#!/bin/bash if [ -z $1 -o -z $2 -o -z $3 -o -z $4 -o -z $5 -o -z $6 -o -z $7 ] then echo 'hehe' else echo 'haha' fi ./test.sh: line 2: [: too many arguments haha
我在HP-UX和Solaris的操作系统上,写了一个这样的小程序,如下,当显示i=3408时,程序中断,并提示arguments too long. 但是我在Linux上是没有的,请问如何解决这个问题呢?是csh的版本吗? #!/bin/csh set i = 1 set a = () while ($i <= 4000) echo $i set a = ($a $i) @ i++ end
书上的程序,一点没改,为什么说“too many arguments”? 谢谢! #!/bin/sh COUNTER=0 while [ $COUNTER -lt 5 ] do COUNTER='expr $COUNTER+1' echo $COUNTER done
我在安装mysql binary 的时候,按照install binary的安装说明,安装完了以后,在最后,执行./mysqld_safe --user/mysql $的时候,抱错,抱错信息如下:(use/local/mysql/data/linuxAS.err) 061225 15:16:00 mysqld started /usr/local/mysql/bin/mysqld: Too many arguments (first extra is '$'). Use --help to get a list of available options 061225 15:16:00 mysqld ended 但是,如果用./mysqld_safe --user/mysql (后...
#!/bin/bash for i in `find ./ -name class` do if [ -z `ls ${i}/cache/` ] ; then echo "$i/cache/ is empty" else rm ${i}/cache/* echo "$i/cache is clear" fi done 执行可以达到功能,但显示 ./rm1.sh: line 19: [: too many arguments ./class/cache is clear 如果是#!/bin/sh 则显示为 [: 39: 12: unexpected operator ./class/cache is clea...
今天编译一很老的程序时出现: error: too few arguments to function 'exit' 查找原因原理是里面有几个函数是exit(); 后来把exit()改成exit(0);问题就解决了。 是不是现在编译器不认exit()了吗? 谢谢。