阅读以下材料，回答下列问题。（28分）材料一：东亚某区域图材料二：石家庄与仁川各月降水量统计资料（单位：mm）月份+ e, _1 U# F* C 1& F8 @3 Z$ C+ [. S 20 U& [7 W, @. h+ S 32 D' M2 ]
a 4) T8 ]- `0 b" U* _6 V 5! @5 A3 ^: b5 a1 I# I" g 67 M; C9 J* Z8 f7 N 7! V# R+ O& J7 M9 a7 N 8* [, f2 D6 M2 g9 G 9" c- _& J; c- f 10& j# Z4 a5 [6 A5 L% I 110 X: ?5 g! [: _' X9 \& G 12- D9 ]9 K9 d4 H! N/ a- P% c0 Q 总量
X( O0 Y, F2 M' g
X 石家庄# L9 `* C+ O5 A( Q5 F% V 3.9
\* f' Z) @5 ]& e1 ]8 Q9 C 7.4
T* S# W* Z: D# O. H 11.31 P4 h. A" M" R 17.8. b* O4 Q! a! d" f( N! G. ? 36.9. R" T- i% A' h* ]$ X 56.75 _7 J6 @: `% T" \' U% G 141.76 E, T( A' ^, g1 N 148.30 h- F+ f) O7 ` 48.1
a6 j/ ^1 [2 J2 T2 h% i 27.3. j
K- a& _9 f9 Y) i 13.2* X; H1 E3 `) N 5.1- I+ H' h+ M) _
a# b8 ?' B 517．7; C" V$ a* N- J 仁川. J; R8 S4 C# ]! K* `+ `$ ^ 58.6) C$ M9 I6 [! Z. L3 G0 W 63.6* a5 B; g7 N' K7 ]/ O1 N8 Q 65.8' D- j% ?1 ^+ b8 ^6 c7 K 77.0' Z; h2 H% ?8 S- O" P 82.2. ^: N9 D& U8 c. L! @4 Y; ? 113.3- `# h& U: O$ W. A9 V* U5 J 202.99 c. X$ [9 Z, [# B3 a8 E 212.05 _# ]/ I1 U& e) A9 A* C 97.6& F3 d9 e7 \: f) C# P+ V- ]( L# H5 V 69.3' B- [1 G; V* h 53.0! c& L. A) a& D& _6 ]# E 54.9: a
a( X' ]7 F) ?9 i2 U8 _* _1 S 1152" S) P+ P3 D/ b3 E0 R$ V （1）日至10月4日，第17届亚运会在韩国仁川（126°37′E、37°28′N）举行。仁川采用
（时区）的区时；闭幕式时，仁川的昼夜长短状况是
。（4分）（2）太行山东西两侧的地形成因，主要是外力作用的结果，其外力作用主要表现为：黄土高原
；华北平原
。（4分）（3）石家庄和仁川年降水量差异的主导因素是
。（2分）威海的降水季节分配与仁川相似，夏季和冬季降水较多，但秋季却经常出现干旱，试分析形成秋旱的主要原因。（4分）（4）黄河下游由于
而成为“地上河”，（2分）航运价值不大，试分析其主要原因。（6分）（5）仁川是韩国第三大城市和重要港口，钢铁、汽车制造等工业和第三产业都十分发达，简述仁川发展钢铁、汽车等工业的有利条件。（6分）
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跟谁学学生版:genshuixue_student精品好课等你领在线咨询下载客户端关注微信公众号&&&分类：阅读以下材料，回答下列问题。（28分）材料一：东亚某区域图材料二：石家庄与仁川各月降水量统计资料（单位：mm）月份+ e, _1 U# F* C 1& F8 @3 Z$ C+ [. S 20 U& [7 W, @. h+ S 32 D' M2 ]
a 4) T8 ]- `0 b" U* _6 V 5! @5 A3 ^: b5 a1 I# I" g 67 M; C9 J* Z8 f7 N 7! V# R+ O& J7 M9 a7 N 8* [, f2 D6 M2 g9 G 9" c- _& J; c- f 10& j# Z4 a5 [6 A5 L% I 110 X: ?5 g! [: _' X9 \& G 12- D9 ]9 K9 d4 H! N/ a- P% c0 Q 总量
