如何从中导入给定电子邮件ID/pwd的联系人
使用python/django应用程序.请指教?
我试图运行python应用程序并根据指定的间隔执行操作.下面的代码不断消耗100%的CPU.
def action_print():
print "hello there"
interval = 5
next_run = 0
while True:
while next_run > time.time():
pass
next_run = time.time() + interval
action_print()
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我想避免让进程进入休眠状态,因为会有更多的动作以不同的间隔执行.
请指教
我想知道是否有人可以帮助$_POST再次发布价值.假设我将表单值发布到post.php,我可以通过$ _POST或$ _REQUEST变量访问数据.但是如何将$ _POST发布到另一个url让我们说post_one.php并在那里访问数据?
我在使用jQuery 1.4.2更新表格单元格值时遇到问题.这一切都适用于Firefox和Safari,但IE8和IE9根本没有做任何事情.没有警告,错误或任何会给我一些提示在哪里寻找它的东西.
表如下:
<table id="test">
<tr id="1">
<td id="name">sample name</td>
<td id="schedule">sample value</td>
<td id="day">sample value</td>
</tr>
<tr id="2">
<td id="name">sample name</td>
<td id="schedule">sample value</td>
<td id="day">sample value</td>
</tr>
<tr id="3">
<td id="name">sample name</td>
<td id="schedule">sample value</td>
<td id="day">sample value</td>
</tr>
</table>
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我正在执行ajax调用并获取json数据:
{"Test": [
{"id":"1", "name":"John", "day":"Monday"},
{"id":"2", "name":"Marry", "day":"Thursday"}
]}
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一旦收到数据,就会有一个循环,它遍历json数据集并使用接收的数据更新相应的列,如下所示:
$.each(json.Tests, function(){
/* update test with details */
var test = this.hash;
/*set values for each test */
$("table#test tr[id=" + test + "]").find("#name").html(this.name);
$("table#test tr[id=" + …Run Code Online (Sandbox Code Playgroud) 我花了几个小时试图弄清楚如何将多个参数传递给python脚本,该脚本应该由subprocess.Popen执行而没有任何运气.
脚本:
command = ['/usr/bin/python', '/tmp/script.py mcl=NULL mtp=data mnm=DS4INST \
mno=NULL mse=NULL mce=cll01']
result = subprocess.Popen(command, stdout = subprocess.PIPE, \
stderr = subprocess.PIPE)
out, err = result.communicate()
print out, err
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我收到以下错误消息:
python: can't open file '/tmp/script.py mcl=NULL mtp=data mnm=DS4INST mno=NULL \
mse=NULL mce=cll01': [Errno 2] No such file or directory
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但是,当我直接从shell执行脚本时
/usr/bin/python /tmp/script.py mcl=NULL mtp=data mnm=DS4INST mno=NULL \
mse=NULL mce=cll01
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我收到所需的输出并且未生成错误消息.
请指教.
下面的代码示例在加密/解密操作后没有返回原始文本,我正在尝试找出原因
from Crypto.Cipher import AES
text = """This is plain text
to use.
It should be exqctly 128 characters long to avoid padding and it is split
with new lines as in
file"""
password = "password........"
block = 32
mode = AES.MODE_CBC
enc = AES.new(password, mode)
encrypted = enc.encrypt(text)
print "ORIGINAL: " + text
print "ENCRYPTED: " + str(encrypted)
print "DECRYPTED: " + str(enc.decrypt(encrypted))
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谁能告诉为什么文本的第一部分格式不正确?
print("enter start() to start the program")
def start():
print("This script converts GBP into any currency based on the exchange rate...")
print(" ") #enters a line
exchangeRate = int(input("Enter the exchange rate (Eg: 0.80)"))
print("how much would you like to convert???")
gpb = int(input())
print(gpb*exchangeRate)
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如果我将汇率设为0.81而我输入£1则总是返回0.
我试图用字典中查找的值替换列表元素值我该怎么做?
list = [1, 3, 2, 10]
d = {'id': 1, 'val': 30},{'id': 2, 'val': 53}, {'id': 3, 'val': 1}, {'id': 4, 'val': 9}, {'id': 5, 'val': 2}, {'id': 6, 'val': 6}, {'id': 7, 'val': 11}, {'id': 8, 'val': 89}, {'id': 9, 'val': 2}, {'id': 10, 'val': 4}
for i in list:
for key, v in d.iteritems():
???
???
所以最后我期待:
list = [30, 1, 53, 4]
谢谢
我正在尝试使用高级python调度程序安排con样式作业.每当我使用不需要任何参数的函数时,一切都很好,但是我无法安排传递一个或多个参数的相同作业.您能否提一些建议?
from apscheduler.scheduler import Scheduler
def job_def(var1, var2):
print "%s - %s" % (str(var1), str(var2))
s = Scheduler()
s.add_cron_job(job_def,['hello', 'there'], seconds='*/30')
s.start()
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错误:
Traceback (most recent call last): File "<stdin>", line 1, in <module>
File "/usr/local/lib/python2.7/dist-packages/APScheduler-2.0.2-py2.7.egg/apscheduler scheduler.py", line 346, in add_cron_job start_date=start_date)
File "/usr/local/lib/python2.7/dist-packages/APScheduler-2.0.2-py2.7.egg/apscheduler/triggers/cron/__init__.py", line 44, in __init__ field = field_class(field_name, exprs, is_default)
File "/usr/local/lib/python2.7/dist-packages/APScheduler-2.0.2-py2.7.egg/apscheduler/triggers/cron/fields.py", line 29, in __init__ self.compile_expressions(exprs)
File "/usr/local/lib/python2.7/dist-packages/APScheduler-2.0.2-py2.7.egg/apscheduler/triggers/cron/fields.py", line 56, in compile_expressions self.compile_expression(expr)
File "/usr/local/lib/python2.7/dist-packages/APScheduler-2.0.2-py2.7.egg/apscheduler/triggers/cron/fields.py", line 69, in compile_expression (expr, self.name))
Run Code Online (Sandbox Code Playgroud) 今天早上我的插入声明有点麻烦.是的,我使用的是不推荐使用的mysql_query函数.我的insert语句如下所示:
$query3 = "INSERT INTO ".$db_prefix ." offer_det
(fname, lname, 10k, 14k, 18k, 21k, 22k, 24k, 925, coins, bars)
VALUES '".$fname."', '".$lname."', '".$_10k."', '".$_14k."',
'".$_18k."', '".$_21k."', '".$_22k."', '".$_24k."',
'".$_925."', '".$coins."', '".$bars."')";
$result3 = mysql_query($query3);
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我的PHP表单值是insert语句的第一部分中列出的所有变量,'fname'等.
我的变量设置为从帖子中拉出,并列为插入插入的值.
我必须在它们开始之前将变量更改为下划线,我猜PHP不喜欢这样.
我的问题:
对于数据类型fname和lname为varchar,而对于10K通过酒吧的十进制(7,3).
单击"+"时,js/jquery代码会在下面列出切换子项的内容?
<ul>
<li id="1" parent="0"><a href="#">1</a></li>
<li id="2" parent="0"><a href="#"> + </a><a href="#">2</a>
<ul parent="2">
<li id="23" parent="2"><a href="#">2.1</a></li>
<li id="50" parent="2"><a href="#"> + </a><a href="#">2.2</a>
<ul parent="50">
<li id="123" parent="50"><a href="#">2.50.1</a></li>
</ul>
</li>
</ul>
</li>
</ul>
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