Question

我想将<代码>(“USERID=XYZ”)“USERPWD=123”改为。审判



(apply # concatenate  string  ("USERID=XYZ" "USERPWD=123"))


编号


但我不知道如何插入和安放; 以下职能行之有效,但似乎很复杂。

(defun join (list &optional (delim "&"))
    (with-output-to-string (s)
        (when list
            (format s "~A" (first list))
            (dolist (element (rest list))
               (format s "~A~A" delim element)))))

Answer 1

http://www.lispwork.com/documentation/HyperSpec/Body/22_cg.htm“rel=“noreferer”>FORMAT。

。 denote iteration, ~A denotes aesthetic publication, and ~^ (aka Tilde Circumflex in the docs) denotesprint the , only when which follows.

* (format nil "~{~A~^, ~}"  ( 1 2 3 4 ))

"1, 2, 3, 4"
*

Answer 2

这一解决办法使我们得以使用<条码>FORMAT,以形成一种扼杀,并具有一个可变的限定。目的是不为这项职能的每一呼吁安排一个新的<>格式>。一位良好的共同文献汇编员还可能想汇编一个固定的format string,当“format string ”在运行时间建造时被打败。见宏观formatter 。

(defun %d (stream &rest args)
  "internal function, writing the dynamic value of the variable
DELIM to the output STREAM. To be called from inside JOIN."
  (declare (ignore args)
           (special delim))
  (princ delim stream))

(defun join (list delim)
  "creates a string, with the elements of list printed and each
element separated by DELIM"
  (declare (special delim))
  (format nil "~{~a~^~/%d/~:*~}" list))

解释:

"~{      iteration start
 ~a      print element
 ~^      exit iteration if no more elements
 ~/%d/   call function %d with one element
 ~:*     move one element backwards
 ~}"     end of iteration command

<>代码>%d只是内部功能,不应在join外打电话。作为标识,它有<条码>%。

rel=“nofollow noreferer”> :>code>foo,从format string 。

各种变量<代码>delim被宣布为特别的,因此可对被转让为%d功能的划界人具有价值。由于我们可以将<条码>%d功能从FORMAT改为限定词,我们需要从其他地方——这里从<条码>中引入的有活力的具有约束力的动态中获取。

职能<代码>%d的唯一目的是写一个限定词——它忽视了<代码>format通过的论点——它只使用<代码>的<的论点。

Answer 3

Melpa 主办一揽子项目。 (“长途失修的操纵图书馆”)提供许多简便的公用事业,包括一个护卫团:

(require  s)                                                                                                                                                                                          
(s-join ", " (list "a" "b" "c"))                                                                                                                                                                      
"a, b, c"

对于其价值,在立方略为细微,它从<条码>fns.c,即<条码>,即<条码>。

(mapconcat  identity (list "a" "b" "c") ", ")                                                                                                                                                         
"a, b, c"

Answer 4

当事人很晚,但reduce 罚款:

(reduce (lambda (acc x)
          (if (zerop (length acc))
              x
              (concatenate  string acc "&" x)))
        (list "name=slappy" "friends=none" "eats=dogpoo")
        :initial-value "")

Answer 5

假设一个插图和单一特性限制清单,以下人员应高效地开展工作,经常在短名单上援引:

(defun join (list &optional (delimiter #&))
  (with-output-to-string (stream)
    (join-to-stream stream list delimiter)))

(defun join-to-stream (stream list &optional (delimiter #&))
  (destructuring-bind (&optional first &rest rest) list
    (when first
      (write-string first stream)
      (when rest
        (write-char delimiter stream)
        (join-to-stream stream rest delimiter)))))

Answer 6

图书馆:

(ql:quickload "str")
(str:join "&"  ("USERID=XYZ" "USERPWD=123"))

It uses format like explained in the other answers:

(defun join (separator strings)
" "
(let ((separator (replace-all "~" "~~" separator)))
  (format nil
        (concatenate  string "~{~a~^" separator "~}")
        strings)))

(作者们,这样简单一些。)

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