X( O0 Y, F2 M' g
X 石家庄# L9 `* C+ O5 A( Q5 F% V 3.9
\* f' Z) @5 ]& e1 ]8 Q9 C 7.4
T* S# W* Z: D# O. H 11.31 P4 h. A" M" R 17.8. b* O4 Q! a! d" f( N! G. ? 36.9. R" T- i% A' h* ]$ X 56.75 _7 J6 @: `% T" \' U% G 141.76 E, T( A' ^, g1 N 148.30 h- F+ f) O7 ` 48.1
a6 j/ ^1 [2 J2 T2 h% i 27.3. j
K- a& _9 f9 Y) i 13.2* X; H1 E3 `) N 5.1- I+ H' h+ M) _
a# b8 ?' B 517．7; C" V$ a* N- J 仁川. J; R8 S4 C# ]! K* `+ `$ ^ 58.6) C$ M9 I6 [! Z. L3 G0 W 63.6* a5 B; g7 N' K7 ]/ O1 N8 Q 65.8' D- j% ?1 ^+ b8 ^6 c7 K 77.0' Z; h2 H% ?8 S- O" P 82.2. ^: N9 D& U8 c. L! @4 Y; ? 113.3- `# h& U: O$ W. A9 V* U5 J 202.99 c. X$ [9 Z, [# B3 a8 E 212.05 _# ]/ I1 U& e) A9 A* C 97.6& F3 d9 e7 \: f) C# P+ V- ]( L# H5 V 69.3' B- [1 G; V* h 53.0! c& L. A) a& D& _6 ]# E 54.9: a
a( X' ]7 F) ?9 i2 U8 _* _1 S 1152" S) P+ P3 D/ b3 E0 R$ V （1）日至10月4日，第17届亚运会在韩国仁川（126°37′E、37°28′N）举行。仁川采用
（时区）的区时；闭幕式时，仁川的昼夜长短状况是
。（4分）（2）太行山东西两侧的地形成因，主要是外力作用的结果，其外力作用主要表现为：黄土高原
；华北平原
。（4分）（3）石家庄和仁川年降水量差异的主导因素是
。（2分）威海的降水季节分配与仁川相似，夏季和冬季降水较多，但秋季却经常出现干旱，试分析形成秋旱的主要原因。（4分）（4）黄河下游由于
而成为“地上河”，（2分）航运价值不大，试分析其主要原因。（6分）（5）仁川是韩国第三大城市和重要港口，钢铁、汽车制造等工业和第三产业都十分发达，简述仁川发展钢铁、汽车等工业的有利条件。（6分）
阅读以下材料，回答下列问题。（28分）材料一：东亚某区域图材料二：石家庄与仁川各月降水量统计资料（单位：mm）月份+ e, _1 U# F* C 1& F8 @3 Z$ C+ [. S 20 U& [7 W, @. h+ S 32 D' M2 ]
a 4) T8 ]- `0 b" U* _6 V 5! @5 A3 ^: b5 a1 I# I" g 67 M; C9 J* Z8 f7 N 7! V# R+ O& J7 M9 a7 N 8* [, f2 D6 M2 g9 G 9" c- _& J; c- f 10& j# Z4 a5 [6 A5 L% I 110 X: ?5 g! [: _' X9 \& G 12- D9 ]9 K9 d4 H! N/ a- P% c0 Q 总量
X( O0 Y, F2 M' g
X 石家庄# L9 `* C+ O5 A( Q5 F% V 3.9
\* f' Z) @5 ]& e1 ]8 Q9 C 7.4
T* S# W* Z: D# O. H 11.31 P4 h. A" M" R 17.8. b* O4 Q! a! d" f( N! G. ? 36.9. R" T- i% A' h* ]$ X 56.75 _7 J6 @: `% T" \' U% G 141.76 E, T( A' ^, g1 N 148.30 h- F+ f) O7 ` 48.1
a6 j/ ^1 [2 J2 T2 h% i 27.3. j
