bash 中的 printf 命令

Question

printf "%50s\n" " I ate 4 eggs"

在此示例中，我想从变量中分配 50。如何在 bash 的 for 循环中使用该变量？

任何想法？

Answer 1

至少对于bash，您可以使用 C 样式printf *转换修饰符传递可变字段宽度：

width=50
printf '%*s\n' $width "I ate 4 eggs"
                                      I ate 4 eggs

man 3 printf详情请参阅。

如何在 for 循环中使用它取决于您想要的输出是什么。

---

根据您的评论，您可以使用

for i in {45..50}; do 
  for j in {1..13}; do 
    printf '%*.*s\n' $i $j "I ate 4 eggs"
  done
done

提供字段宽度和精度作为变量，导致

                                   I
                                   I
                                  I a
                                 I at
                                I ate
                               I ate
                              I ate 4
                             I ate 4
                            I ate 4 e
                           I ate 4 eg
                          I ate 4 egg
                         I ate 4 eggs
                         I ate 4 eggs
                                     I
                                    I
                                   I a
                                  I at
                                 I ate
                                I ate
                               I ate 4
                              I ate 4
                             I ate 4 e
                            I ate 4 eg
                           I ate 4 egg
                          I ate 4 eggs
                          I ate 4 eggs

等等。

Answer 2

您可以将 50 分配给变量，如下所示：

variable=50

您可以在 for 循环中使用该变量，如下所示：

for i in foo bar baz; do
    printf "%${variable}s\n" "I ate 4 eggs"
done

---

如果您尝试增加 printf 语句上的填充，您可以执行以下操作：

$ for i in {45..50}; do printf "%${i}s\n" "I ate 4 eggs"; done
                                 I ate 4 eggs
                                  I ate 4 eggs
                                   I ate 4 eggs
                                    I ate 4 eggs
                                     I ate 4 eggs
                                      I ate 4 eggs