我有下表：
<table border="1">
<tr>
<th scope="col">Header</th>
<th scope="col">Header</th>
<th scope="col">Header</th>
</tr>
<tr>
<th scope="row"> </th>
<td> </td>
<td>Split this one into two columns</td>
</tr>
</table>
我希望将包含“将此分成两列”的单元格拆分为两个单元格/列。我该怎么做？
答案 0 :(得分：34)
喜欢这个http://jsfiddle.net/8ha9e/1/
将colspan="2"添加到第3个<th>，然后在第二行中添加4个<td>。
<table border="1">
<tr>
<th scope="col">Header</th>
<th scope="col">Header</th>
<th scope="col" colspan="2">Header</th>
</tr>
<tr>
<th scope="row"> </th>
<td> </td>
<!-- The following two cells will appear under the same header -->
<td>Col 1</td>
<td>Col 2</td>
</tr>
</table>
答案 1 :(得分：13)
您有两种选择。
<colspan>来拉伸两列或更多列的单元格。<table>的{{1}}内插入一个包含2列的td。答案 2 :(得分：13)
我来到这里是因为我遇到了类似的问题。
@MrMisterMan的答案，以及其他人，真的很有帮助，但边界正在击败我的游戏。所以，我做了一些研究，发现使用 rowspan 。
这就是我的所作所为，我想这可能有助于其他人面对类似的事情。
<table style="width: 100%; margin-top: 10px; font-size: 0.8em;" border="1px">
<tr align="center" >
<th style="padding:2.5px; width: 10%;" rowspan="2">Item No</th>
<th style="padding:2.5px; width: 55%;" rowspan="2">DESCRIPTION</th>
<th style="padding:2.5px;" rowspan="2">Quantity</th>
<th style="padding:2.5px;" colspan="2">Rate per Item</th>
<th style="padding:2.5px;" colspan="2">AMOUNT</th>
</tr>
<tr>
<th>Rs.</th>
<th>P.</th>
<th>Rs.</th>
<th>P.</th>
</tr>
</table>
答案 3 :(得分：3)
是你在找什么？
<table border="1">
<tr>
<th scope="col">Header</th>
<th scope="col">Header</th>
<th scope="col" colspan="2">Header</th>
</tr>
<tr>
<th scope="row"> </th>
<td> </td>
<td>Split this one</td>
<td>into two columns</td>
</tr>
</table>
答案 4 :(得分：2)
将要分割的<td>更改为如下所示：
<td><table><tr><td>split 1</td><td>split 2</td></tr></table></td>
答案 5 :(得分：2)
使用此示例，您可以使用colspan属性
<TABLE BORDER>
<TR>
<TD>Item 1</TD>
<TD>Item 1</TD>
<TD COLSPAN=2>Item 2</TD>
</TR>
<TR>
<TD>Item 3</TD>
<TD>Item 3</TD>
<TD>Item 4</TD>
<TD>Item 5</TD>
</TR>
</TABLE>
http://hypermedia.univ-paris8.fr/jean/internet/ex_table.html的更多示例。
答案 6 :(得分：0)
请尝试以下方式。
<table>
<tr>
<th>Month</th>
<th>Savings</th>
</tr>
<tr>
<td>January</td>
<td>$100</td>
</tr>
<tr>
<td>February</td>
<td>$80</td>
</tr>
<tr>
<td colspan="2">Sum: $180</td>
</tr>
</table>
答案 7 :(得分：0)
https://jsfiddle.net/SyedFayaz/ud0mpgoh/7/
import random
import binpacking
# DISTRIBUTES A DICTIONARY OF LENGTHS (KEY-VALUE = ID-LENGTH PAIR) TO A MINIMAL NUMBER IF BINS WHICH CAN HAVE DIFFERENT VOLUME.