K- a& _9 f9 Y) i 13.2* X; H1 E3 `) N 5.1- I+ H' h+ M) _
a# b8 ?' B 517．7; C" V$ a* N- J 仁川. J; R8 S4 C# ]! K* `+ `$ ^ 58.6) C$ M9 I6 [! Z. L3 G0 W 63.6* a5 B; g7 N' K7 ]/ O1 N8 Q 65.8' D- j% ?1 ^+ b8 ^6 c7 K 77.0' Z; h2 H% ?8 S- O" P 82.2. ^: N9 D& U8 c. L! @4 Y; ? 113.3- `# h& U: O$ W. A9 V* U5 J 202.99 c. X$ [9 Z, [# B3 a8 E 212.05 _# ]/ I1 U& e) A9 A* C 97.6& F3 d9 e7 \: f) C# P+ V- ]( L# H5 V 69.3' B- [1 G; V* h 53.0! c& L. A) a& D& _6 ]# E 54.9: a
a( X' ]7 F) ?9 i2 U8 _* _1 S 1152" S) P+ P3 D/ b3 E0 R$ V （1）日至10月4日，第17届亚运会在韩国仁川（126°37′E、37°28′N）举行。仁川采用
（时区）的区时；闭幕式时，仁川的昼夜长短状况是
。（4分）（2）太行山东西两侧的地形成因，主要是外力作用的结果，其外力作用主要表现为：黄土高原
；华北平原
。（4分）（3）石家庄和仁川年降水量差异的主导因素是
。（2分）威海的降水季节分配与仁川相似，夏季和冬季降水较多，但秋季却经常出现干旱，试分析形成秋旱的主要原因。（4分）（4）黄河下游由于
而成为“地上河”，（2分）航运价值不大，试分析其主要原因。（6分）（5）仁川是韩国第三大城市和重要港口，钢铁、汽车制造等工业和第三产业都十分发达，简述仁川发展钢铁、汽车等工业的有利条件。（6分）
科目:难易度:最佳答案（1）东8区
昼短夜长（2）风力沉积
流水沉积（3）海陆位置
秋季，东南季风减弱，暖湿空气衰退；北方冷空气加强，天气晴朗，秋高气爽，降水不多而蒸发较大（4）泥沙沉积
夏季水量不足，时常断流；冬季封航，还受凌汛影响；河水含沙量大，水位较浅（5）经济发达，消费市场广阔；发达的海洋运输，有利于原料的输入和产品的输出；工业基础好，技术先进，劳动力丰富解析试题分析：（1）仁川的经度为126°37′E，其所在的时区为东8区，故仁川采用的为东8区的区时。闭幕式时间为日，此时太阳直射南半球，而仁川在北半球，此时北半球昼短夜长。（2）黄土高原是大风将中亚、蒙古的沙子吹到黄土高原，遇到太行山、秦岭的阻挡而沉积下来。华北平原主要是黄河携带泥沙，在流速减缓时大量泥沙沉积而形成。（3）由表可知石家庄年降水量多于仁川，两地纬度相当，仁川位于沿海，受海洋影响较大，故两地年降水量差异的主导因素为海陆位置。威海夏季时受东南季风从海洋带来暖湿气流影响降水较多，冬季时地处西北季风从渤海带来暖湿气流的迎风坡，降水多；秋季时东南季风减弱，降水减少，此时北方多晴朗天气，降水少而蒸发大。（4）“地上河”主要是黄河携带泥沙沉积而形成。河流航运价值要从流量、流速、通航时间、水位及货运量来分析。而此地航运价值低的主要原因是，位于我国北方地区，年降水量少，夏季时水量不足，还会出现断流；冬季时，该河段会结冰，还会产生凌汛，使得通航时间短；黄河含沙量大，水量少，使得水位低，也影响了通航。（5）工业发展的区位条件要从资源、市场、劳动力、交通、技术、政策来分析。材料中提到仁川是韩国第三大城市和重要港口，故当地海运发达；经济也发达，消费市场广阔；由于该国煤铁资源匮乏，可以利用发达的海洋运输，便于于原料的输入和产品的输出；此外当地工业基础好，技术先进；仁川人口多，为工业的发展提供了丰富的劳动力。考点：区时、昼夜长短的判断、外力作用的表现、河流航运价值的分析及工业区位因素的考查。知识点:&&基础试题拔高试题热门知识点最新试题
关注我们官方微信关于跟谁学服务支持帮助中心bit manipulations
int(x, base=10)
Convert a number or string x to an integer, or return 0 if no arguments are given.
