blocks|key|1112686|text|这会将整数转换为带有0x前缀的两位十六进制字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112687|strHex+=+"0x%250.2X"+%25+255|code-block|syntax|javascript|1112688|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This will convert an integer to a 2 digit hex string with the 0x prefix:

<pre><code>strHex = "0x%0.2X" % 255
</code></pre>

blocks|key|1112645|text|那hex()呢？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1112646|hex(255)++#+0xff|code-block|syntax|javascript|1112647|如果你真的想把\放在前面，你可以这样做：|1112648|print+'\\'+%2B+hex(255)[1:]|1112649|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/functions.html^0|1|5|1|5|0|0|0|7|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

What about <a href="http://docs.python.org/library/functions.html" rel="noreferrer"><code>hex()</code></a>?

<pre><code>hex(255) # 0xff
</code></pre>

If you really want to have <code>\</code> in front you can do:

<pre><code>print '\\' + hex(255)[1:]
</code></pre>

blocks|key|1112589|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112590|"0x%25x"+%25+255+#+=>+0xff|code-block|syntax|javascript|1112591|或|1112592|"0x%25X"+%25+255+#+=>+0xFF|1112593|Python文档说：“把这个放在枕头下面：http://docs.python.org/library/index.html”|offset|length|1112594|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/index.html^0|0|0|0|0|L|15|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Try:

<pre><code>"0x%x" % 255 # =&gt; 0xff
</code></pre>

or

<pre><code>"0x%X" % 255 # =&gt; 0xFF
</code></pre>

Python Documentation says: "keep this under Your pillow: <a href="http://docs.python.org/library/index.html" rel="noreferrer">http://docs.python.org/library/index.html</a>"

blocks|key|1327910|text|如果您想打包一个值<255+(一个字节无符号，uint8_t)的结构，并最终得到一个只有一个字符的字符串，那么您可能正在寻找format+B+instead+of+c。C将字符转换为字符串(本身没有太大用处)，而B则转换为整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1327911|struct.pack('B',+65)|code-block|syntax|javascript|1327912|(是的，65是\x41，而不是\x65。)|1327913|struct类还可以方便地处理用于通信或其他用途的字节顺序。|1327914|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/struct.html#format-characters^0|1Q|L|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

If you want to pack a struct with a value &lt;255 (one byte unsigned, uint8_t) and end up with a string of one character, you're probably looking for the <a href="http://docs.python.org/library/struct.html#format-characters" rel="noreferrer">format B instead of c</a>. C converts a character to a string (not too useful by itself) while B converts an integer.

<pre><code>struct.pack('B', 65)
</code></pre>

(And yes, 65 is \x41, not \x65.)

The struct class will also conveniently handle endianness for communication or other uses.

blocks|key|1112838|text|请注意，对于较大的值，hex()仍然有效(其他一些答案不起作用)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1112839|x+=+hex(349593196107334030177678842158399357)
print(x)|code-block|syntax|javascript|1112840|Python+2：0x4354467b746f6f5f736d616c6c3f7dL|1112841|Python+3：0x4354467b746f6f5f736d616c6c3f7d|1112842|对于解密的RSA消息，可以执行以下操作：|1112843|import+binascii

hexadecimals+=+hex(349593196107334030177678842158399357)

print(binascii.unhexlify(hexadecimals[2:-1]))+#+python+2
print(binascii.unhexlify(hexadecimals[2:]))+#+python+3|1112844|entityMap^0|B|5|0|0|9|X|0|9|W|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Note that for large values, <code>hex()</code> still works (some other answers don't):

<pre><code>x = hex(349593196107334030177678842158399357)
print(x)
</code></pre>

Python 2: <code>0x4354467b746f6f5f736d616c6c3f7dL</code> 
Python 3: <code>0x4354467b746f6f5f736d616c6c3f7d</code>

For a decrypted RSA message, one could do the following:

<pre><code>import binascii

hexadecimals = hex(349593196107334030177678842158399357)

print(binascii.unhexlify(hexadecimals[2:-1])) # python 2
print(binascii.unhexlify(hexadecimals[2:])) # python 3
</code></pre>

blocks|key|1328198|text|根据format-examples，使用format()，我们可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1328199|>>>+#+format+also+supports+binary+numbers
>>>+"int:+{0:d};++hex:+{0:x};++oct:+{0:o};++bin:+{0:b}".format(42)
'int:+42;++hex:+2a;++oct:+52;++bin:+101010'
>>>+#+with+0x,+0o,+or+0b+as+prefix:
>>>+"int:+{0:d};++hex:+{0:#x};++oct:+{0:#o};++bin:+{0:#b}".format(42)
'int:+42;++hex:+0x2a;++oct:+0o52;++bin:+0b101010'|code-block|syntax|javascript|1328200|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.4/library/string.html#format-examples^0|K|8|2|F|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

