bash 中带有嵌入引号的字符串扩展
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String expansion in bash with embedded quotes
提问by dreftymac
Problem:
问题：
The following shell script code does not produce the expected result:
以下 shell 脚本代码不会产生预期的结果：
# MYSQL, MyUSER MyHost etc ... all defined above as normal
TARG_DB="zztest";
DB_CREATE="$($MYSQL -u $MyUSER -h $MyHOST -p$MyPASS -Bse 'create database $TARG_DB')";
Expected outcome:
预期结果：
A new database created with the name zztest
使用名称创建的新数据库 zztest
Actual outcome:
实际结果：
A new database created with the name $TARG_DB
使用名称创建的新数据库 $TARG_DB
Question:
题：
How can this example code be changed such that $TARG_DBis interpolated or expanded, giving the expected outcome?
如何更改此示例代码以进行$TARG_DB插值或扩展，从而给出预期结果？
回答by Kevin
Because $TARG_DBis within single quotes within the subshell, it's taken literally. Use double quotes there, they won't mess up the subshell. e.g.
因为$TARG_DB在子shell 中的单引号内，它是按字面意思理解的。在那里使用双引号，它们不会弄乱子外壳。例如
$ tmp="a b c"
$ echo $tmp
a b c
$ echo $(echo $tmp)
a b c
$ echo $(echo "$tmp")
a b c
$ echo $(echo '$tmp')
$tmp
$ echo "$(echo '$tmp')"
$tmp
$ echo "$(echo "$tmp")"
a b c
回答by SiegeX
Don't wrap it in single-quotes.
不要用单引号括起来。
There is no need to quote the command substitution $()so you can remove those and use double-quotes inside it.
无需引用命令替换，$()因此您可以删除它们并在其中使用双引号。
DB_CREATE=$($MYSQL -u $MyUSER -h $MyHOST -p$MyPASS -Bse "create database $TARG_DB");
回答by sgibb
TARG_DB="zztest";
DB_CREATE="$(${MYSQL} -u ${MyUSER} -h ${MyHOST} -p${MyPASS} -Bse \"create database ${TARG_DB}\")";
回答by Mithrandir
The 'create database $TARG_DB'part is wrong! Single quotes supress variable substitution in bash. MySQL "sees" the literal text "$TARG_DB" and creates a database with that name. Put your sql statement in double quotes.
该'create database $TARG_DB'部分是错的！单引号禁止 bash 中的变量替换。MySQL“看到”文字文本“$TARG_DB”并创建一个具有该名称的数据库。将您的 sql 语句放在双引号中。