>>> int('', 2)
>>> int('0xff',16)
>>> int('ff', 16)
hex string
To convert to hex string:
>>> "0x%x" % (int('', 2))
>>> chr(int('))
>>> int(')
>>> chr(int('))
>>> int('', 2)
>>> chr(int('', 2))
>>> ord('t')
bitstring classes
provides four classes:
BitStream and BitArray and their immutable versions ConstBitStream and Bits:
Bits (object): This is the most basic class. It is immutable and so its contents can't be changed after creation.
BitArray (Bits): This adds mutating methods to its base class.
ConstBitStream (Bits): This adds methods and properties to allow the bits to be treated as a stream of bits, with a bit position and reading/parsing methods.
BitStream (BitArray, ConstBitStream): This is the most versative class, having both the bitstream methods and the mutating methods.
>>> from bitstring import BitStream, BitArray
>>> c = BitArray(hex='')
BitArray('0x')
binary string
>>> d = BitArray(bin='0011 00')
BitArray('0b001100')
>>> a = BitArray(uint=45, length=12)
>>> b = BitArray(int=-1, length=7)
(BitArray('0x02d'), BitArray('0b1111111'))
from raw byte
>>> a = BitArray(bytes=b'\x00\x01\x02\xff', length=28, offset=1)
BitArray('0x000205f')
>>> b = BitArray(bytes=open('video.mp4', 'rb').read())
BitArray('0xd6f6dfe6d15acd941e08effcb8ebe2cb22c05ffffcfdc45e9bde6d948b3eeefd20...') # length=
class bitstring.ConstBitStream([auto, length, offset, **kwargs])
>>> from bitstring import ConstBitStream
>>> s = ConstBitStream('0x123456')
ConstBitStream('0x123456')
>>> s.peek(12)
ConstBitStream('0x123')
>>> s.peek('hex:12')
peek reads from the current bit position pos in the bitstring according to the fmt string or integer
and returns the result. The bit position is unchanged.
It reads from current bit position pos in the bitstring according the the format string and returns a single result.
n bits as a signed integer.
n bits as an unsigned integer.
n bits as a hexadecimal string.
n bits as a binary string.
n bits as a new bitstring.
bytes:n n bytes as bytes object.
The sample run below shows it advances 4 bits each time we read a hex number:
>>> s = ConstBitStream('0x1234')
ConstBitStream('0x1234')
>>> s.read('hex:4')
>>> s.read('hex:4')
>>> s.read('hex:4')
>>> s.read('hex:4')
If we read 4 bits and output it as a binary format:
>>> s = ConstBitStream('0x1234')
>>> s.read('bin:4')
>>> s.read('bin:4')
>>> s.read('bin:4')
>>> s.read('bin:4')
Output as an unsigned integer as we read in 4 bits each time and advance:
>>> s = ConstBitStream('0x1234')
>>> s.read('uint:4')
>>> s.read('uint:4')
>>> s.read('uint:4')
>>> s.read('uint:4')
pos and bitpos
pos/bitpos is a read and write property for setting and getting the current bit position in the bitstring. Can be set to any value from 0 to len.
>>> s = ConstBitStream('0x1234')
>>> s.read('uint:4')
>>> s.pos += 4
>>> s.read('uint:4')
>>> s.read('uint:4')
The following code reads in
one byte at a time, convert it to character, and then puts it into a list. It reads only the first 180 bytes.