def packing2dOptimizer(b, bin_bucket):
# OPTIMIZED BINS WILL BE STORED HERE
bin_list = []
bin_bucket.append(0)
bin_bucket.sort(reverse=True)
# LIST OF DIFFERENCES BETWEEN BIN SIZES
difference = [0]
if len(bin_bucket) > 1:
for index in range(1, len(bin_bucket)):
difference.append(bin_bucket[0] - bin_bucket[index])
# FIND THE EXACT BIN WIDTH AS PER USAGE OF PACKINGS
def checkFinalSize(total):
bin_width = 0
for i in bin_bucket:
if total <= i: bin_width = i
else: break
return bin_width
# PACK ALL THE VALUES IN BINS WITH MAXIMUM SIZE
packer_list = binpacking.to_constant_volume(b,bin_bucket[0])
# CREATE A DETAILED DATA OF BINPACKING
counter = 0
for packer in packer_list:
bin_usage = sum(packer.values())
bin_width = checkFinalSize(bin_usage)
bin_waste = bin_width - bin_usage
bin_data = {
'bin_id': counter,
'bin_width': bin_width,
'bin_usage': bin_usage,
'bin_waste': bin_waste,
'pack_data': packer
}
counter += 1
bin_list.append(bin_data)
# PRINT THE LIST
area_used = 0
area_covered = 0
area_waste = 0
for i in bin_list:
print("Bin", i['bin_width'],":",i['pack_data'], i['bin_waste'])
area_used += i['bin_width']
area_covered += sum(i['pack_data'].values())
area_waste += i['bin_waste']
print()
print("Area Used:", area_used)
print("Area Covered:", area_covered)
print("Area Waste:", area_waste)
print()
# TAKES TWO PACKED BINS AND THE AVAILABLE SIZES OF ALL BINS AS ARGUMENT
def optimizer(small_bin, big_bin):
dict1 = small_bin['pack_data']
dict2 = big_bin['pack_data']
# CREATE A NEW DICTIONARY OF DATA OF THE BOTH PACKED BINS
new_dict = dict1.copy()
new_dict.update(dict2)
max_val = max(list(new_dict.values()))
# FIND A SUITABLE BIN WITH LOWEST SIZE AND WASTAGE
for size in bin_bucket:
if size >= max_val:
new_bins = binpacking.to_constant_volume(new_dict,size)
if len(new_bins) <= 2:
final_bins = new_bins
# UPDATE BOTH PACKED BIN VALUES AND RETURNS
big_bin['pack_data'] = final_bins[0]
small_bin['pack_data'] = final_bins[1]
return small_bin, big_bin
# OPTIMIZE THE BINS TO MINIMIZE THE WASTE
bin_bucket.sort(reverse=True)
for small_bin in bin_list:
if small_bin['bin_width'] == bin_bucket[-2] and small_bin['bin_waste'] > 0:
for big_bin in bin_list:
if big_bin['bin_width'] > small_bin['bin_width']:
pack_data = list(big_bin['pack_data'].values())
for length in pack_data:
if length <= small_bin['bin_waste']:
# CALL THE OPTIMIZER
small_bin, big_bin = optimizer(small_bin, big_bin)
# UPDATE THE BIN DATA
bin_usage = sum(small_bin['pack_data'].values())
bin_width = checkFinalSize(bin_usage)
small_bin['bin_width'] = bin_width
small_bin['bin_usage'] = bin_usage
small_bin['bin_waste'] = bin_width - bin_usage
bin_usage = sum(big_bin['pack_data'].values())
bin_width = checkFinalSize(bin_usage)
big_bin['bin_width'] = bin_width
big_bin['bin_usage'] = bin_usage
big_bin['bin_waste'] = bin_width - bin_usage
break
# SORT THE OPTIMIZED BINS BY BIN SIZE
bin_list = sorted(bin_list, key = lambda i: i['bin_width'], reverse = True)
# PRINT THE LIST
area_used = 0
area_covered = 0
area_waste = 0
for i in bin_list:
print("Bin", i['bin_width'],":",i['pack_data'], i['bin_waste'])
area_used += i['bin_width']
area_covered += sum(i['pack_data'].values())
area_waste += i['bin_waste']
print()
print("Area Used:", area_used)
print("Area Covered:", area_covered)
print("Area Waste:", area_waste)
return bin_list
def main():
# LIST OF DATA THAT WILL BE SELECTED RANDOMLY
random_data = [ i for i in range(100, 3200,100) ]
# DIFFERENT SIZE OF VALUES WITH UNIQUE ID ASSOCIATED WITH IT
b = {
'a': random.choice(random_data),
'b': random.choice(random_data),
'c': random.choice(random_data),
'd': random.choice(random_data),
'e': random.choice(random_data),
'f': random.choice(random_data),
'g': random.choice(random_data),
'h': random.choice(random_data),
'i': random.choice(random_data),
'j': random.choice(random_data),
'k': random.choice(random_data),
'l': random.choice(random_data),
'm': random.choice(random_data),
'n': random.choice(random_data),
'o': random.choice(random_data),
'p': random.choice(random_data),
'q': random.choice(random_data),
'r': random.choice(random_data),
's': random.choice(random_data),
't': random.choice(random_data),
'u': random.choice(random_data),
'v': random.choice(random_data),
'w': random.choice(random_data),
'x': random.choice(random_data),
'y': random.choice(random_data),
'z': random.choice(random_data),
}
# LIST OF DIFFERENT SIZES OF BINS
bin_bucket = [2400, 2700, 3100]
bin_list = packing2dOptimizer(b, bin_bucket)
main()