bash 中意外标记“elif”附近的语法错误
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Syntax error near unexpected token "elif" in bash
提问by t3hcakeman
Here's my code to display some dialogs.
这是我显示一些对话框的代码。
#!/bin/bash
output=$(zenity --list --text="Choose action" --column= --hide-header "Hidden Files" "Desktop")
if [ $output = "Hidden Files"]
then
output2=$(zenity --list --text="Do what?" --column= --hide-header "Show" "Hide")
if [ $output2 = "Show"]
then
echo showing files
else
echo hiding files
elif [ $output = "Desktop"];then
output3=$(zenity --list --text="Do what?" --column= --hide-header "Show" "Hide")
if [ $output2 = "Show"]
then
echo showing files
else
echo hiding files
else
exit
fi
I get this error after the first dialog:
第一次对话后我收到此错误：
systool.sh: line 12: syntax error near unexpected token `elif'
systool.sh: line 12: `elif [ $output = "Desktop"];then'
Whats wrong?
怎么了？
回答by dogbane
You need to close both of your inner if-else statements with a fi.
您需要使用fi.
For example:
例如：
if [ $output2 = "Show"]
then
echo showing files
else
echo hiding files
fi
You also need a space before the closing ]in your if conditions. For example:
您还需要]在 if 条件中的结束之前留一个空格。例如：
if [ $output2 = "Show" ]
回答by Aaron Digulla
You're missing a fiafter both echo hiding files
你错过了fi两个echo hiding files
回答by Michael Krelin - hacker
Your inner ifs have no corresponding fis.
您的内部ifs 没有相应的fis。