from bitstring import ConstBitStream
nbytes = 180
with open('video.mp4', 'rb)') as vfile:
packet = ConstBitStream(bytes = vfile.read(nbytes), length = nbytes*8)
while(packet.pos & nbytes*8):
#byte = packet.read(8).hex
#info.append(chr(int(byte, 16)))
byte = packet.read(8).uint
info.append(chr(byte))
print info
The output looks like this:
['\x00', '\x00', '\x00', ' ', 'f', 't', 'y', 'p', 'i', 's', 'o', 'm',
'\x00', '\x00', '\x02', '\x00', 'i', 's', 'o', 'm', 'i', 's', 'o', '2',
'a', 'v', 'c', '1', 'm', 'p', '4', '1', '\x00', '\x00', '\x00', '\x08',
'f', 'r', 'e', 'e', '\x04', '\x8e', '\x01', '\xfe', 'm', 'd', 'a', 't',
'\x00', '\x00', '\x00', '\x00', '\x00', '\x1d', 'g', 'd', '\x00',
'\x15', '\xac', '\xd9', 'A', '\xe0', '\x8e', '\xff', '\xc0', '"',
'\x00', '!', '\xc4', '\x00', '\x00', '\x03', '\x00', '\x04', '\x00',
'\x00', '\x03', '\x00', '\xca', '&', 'X', '\xb6', 'X', '\x00', '\x00',
'\x00', '\x06', 'h', '\xeb', '\xe2', '\xcb', '"', '\xc0', '\x00',
'\x00', '\x02', '\xd3', '\x06', '\x05', '\xff', '\xff', '\xcf', '\xdc',
'E', '\xe9', '\xbd', '\xe6', '\xd9', 'H', '\xb7', '\x96', ',', '\xd8',
' ', '\xd9', '#', '\xee', '\xef', 'x', '2', '6', '4', ' ', '-', ' ',
'c', 'o', 'r', 'e', ' ', '1', '4', '0', ' ', 'r', '2', ' ', '1', 'c',
'a', '7', 'b', 'b', '9', ' ', '-', ' ', 'H', '.', '2', '6', '4', '/',
'M', 'P', 'E', 'G', '-', '4', ' ', 'A', 'V', 'C', ' ', 'c', 'o', 'd',
'e', 'c', ' ', '-', ' ', 'C', 'o', 'p', 'y', 'l', 'e', 'f', 't']
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Artificial Neural Networks (ANN)7 - PDOException: SQLSTATE[HY000]: General error: 1366 Incorrect string value --- when trying to run cron - Drupal Answers
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I am using search file attachments module, after adding some content with attached files, when I try to run cron the whole site is not indexing. But when I add content with just title and body it is working file, if I attached a file to node then I am unable to search the file contents.
Log report says
PDOException: SQLSTATE[HY000]: General error: 1366 Incorrect string
value: '\xF0\x9D\x90\xBF ...' for column 'data' at row 1: INSERT INTO
{search_dataset} (sid, type, data, reindex) VALUES
(:db_insert_placeholder_0, :db_insert_placeholder_1,
:db_insert_placeholder_2, :db_insert_placeholder_3); Array (
[:db_insert_placeholder_0] => 26 [:db_insert_placeholder_1] => file
[:db_insert_placeholder_2] => filename 6 hl bu slides r2 may 2 2012pdf
content head loss back up slides head loss calcula3ons casa grande
requires a correla3on to determine head loss over the range of
relevant condi3ons debris loads fiber par3culate microporous chemical
flow rate temperature npsh margin slide 2 porous media head loss
correla3ons porous media head loss correla3ons follows the classical
porous media flow equa3ons dp a u b u2 dl where a coefficient for viscous
term b coefficient for iner3a term nureg cr 6224 correla3on is a semi
theore3cal correla3on developed based on flat plate ver3cal loop head
loss tes3ng with nukon fiberglass fibers nominally 7 micron diameter and
bwr suppression pool sludge iron oxide nominally 10 micron diameter
nureg cr 6224 experimental data were performed at fluid temperatures
ranging from 60of to 125of debris bed thicknesses ranging from 125 in
to 4 in and approach veloci3es ranging from 15 c s to 15 c s slide 3
nureg cr 6224 correla3on where δh head loss c water sv surface to
volume ra3o of the debris c2 c3 u dynamic viscosity of water lbm c sec
u fluid approach velocity c sec ρ density of water lbm c3 αm mixed
debris bed solidity one minus the porosity δlm actual mixed debris bed
thickness in λ
c water in lbm c2 sec2 conversion factor for
english units m m m vmmv lssh δ λ δ 23512 u 1 660u57153 ρ α α uαα
slide 4 nureg cr 6224 correla3on the very low approach velocity at stp
1 c sec suggests that the head loss will be dominated by the viscous
term the viscous term of the nureg cr 6224 correla3on is based on
experimental work by c n davies proceedings of ins3tute of mechanical
