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 - Remove keys with empty or null values in qs.stringify when constructing query params in URL - Stack Overflow

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    javascript - Remove keys with empty or null values in qs.stringify when constructing query params in URL - Stack Overflow

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    I'm constructing a URL with query string parsing qs.stringify in which I want only non-empty or non-null values to be present in the URL.

    The result I am getting is below:

    localhost:3000/user?name=john&age=

    But, I don't want the age param to be included in the url as the value for age is null.

    Below is an excerpt from my code:

    const url = `/user?${qs.stringify({ name, age })}`;

    Could anyone please help?

    I'm constructing a URL with query string parsing qs.stringify in which I want only non-empty or non-null values to be present in the URL.

    The result I am getting is below:

    localhost:3000/user?name=john&age=

    But, I don't want the age param to be included in the url as the value for age is null.

    Below is an excerpt from my code:

    const url = `/user?${qs.stringify({ name, age })}`;

    Could anyone please help?

    • javascript
    • node.js
    Share Improve this question edited Dec 18, 2020 at 21:36 Aplet123 35.5k1 gold badge39 silver badges64 bronze badges asked Dec 18, 2020 at 21:33 scriobhscriobh 89012 silver badges27 bronze badges
    Add a ment  | 

    7 Answers 7

    Reset to default 5

    You can make a function using Object.fromEntries and Object.entries to filter out non-null and non-empty values:

    function filterNonNull(obj) {
    return Object.fromEntries(Object.entries(obj).filter(([k, v]) => v));
}

const url = `/user?${qs.stringify(filterNonNull({ name, age }))}`;

    qs package added skipNulls option since v5.1.0

    To pletely skip rendering keys with null values, use the skipNulls flag:

    var nullsSkipped = qs.stringify({ a: 'b', c: null}, { skipNulls: true });
assert.equal(nullsSkipped, 'a=b');

    Don't worry, the qs package supports skip undefined values as well.

    console.log(qs.stringify({a: 1, b: undefined}))
// => "a=1"

    runkit

    But qs does NOT support omitting the below zero values: 0, ''

    var qs = require("qs")
console.log(qs.stringify({a:1,b:''})) // "a=1&b="

    You can remove nullish values in filter option of qs.stringify via fallback to undefined for nullish values:

    const values = qs.stringify({a: 1, b: ''}, { filter: (prefix, value) => value || undefined });
assert.equal(values, 'a=1'); // true

    you can try this way

    const url = `/user?${qs.stringify(Object.entries(obj).forEach( item => ((item[1] === null) || (item[1] === undefined ) ? delete obj[item[0]] : item));)}`;

    You also can use reg exp const url = ``/user?${qs.stringify({ name, age })}``.replace(/[^=&]+=(&|$)/g,"").replace(/&$/,"")

    You should allow only those properties which are not null; Easiest way to do that is below

    const name=null,age=23;
const obj = {
name,
age}
let newObj = {
}
for (prop in obj){
if(obj[prop]){
newObj[prop] = obj[prop];
}
}
console.log(newObj);

    const params = {
  a: 1, 
  b: '', 
  c: '', 
  d: 4, 
  e: undefined, 
  f: null
}
queryString.stringify(params, {
  skipEmptyString: true,
  skipNull: true
})

//=> 'a=1&d=4'
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