
if ((type === 'add' || type === 'copy') && isSave) {
this.addProductModel(values);
} else {
const status = type === 'wh' ? null : values.status;
this.updateProduct({ ...values, status });
}
if (this.state.activeKey === 6) {
if (values.fpmsBaseInfo && values.fpmsBaseInfo.payType) {
const payTypeValues = values.fpmsBaseInfo.payType;
if (!Array.isArray(payTypeValues)) {
const payTypeArray = payTypeValues.split(',');
values.fpmsBaseInfo.payType = payTypeArray;
}
}
this.props.productInput(values, () => {
window.location.href = '/productFactory/product/template';
this.props.queryProducts();
});
}
} else {
this.setState(prevState => ({
activeKey: prevState.activeKey + 1
}));
}
}
通过合并条件，如果isSave为真或者activeKey为6时，会执行相应的提交逻辑。
内容由零声教学AI助手提供，问题来源于学员提问
![驱动代码
void kalRxTaskletSchedule(struct GLUE_INFO *pr)
{
static unsigned int num = 0;
tasklet_hi_schedule(&pr->rRxTask[(num++)%NR_CPUS]);
// tasklet_hi_schedule(&pr->rRxTask);
DBGLOG(HAL, ERROR,](https://linuxcpp.0voice.com/zb_users/upload/2023/05/202305162226144313964.png)
