shell脚本获取mysql插入数据自增长id的值

时间：2024-07-03 08:33:38

shell脚本获取mysql插入数据自增长id的值

在shell脚本中我们可以通过last_insert_id()获取id值，但是，需要注意的是，该函数必须在执行插入操作的sql语句之后，立即调用，否则获取的值就为0，LAST_INSERT_ID 是与table无关的，如果向表a插入数据后，在向表b插入数据，LAST_INSERT_ID会改变。当然还有其他方法:

1. select max(id) from tablename；

2. select @@IDENTITY;

3. SHOW TABLE STATUS;
具体的差别和各自具体的特征，本文不再叙述，自行百度即可.

本文主要以LAST_INSERT_ID()为例：

错误代码示例如下:

echo

$($MYSQL -e "INSERT INTO problem_logger VALUES (null,$REPORT_DATE,$FIXED_DATE,'$PROB_SYMPTOMS','$PROB_SOLUTIONS');")

id=$($MYSQL -e "SELECT LAST_INSERT_ID() id")

id=`echo $id | gawk '{print $2}'`

$MYSQL <<EOF

SELECT * FROM problem_logger where id_number=$id\G

EOF

#######上述代码获取到的id值就为0，而不是我们期望的值

正确的代码如下:
echo

$($MYSQL -e "INSERT INTO problem_logger VALUES (null,$REPORT_DATE,$FIXED_DATE,'$PROB_SYMPTOMS','$PROB_SOLUTIONS');")

	id=$($MYSQL -e "SELECT LAST_INSERT_ID() id")

id=`echo $id | gawk '{print $2}'`

$MYSQL <<EOF

SELECT * FROM problem_logger where id_number=$id\G

EOF

完整的访问数据库，插入数据，并获取自增id值的shell脚本如下:

########备注:1. 数据库名称:Problem_Trek      操作的表名称:problem_logger

#!/bin/bash

#

# Record_Problem - records system problems in database

#

###########################################################

#

# Determine mysql location & put into variable

#

MYSQL=which mysql" Problem_Trek -u root"

#

###########################################################

#

# Create Record Id & Report_Date

#

#ID_NUMBER=date +%y%m%d%H%M

#

REPORT_DATE=date +%y%m%d

#

############################################################

#

# Acquire information to put into table

#

echo

echo -e "Birefly describe the problem & its symptoms: \c"

#

read ANSWER

PROB_SYMPTOMS=$ANSWER

#

# Set Fixed Date & Problem Solution to null for now

#

FIXED_DATE=0

PROB_SOLUTIONS=""

#

#############################################################

#

# Insert acquired information into table

#

echo

echo "Problem recorded as follows:"

echo

id=$($MYSQL -e "INSERT INTO problem_logger VALUES (null,$REPORT_DATE,$FIXED_DATE,'$PROB_SYMPTOMS','$PROB_SOLUTIONS');SELECT LAST_INSERT_ID() id")

id=echo $id | gawk '{print $2}'

$MYSQL <<EOF

SELECT * FROM problem_logger where id_number=$id\G

EOF

#

#############################################################

#

# Check if want to enter a solution now

#

echo

echo -e "Do you have a solution yet?(y/n) \c"

read ANSWER

#

case $ANSWER in

y|Y|YES|yes|Yes|yEs|yeS|YEs|yES)

./Update_Problem.sh $id

#

;;

*)

# if answer is anything but yes, just exit script

;;

esac

#

############################################################

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