With <code>format()</code>, as per <a href="https://docs.python.org/3.4/library/string.html#format-examples" rel="noreferrer">format-examples</a>, we can do:

<pre><code>&gt;&gt;&gt; # format also supports binary numbers
&gt;&gt;&gt; "int: {0:d}; hex: {0:x}; oct: {0:o}; bin: {0:b}".format(42)
'int: 42; hex: 2a; oct: 52; bin: 101010'
&gt;&gt;&gt; # with 0x, 0o, or 0b as prefix:
&gt;&gt;&gt; "int: {0:d}; hex: {0:#x}; oct: {0:#o}; bin: {0:#b}".format(42)
'int: 42; hex: 0x2a; oct: 0o52; bin: 0b101010'
</code></pre>

blocks|key|1112912|text|这对我来说效果最好|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112913|"0x%2502X"+%25+5++#+=>+0x05
"0x%2502X"+%25+17+#+=>+0x11|code-block|syntax|javascript|1112914|如果您想要一个宽度更大的数字(2用于2个十六进制打印字符)，请更改(2)，这样3将为您提供以下结果|1112915|"0x%2503X"+%25+5++#+=>+0x005
"0x%2503X"+%25+17+#+=>+0x011|1112916|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This worked best for me

<pre><code>"0x%02X" % 5 # =&gt; 0x05
"0x%02X" % 17 # =&gt; 0x11
</code></pre>

Change the (2) if you want a number with a bigger width (2 is for 2 hex printned chars) so 3 will give you the following

<pre><code>"0x%03X" % 5 # =&gt; 0x005
"0x%03X" % 17 # =&gt; 0x011
</code></pre>

blocks|key|1328211|text|(int_variable).to_bytes(bytes_length,+byteorder='big'%7C'little').hex()|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1328212|例如：|unstyled|1328213|>>>+(434).to_bytes(4,+byteorder='big').hex()
'000001b2'
>>>+(434).to_bytes(4,+byteorder='little').hex()
'b2010000'|1328214|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>(int_variable).to_bytes(bytes_length, byteorder='big'|'little').hex()
</code></pre>

For example:

<pre><code>&gt;&gt;&gt; (434).to_bytes(4, byteorder='big').hex()
'000001b2'
&gt;&gt;&gt; (434).to_bytes(4, byteorder='little').hex()
'b2010000'
</code></pre>

blocks|key|1113134|text|此外，您可以将任何基数的任何数字转换为十六进制。在这里使用这一行代码它很容易使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1113135|hex(int(n,x)).replace("0x","")|offset|length|style|CODE|1113136|您有一个字符串n，它是您的数字，x是该数字的基数。首先，将其更改为整数，然后更改为十六进制，但十六进制的第一个值为0x，因此使用replace将其删除。|1113137|entityMap^0|0|0|U|0|7|1|G|1|1L|2|1S|7|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$]]

Also you can convert any number in any base to hex. Use this one line code here it's easy and simple to use:

<code>hex(int(n,x)).replace("0x","")</code>

You have a string <code>n</code> that is your number and <code>x</code> the base of that number. First, change it to integer and then to hex but hex has <code>0x</code> at the first of it so with <code>replace</code> we remove it.

blocks|key|1327963|text|我希望将一个随机整数转换为一个开头带有#的六位十六进制字符串。为了得到这个，我用了|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327964|"#%256x"+%25+random.randint(0xFFFFFF)|code-block|syntax|javascript|1327965|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I wanted a random integer converted into a six-digit hex string with a # at the beginning. To get this I used

<pre><code>"#%6x" % random.randint(0xFFFFFF)
</code></pre>

blocks|key|1328117|text|作为另一种表示，您可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1328118|[in]+'%25s'+%25+hex(15)
[out]'0xf'|code-block|syntax|javascript|1328119|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

As an alternative representation you could use

<pre><code>[in] '%s' % hex(15)
[out]'0xf'
</code></pre>

I want to take an integer (that will be &lt;= 255), to a hex string representation

e.g.: I want to pass in <code>65</code> and get out <code>'\x41'</code>, or <code>255</code> and get <code>'\xff'</code>.

I've tried doing this with the <code>struct.pack('c',</code>65<code>)</code>, but that chokes on anything above <code>9</code> since it wants to take in a single character string.

How to convert an int to a hex string?

Python

我想将一个整数(将是<= 255)转换为十六进制字符串表示形式例如:我想传入65并获得'\x41'，或者255并获得'\xff'。我试过用struct.pack('c',65)来做这件事，但它在9以上的任何东西上都会卡住，因为它想要接受单个字符串。

问如何将int转换为十六进制字符串？
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问如何将int转换为十六进制字符串？
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