如何在回聲時避免空格被分割成多行

Question

我有一個很長的字符串可以通過echo命令打印。通過這樣做，我希望它完美縮進。 我想這和它是完全工作正常

echo "This is a very long string. An"\ 
"d it is printed in one line" 

Output: 
This is a very long string. And it is printed in one line 

但是，當我嘗試正確縮進它作爲echo語句也縮進。它增加了額外的空間。

echo "This is a very long string. An"\ 
    "d it is printed in one line" 

Output: 
This is a very long string. An d it is printed in one line 

我找不到任何可以完美實現這一點的工作響應。

Answer 1

這裏的問題是，你是給兩個參數echo，它的默認行爲是將它們打印背部採用了空間之間：

$ echo "a"    "b" 
a b 
$ echo "a" "b" 
a b 
$ echo "a"\ 
>   "b" 
a b 

---

如果你想在有完全控制權你要打印，使用陣列，printf：

lines=("This is a very long string. An" 
     "d it is printed in one line") 
printf "%s" "${lines[@]}" 
printf "\n" 

這將返回：

This is a very long string. And it is printed in one line 

或者爲suggested by 123 in comments，使用echo還與陣列IFS設置爲null：

# we define the same array $lines as above 

$ IFS="" 
$ echo "${lines[*]}" 
This is a very long string. And it is printed in one line 
$ unset IFS 
$ echo "${lines[*]}" 
This is a very long string. An d it is printed in one line 
#       ^
#        note the space 

從Bash manual → 3.4.2. Special Parameters：

*

（$ ）擴展到位置參數，從一開始。當擴展不在雙引號內時，每個位置參數擴展爲單獨的單詞。在執行的上下文中，這些單詞受到進一步的單詞分割和路徑名擴展。當擴展出現在雙引號內時，它將擴展爲一個單詞，每個參數的值由IFS特殊變量的第一個字符分隔。也就是說，「$」相當於「$ 1c $ 2c ...」，其中c是IFS變量值的第一個字符。 如果IFS未設置，則參數由空格分隔。如果IFS爲空，則參數在沒有介入分隔符的情況下連接。

有趣的閱讀：Why is printf better than echo?。