mysql的“where子句”bash脚本中的未知列

这是我的控制台命令的第一行和输出：

source request.sh a a "ZP07"
a a ZP07
USE intro_ict; SELECT idbuildings from buildings WHERE address = ZP07;
mysql: [Warning] Using a password on the command line interface can be insecure.
the tables in the database are: buildings
room
sensordata
mysql: [Warning] Using a password on the command line interface can be insecure.
ERROR 1054 (42S22) at line 1: Unknown column 'ZP07' in 'where clause'
the selected building ZP07 has idbuilding

这是request.sh：

  1 #!/bin/bash
  2
  3 #args = startdate ($1) enddate ($2) address($3) roomnr($4)
  4 echo "$1 $2 $3 $4"
  5
  6 sql1="USE intro_ict; SELECT idbuildings from buildings WHERE address = $3;"
  7 tables="USE intro_ict; SHOW TABLES;"
  8 echo "$sql1"
  9
 10 echo "the tables in the database are: $(mysql -h 127.0.0.1 -P 3306 -u root -padmin -se "$tables")"
 11 idbuilding=$(mysql -h 127.0.0.1 -P 3306 -u root -padmin -se "USE intro_ict; SELECT idBuildings FROM Buildings WHERE (address = $3);")
 12
 13 echo "the selected building $3 has idbuilding $idbuilding"
 14 #mysql -h 127.0.0.1 -P 3306 -u root -padmin -se "USE intro_ict; SELECT * FROM buildings;"
 15
 16

我想使用bash脚本将数据从mysql数据库显示到控制台。我用不同的引语研究了这个问题的其他解决方案，`但没有任何效果。我正在使用ubuntu for windows，我可以在mysqlworkbench中看到buildings.address列肯定在那里

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我不知道该试什么了。

有问题的条款：

WHERE address = ZP07

表示查找一个名为address的列，该列等于另一个名为ZP07的列

我猜：

ZP07实际上是一个存储在地址列中的值，并且。。。 表中没有名为ZP07的列。。。 解释错误消息：未知列“ZP07” 要解决此问题，您需要在ZP07值周围加上单引号，例如：

WHERE address = 'ZP07'

这可以通过在shell脚本中添加3美元左右的单引号来实现，例如：

sql1="USE intro_ict; SELECT idbuildings from buildings WHERE address = '$3';"

# and

idbuilding=$(mysql ... "USE intro_ict; SELECT idBuildings FROM Buildings WHERE (address = '$3');")

注意：OP可能需要重新考虑重写idbuilding=$mysql。。。部分使用sql1变量，这样SELECT语句就不必在2个不同的位置进行维护，例如：

idbuilding=$(mysql -h 127.0.0.1 -P 3306 -u root -padmin -se "${sql1}")

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请从代码中删除行号。