调用bash脚本会在if语句和变量赋值时生成错误

Question

I'm trying to execute a very simple program (round numbers to lowest integer divisible by 15) but am getting an error:

$min = date +"%M";

if [ $min%15 != 0 ]
  then
    $min - $min%1
fi

echo $min;

I call it with sh cache.sh

I feel I've followed the syntax I've learned here but I'm getting line 9: syntax error: unexpected end of file What have I got wrong here?

Answer 1

That script is not valid bash syntax. I would start by finding some working examples, and perhaps an entire tutorial. You might start with William Shotts' book, which is available online .

Some notes about your attempt:

1. The $ is used to request replacement of a variable ¹ by its value. It is not a sigil that is part of the variable name, as it is in Perl or PHP. So it is not used on the left-hand-side of an assignment.

2. The shell is primarily used to run other executables, and interprets everything through that lens. If a command line looks like an invocation of another program, the shell will try to run that other program, rather than do anything shell-scripty. Therefore, the command min = date +"%M" will cause the shell to look for a program named min and execute it with three command-line arguments: = , date , and +%M .
   In order for an assignment to be recognized as such, there cannot be any space around the = .

3. Without spaces, min=date +"%M" is still not right, however. The shell will just temporarily assign the literal string "date" to the variable min and then try to run a command called +%M .
   If a value has spaces in it, you need quotation marks around it ² .

4. Even with quotes, however, min="date +%M" would assign to min the literal string "date +%M". If you actually want to run the command date +"%M" and use its output as a value, then you have to request that using the command-substitution syntax, $(...) . Here our friend the dollar sign is again requesting replacement by a dynamic value, but the parentheses make it a different type of request; instead of a variable's value, the expression is replaced by the output of a command.

5. Because of the parsing issues noted above, the built-in arithmetic operations only work in certain contexts. Two ways to create a valid arithmetic context are the (( ... )) special forms and the let command.

6. Finally, even if your script were syntactically valid, it is semantically incorrect if your goal is to round down to the nearest multiple of 15. The remainder after dividing by 1 is always zero, so your script ends by attempting to subtract 0 from min - and does nothing with the result anyway, since there's no assignment back to min . If you want to round down, you have to actually subtract the remainder that you just tested. You could do it like this:

    min=$(date +%M) let rem=min%15 if (( rem != 0 )); then let min-=rem fi echo $min 

But you could also do it a bit more succinctly:

echo $(( min=$(date +%M), min-=min%15 ))

This works without an if because subtracting 0 is harmless. The comma just lets us put two expressions inside a single set of ((...)) . The second expression min-=min%15 is a modifying assignment - it means the same thing as min=min-min%15 , but saves us one instance of typing out "min". Putting our friend the replacement-requesting $ in front of (( ... )) causes the whole expression to be replaced by its value, so that echo gets something to print out. The value of a list of expressions is the value of the last expression, and the value of an assignment is the same as the value that was assigned, so the result that is echo ed is the same as the final value of $min : the closest multiple of 15 minutes after the hour.

¹ In shell terminology, variables are actually called "parameters". Just something to bear in mind when reading documentation.

² You actually don't need quotation marks around the %M in your command for this reason. Everything in the shell is automatically a string; you don't need the quotes to make it one. However, they don't hurt, and putting quotation marks around things is a good habit to have, since it keeps your code from being broken by unexpected special characters in input values.

Answer 2

Your script has many syntax issues. In shell assignment is is

var='val'

instead of

$var='val'

Also there is no space around =

Your correct script can be:

min=$(date +"%M")

if (( min % 15 != 0 ))
then
    echo "not fully divisible by 15"
fi

echo $min

Answer 3

It looks like you converted this from some other language. Here's a working bash version.

#!/bin/bash
min=$(date +"%M")

if [ $(($min % 15)) != 0 ] ; then
    min=$(( min - min % 1 ))
fi

echo $min;

local output:

~/tmp › sh ./test.sh
34

Answer 4

 min=`date +"%M"`;

 if [ $min%15 != 0 ]

 then

 min=$((min - min%1))

 fi

 echo $min;