Question

如何指定换行符分隔的字符串，例如三行到三个变量？

class MyBasePopUp: UIViewController, UIGestureRecognizerDelegate {

var tap: UITapGestureRecognizer!
override func viewDidAppear(_ animated: Bool) {

    tap = UITapGestureRecognizer(target: self, action: #selector(onTap(sender:)))
    tap.numberOfTapsRequired = 1
    tap.numberOfTouchesRequired = 1
    tap.cancelsTouchesInView = false
    tap.delegate = self
    self.view.window?.addGestureRecognizer(tap)
}

internal func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldRecognizeSimultaneouslyWith otherGestureRecognizer: UIGestureRecognizer) -> Bool {
    return true
}

internal func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    let location = touch.location(in: self.view)

    if self.view.point(inside: location, with: nil) {
        return false
    }
    else {
        return true
    }
}

@objc private func onTap(sender: UITapGestureRecognizer) {

    self.view.window?.removeGestureRecognizer(sender)
    self.dismiss(animated: true, completion: nil)
}

我也尝试使用# test string s='line 01 line 02 line 03' # this doesn't seem to make any difference at all IFS=$'\n' # first naive attempt read a b c <<< "${s}" # this prints 'line 01||': # everything after the first newline is dropped echo "${a}|${b}|${c}" # second attempt, remove quotes read a b c <<< ${s} # this prints 'line 01 line 02 line 03||': # everything is assigned to the first variable echo "${a}|${b}|${c}" # third attempt, add -r read -r a b c <<< ${s} # this prints 'line 01 line 02 line 03||': # -r switch doesn't seem to make a difference echo "${a}|${b}|${c}" # fourth attempt, re-add quotes read -r a b c <<< "${s}" # this prints 'line 01||': # -r switch doesn't seem to make a difference echo "${a}|${b}|${c}"代替echo ${s} | read a b c，但无法使用它。

这可以用bash完成吗？

Answer 1

读取默认输入分隔符是\ n

{ read a; read b; read c;} <<< "${s}"

-d char：允许指定另一个输入分隔符

例如，输入字符串

中没有字符SOH（1 ASCII）

IFS=$'\n' read -r -d$'\1' a b c <<< "${s}"

编辑：-d可以采用空参数，空格在-d和null参数之间是必需的：

IFS=$'\n' read -r -d '' a b c <<< "${s}"

编辑：在评论任意行数的解决方案之后

function read_n {
    local i s n line
    n=$1
    s=$2
    arr=()
    for ((i=0;i<n;i+=1)); do
        IFS= read -r line
        arr[i]=$line
    done <<< "${s}"
}

nl=$'\n'
read_n 10 "a${nl}b${nl}c${nl}d${nl}e${nl}f${nl}g${nl}h${nl}i${nl}j${nl}k${nl}l"

printf "'%s'\n" "${arr[@]}"

Answer 2

您正在寻找readarray命令，而不是read。

readarray -t lines <<< "$s"

（从理论上讲，这里不需要引用$s。除非您使用的是bash 4.4或更高版本，否则我会引用它，因为先前版本的{{1}中存在一些错误}。）

一旦线条在数组中，您可以根据需要分配单独的变量

bash

bash：将多行字符串读入多个变量

2 个答案: