read_sql_table('xxx')  should equal to  read_sql('select * from xxx')
I have permission to `products` table, and no permission to `areas` table.

## read_sql_table

```python
dfp = pd.read_sql_table('products', db.engine)  
```

Got error:
```
OperationalError: (_mysql_exceptions.OperationalError) (1142, "SHOW command denied to user 'aaa'@'111.111.111.111' for table 'areas'") [SQL: 'SHOW CREATE TABLE `areas`'] (Background on this error at: http://sqlalche.me/e/e3q8)
```

## read_sql

But  `dfp = pd.read_sql('select * from products', db.engine)` is fine.

I just wonder why `pd.read_sql_table('products', db.engine) `  need SHOW ` table 'areas'`

`pd.read_sql_table('products')` should equal to `pd.read_sql('select * from products')`

#### Output of ``pd.show_versions()``

<details>

INSTALLED VERSIONS
------------------
commit: None
python: 3.6.5.final.0
python-bits: 64
OS: Windows
OS-release: 10
machine: AMD64
processor: Intel64 Family 6 Model 45 Stepping 7, GenuineIntel
byteorder: little
LC_ALL: None
LANG: None
LOCALE: None.None

pandas: 0.23.4
pytest: 3.5.1
pip: 19.0.1
setuptools: 39.1.0
Cython: 0.28.2
numpy: 1.14.3
scipy: 1.1.0
pyarrow: 0.12.0
xarray: None
IPython: 7.2.0
sphinx: 1.7.4
patsy: 0.5.0
dateutil: 2.6.1
pytz: 2018.4
blosc: None
bottleneck: 1.2.1
tables: 3.4.3
numexpr: 2.6.5
feather: None
matplotlib: 3.0.1
openpyxl: 2.5.3
xlrd: 1.1.0
xlwt: 1.3.0
xlsxwriter: 1.0.4
lxml: 4.2.1
bs4: 4.6.0
html5lib: 1.0.1
sqlalchemy: 1.2.7
pymysql: None
psycopg2: 2.7.6.1 (dt dec pq3 ext lo64)
jinja2: 2.10
s3fs: 0.1.6
fastparquet: None
pandas_gbq: None
pandas_datareader: None

</details>