engineers london b1 p 185 1952 for stp condi3ons the nureg cr 6224
correla3on could be simplified to slide 5 mmmv lsh δ λ δ u
uαα nureg cr 6224 correla3on suppor3ng compression equa3on nureg cr
6224 based on w l ingmanson et al internal pressure distribu3on in
compressible mats under fluid stress tapi journal vol 42 no 101959 L
L m ? H L ? where α and ? are empirically based currently α 13
and ? 38 alterna3ve clint shaffer 2005 1 L L m 1 ? L H ? where α
? and ф are empirically based currently α 65 ? 38 and ф 35 1 6224
correla3on training session nrc headquarters april 12 2005 slide 6
shaffer compression alterna3ve 1 slide 7 1 6224 correla3on training
session nrc headquarters april 12 2005 head loss correla3on refinements
perform ver3cal loop tes3ng to acquire head loss data at stp specific
condi3ons ie low approach velocity and representa3ve fiber par3culate
loadings adjust correla3on suppor3ng equa3ons to best fit the
experimental results determine from integrated chemical effects tests
the impact of chemical precipitates on head loss slide 8 strainer
geometry calculate strainer area and gap dimensions based on strainer
drawings calculate average approach velocity based on total strainer
area calculate inters33al volume based on gap dimensions calculate
increased approach velocity for large debris loads based on
circumscribed strainer area slide 9 strainer dimensions strainer area
per train 18185 c2 circumscribed area per train 4190 c2 inters33al
volume per train 818 c3 photos of stp pci strainer slide 10 flow rate
and temperature input total flow rate through each eccs strainer for
the specific case analyzed maximum of 7020 gpm per train at stp based
on 1620 gpm per hhsi pump 2800 gpm per lhsi pump and 2600 gpm per cs
pump calculate debris accumula3on on each strainer based on rela3ve
flow split calculate pool fluid density and viscosity for a given pool
temperature slide 11 npsh margin input npsh margin for each safety
injec3on and containment spray pump compare calculated debris bed head
loss to the pump npsh margin to determine whether the pump would fail
npsh required lhsi pumps 165 c hhsi pumps 161 c cs pumps 164 c npsh
available excluding clean strainer and debris losses start of
recircula3on 267 f 22 c 24 hours 171 f 42 c 30 days 128 f 51 c slide
12 nureg cr 6224 head loss correla3on mixed debris bed solidity the
mixed debris bed solidity αm is given by where αo the solidity of the
original fiber blanket ie the as fabricated solidity η mp mf the
par3culate to fiber mass ra3o in the debris bed m σ mi is the total
par3culate mass lbm ρf the fiber density lbm c3 ρp the average
par3culate material density lbm c3 σ ρivi σ vi c the head loss induced
volumetric compression of the debris inches inch slide 13 nureg cr
6224 head loss correla3on sv averaging for mixed debris bed the
averaged surface to volume ra3o for a mixed debris bed is given by
where svn sv of the nth cons3tuent vn volume of the nth cons3tuent
slide 14 sv sqrt σ svn2 vn σ vn [:db_insert_placeholder_3] => 0 ) in
search_index() (line 705 of
/var/www//pilogix/praapps/modules/search/search.module).
Can any one know what is happening and how to fix this?
5,09561739
This is a completely wild guess, but here goes:
You run mysql
Your tables are "UTF-8"
The data you are inserting into the index contains "unusual" UTF-8 characters.
If those three are true, then there's your problem.
MySQL "cheats" on their UTF-8 implementation. Drupal will always instruct MySQL to create UTF-8 tables, but by default,
doesn't cover the entire UTF-8.
You will need to manually convert your tables to utf8mb4, and then also manually "upgrade"
the connection to mysql whenever you handle this data. I stuck some code
at one point that partly addresses this, but it's not generally applicable at this time.
Or you switch to , that actually gets this right by default.
There's an
for resolving this.
24.2k865130
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