te')); return $arr; } /* 遍历用户所有主题 * @param $uid 用户ID * @param int $page 页数 * @param int $pagesize 每页记录条数 * @param bool $desc 排序方式 TRUE降序 FALSE升序 * @param string $key 返回的数组用那一列的值作为 key * @param array $col 查询哪些列 */ function thread_tid_find_by_uid($uid, $page = 1, $pagesize = 1000, $desc = TRUE, $key = 'tid', $col = array()) { if (empty($uid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('uid' => $uid), array('tid' => $orderby), $page, $pagesize, $key, $col); return $arr; } // 遍历栏目下tid 支持数组 $fid = array(1,2,3) function thread_tid_find_by_fid($fid, $page = 1, $pagesize = 1000, $desc = TRUE) { if (empty($fid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('fid' => $fid), array('tid' => $orderby), $page, $pagesize, 'tid', array('tid', 'verify_date')); return $arr; } function thread_tid_delete($tid) { if (empty($tid)) return FALSE; $r = thread_tid__delete(array('tid' => $tid)); return $r; } function thread_tid_count() { $n = thread_tid__count(); return $n; } // 统计用户主题数 大数量下严谨使用非主键统计 function thread_uid_count($uid) { $n = thread_tid__count(array('uid' => $uid)); return $n; } // 统计栏目主题数 大数量下严谨使用非主键统计 function thread_fid_count($fid) { $n = thread_tid__count(array('fid' => $fid)); return $n; } ?>javascript - vuex subscribe to individual mutation - Stack Overflow
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javascript - vuex subscribe to individual mutation - Stack Overflow

programmeradmin3浏览0评论

Is it possible to subscribe to an individual mutation?

Instead of:

this.$store.subscribe((mutation, state) => {
   if(mutation === 'someMutation'){
       doSomething()
   }
})

I would like something like this:

this.$store.subscribe('someMutation', state => {
    doSomething()
})

Is it possible to subscribe to an individual mutation?

Instead of:

this.$store.subscribe((mutation, state) => {
   if(mutation === 'someMutation'){
       doSomething()
   }
})

I would like something like this:

this.$store.subscribe('someMutation', state => {
    doSomething()
})
Share Improve this question asked Aug 23, 2017 at 17:16 ChrisChris 14.2k23 gold badges93 silver badges182 bronze badges 4
  • u might need to check vuex.vuejs/api/#watch – ctf0 Commented Nov 5, 2018 at 12:15
  • 2 No it will run for all mutations unfortunately: docs – Mike Harrison Commented Jan 8, 2019 at 13:10
  • How about using mapState in the ponent for just the particular data in your store state and then using a watch on that? – Ross Coundon Commented Mar 24, 2019 at 17:06
  • as @MikeHarrison mention in the docs and also as you can see in the vuex code github./vuejs/vuex/blob/dev/src/store.js (line 103). vuex just call any function in sub arrays without checking for mutation. so vuex actually not support it. of course you can go behind the secenes and build your own module/function which will handle it. and use it. – eli chen Commented Jun 21, 2019 at 5:36
Add a ment  | 

3 Answers 3

Reset to default 4

From vuex docs:

The handler is called after every mutation and receives the mutation descriptor and post-mutation state as arguments.

It will react to all mutations, So you only can implement with your own conditional.

How about wrapping the method you have somewhere in Vue prototype?

So instead of having:

this.$store.subscribe((mutation, state) => {
   if(mutation === 'someMutation'){
       doSomething()
   }
})

You would have something like:

Vue.prototype.subscribeMutation = function(someMutation, someFunction) {
       this.$store.subscribe((mutation, state) => {
       if(mutation === someMutation){
           someFunction(state)
       }
    })
}

I haven't tested the code, but you should be able to get the working result easily.

Try something like this:

this.$store.subscribe(mutation => {
                if (mutation.type === 'setData') {
//  do something here
                }
            });
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