bash 用于 if 条件的 Shell 脚本过多参数
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Shell Script Too Many Arguments for if condition
提问by A. Mesut Konuklar
My current script does the following;
我当前的脚本执行以下操作；
It takes integer as a command line argument and starts from 1 to N , it checks whether the numbers are divisible by 3, 5 or both of them. It simply prints out Uc for 3, Bes for 5 and UcBes for 3,5. If the command line argument is empty, it does the same operation but the loop goes to 1 to 20.
它接受 integer 作为命令行参数并从 1 到 N 开始，它检查数字是否可以被 3、5 或两者整除。它只是打印出 3 的 Uc、5 的 Bes 和 3,5 的 UcBes。如果命令行参数为空，它会执行相同的操作，但循环会转到 1 到 20。
I am having this error "Too many arguments at line 11,15 and 19".
我有这个错误“第 11,15 和 19 行参数太多”。
Here is the code:
这是代码：
#!/bin/bash
if [ ! -z ]; then
for i in `seq 1 `
do
if [ [$i % 3] -eq 0 ]; then
echo "Uc"
elif [ i % 5 -eq 0 ]; then
echo "Bes"
elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0?]
then
echo "UcBes"
else
echo "$i"
fi
done
elif [ -z ]
then
for i in {1..20}
do
if [ i % 3 -eq 0 ]
then
echo "Uc"
elif [ i % 5 -eq 0 ]
then
echo "Bes"
elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0?]
then
echo "UcBes"
else
echo "$i"
fi
done
else
echo "heheheh"
fi
回答by sampson-chen
Note that [is actually synonym for the testbuiltin in shell (try which [in your terminal), and not a conditional syntax like other languages, so you cannot do:
请注意，这[实际上是testshell 中内置函数的同义词（which [在您的终端中尝试），而不是像其他语言那样的条件语法，因此您不能这样做：
if [ [$i % 3] -eq 0 ]; then
Moreover, always make sure that there is at least one space between [, ], and the variables that comprise the logical condition check in between them.
此外，始终确保有之间至少有一个空格[，]以及包含在它们之间的逻辑条件检查的变量。
The syntax for evaluating an expression such as modulo is enclosure by $((...)), and the variable names inside need not be prefixed by $:
求值表达式（例如模数）的语法是由 包围$((...))，并且其中的变量名称不需要以 为前缀$：
remainder=$((i % 3))
if [ $remainder -eq 0 ]; then
回答by Julien Vivenot
You should probably use something like :
你可能应该使用类似的东西：
if [ $(($i % 3)) -eq 0 ]
instead of
代替
if [ $i % 3 -eq 0 ]
if [ [$i % 3] -eq 0 ]
回答by William Pursell
Your script could be greatly simplified. For example:
您的脚本可以大大简化。例如：
#!/bin/sh
n=0
while test $(( ++n )) -le ${1:-20}; do
t=$n
expr $n % 3 > /dev/null || { printf Uc; t=; }
expr $n % 5 > /dev/null || { printf Bes; t=; }
echo $t
done
gives slightly different error messages if the argument is not an integer, but otherwise behaves the same.
如果参数不是整数，则给出略有不同的错误消息，否则行为相同。
