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const {run} = useRequest(getName, { manual: true, onError: console.log })
在触发error后,
提示我要添加
throwOnError
,我把useRequest
替换成ahooks
的就不会打印错误了。好像是
umi
的useRequest run
对应的是ahooks useRequest
的runAsync
?有什么方式让它的设置onError之后不设置throwOnError也不会打印错误吗
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