使用bash:

if [ "`mysql -u'USER' -p'PASSWORD' -se'USE $DATABASE_NAME;' 2>&1`" == "" ]; then
    echo $DATABASE_NAME exist
else
    echo $DATABASE_NAME doesn't exist
fi

对于那些使用php与mysqli然后这是我的解决方案。我知道答案已经回答了，但我认为这将是有帮助的答案作为一个mysqli准备的声明。

$db = new mysqli('localhost',username,password);
$database="somedatabase";
$query="SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME=?";
$stmt = $db->prepare($query);
$stmt->bind_param('s',$database);
$stmt->execute();
$stmt->bind_result($data);
if($stmt->fetch())
{
    echo "Database exists.";
}
else
{
    echo"Database does not exist!!!";
}
$stmt->close();

下面的解决方案对我来说很有效:

mysql -u${MYSQL_USER} -p${MYSQL_PASSWORD} \
-s -N -e "SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME='${MYSQL_DATABASE}'"

从壳像bash

if [[ ! -z "`mysql -qfsBe "SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME='db'" 2>&1`" ]];
then
  echo "DATABASE ALREADY EXISTS"
else
  echo "DATABASE DOES NOT EXIST"
fi

Golang解决方案

创建一个测试包并添加:

import "database/sql"

// testing database creation
func TestCreate(t *testing.T){
    Createdb("*Testdb") // This just calls the **sql.DB obect *Testdb 
    db,err := sql.Open("mysql", "root:root@tcp(127.0.0.1:3306)/*Testdb")
    if err != nil{
        panic(err)
    }
    defer db.Close()
    _, err = db.Exec("USE *Testdb")
    if err != nil{
        t.Error("Database not Created")
    }

} 

SELECT COUNT(*) FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'DbName'

1 -存在，0 -不存在