我需要调试使用soap客户端的代码.我在php.net中找到了getLast*方法,但是当我尝试获取最后一次调试请求时,它返回NULL
<?php
$client = new SoapClient("http://www.webservicex.net/ConverPower.asmx?WSDL");
$response = $client->ChangePowerUnit(array(
"PowerValue" => 100,
"fromPowerUnit" => "horsepower",
"toPowerUnit" => "megawatts"
));
echo "====== REQUEST HEADERS =====" . PHP_EOL;
var_dump($client->__getLastRequestHeaders());
echo "========= REQUEST ==========" . PHP_EOL;
var_dump($client->__getLastRequest());
echo "========= RESPONSE =========" . PHP_EOL;
var_dump($response);
?>
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代码执行的结果:
$php soap_test.php
====== REQUEST HEADERS =====
NULL
========= REQUEST ==========
NULL
========= RESPONSE =========
object(stdClass)#2 (1) {
["ChangePowerUnitResult"]=>
float(0.0746)
}
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如何获取最后一个SoapClient请求的正文和标题的内容?
这是我的代码:
class Hero:
def __init__(self, name, age):
self.name = name
self.age = age
def __str__(self):
return self.name + str(self.age)
def __hash__(self):
print(hash(str(self)))
return hash(str(self))
heroes = set()
heroes.add(Hero('Zina Portnova', 16)) # gets hash -8926039986155829407
print(len(heroes)) # gets 1
heroes.add(Hero('Lara Miheenko', 17)) # gets hash -2822451113328084695
print(len(heroes)) # gets 2
heroes.add(Hero('Zina Portnova', 16)) # gets hash -8926039986155829407
print(len(heroes)) # gets 3! WHY?
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为什么会这样?
第一个和第三个对象具有相同的内容和相同的散列但是len()讲述了3个唯一对象?
此问题适用于Rust的预发布版本.
这个年轻的问题很相似.
我试图按io::println功能打印一个符号
fn main() {
io::println('c');
}
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但我得到了下一个错误:
$ rustc pdst.rs
pdst.rs:2:16: 2:19 error: mismatched types: expected `&str` but found `char` (expected &str but found char)
pdst.rs:2 io::println('c');
^~~
error: aborting due to previous error
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如何将char转换为字符串?
UPDATE
直接类型转换不起作用:
let text:str = 'c';
let text:&str = 'c';
let text:@str = 'c';
let text:~str = 'c';
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它返回:
pdst.rs:7:13: 7:16 error: bare `str` is not a type
pdst.rs:7 let text:str = 'c';
^~~
pdst.rs:7:19: 7:22 error: mismatched types: expected …Run Code Online (Sandbox Code Playgroud) Factory Girl是一个方便的rails框架,可以轻松创建测试模型实例.
factory_girl允许您快速定义每个模型的原型,并询问具有对手头测试很重要的属性的实例.
一个例子(也来自主页):
Factory.sequence :email do |n|
"somebody#{n}@example.com"
end
# Let's define a factory for the User model. The class name is guessed from the
# factory name.
Factory.define :user do |f|
# These properties are set statically, and are evaluated when the factory is
# defined.
f.first_name 'John'
f.last_name 'Doe'
f.admin false
# This property is set "lazily." The block will be called whenever an
# instance is generated, and the return value of …Run Code Online (Sandbox Code Playgroud) 我试图检查演员的行为。这是Swift5.5提供的新功能。
我使用官方文档swift.org中的示例代码创建了一个游乐场:
import Foundation
actor TemperatureLogger {
let label: String
var measurements: [Int]
private(set) var max: Int
init(label: String, measurement: Int) {
self.label = label
self.measurements = [measurement]
self.max = measurement
}
}
let logger = TemperatureLogger(label: "Outdoors", measurement: 25)
print(await logger.max)
// Prints "25"
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但我的编译器在这个例子中失败了:
斯威夫特编译器错误:
不支持并发的函数中的“await”
Actor 隔离属性“max”只能从 Actor 内部引用
那么如何访问与参与者隔离的属性呢?
也许这是编译器或示例代码中的错误?
Xcode 版本 13.0 测试版 (13A5154h) Swift 版本 5.5
我编写了一个程序,它具有特征Animal和Dog实现特征的结构.它还有一个AnimalHouse存储动物作为特征对象的结构Box<Animal>.
trait Animal {
fn speak(&self);
}
struct Dog {
name: String,
}
impl Dog {
fn new(name: &str) -> Dog {
return Dog {
name: name.to_string(),
};
}
}
impl Animal for Dog {
fn speak(&self) {
println!{"{}: ruff, ruff!", self.name};
}
}
struct AnimalHouse {
animal: Box<Animal>,
}
fn main() {
let house = AnimalHouse {
animal: Box::new(Dog::new("Bobby")),
};
house.animal.speak();
}
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它返回"Bobby:ruff,ruff!" 正如所料,但如果我尝试克隆house编译器返回错误:
fn main() {
let house …Run Code Online (Sandbox Code Playgroud) 我使用Nokogiri来解析具有相同内容的html页面:
<p class="parent">
Useful text
<br>
<span class="child">Useless text</span>
</p>
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当我打电话给方法时,page.css('p.parent').textNokogiri会返回'有用的文字无用的文字'.但我只需要'有用的文字'.
如何获取没有子节点的节点文本?
我有一个存储在String值中的着色器:
var myShader =
" attribute vec4 a_position;" +
" void main() {" +
" gl_Position = a_position;" +
" }"
glShaderSource(shader, GLsizei(1), myShader, nil)
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函数glShaderSource具有绑定签名:
func glShaderSource(shader: GLuint, count: GLsizei, string: UnsafePointer<UnsafePointer<GLchar>>, length: UnsafePointer<GLint>)
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当我尝试直接将着色器String传递给glShaderSource时,我收到错误消息:
'String' is not convertible to 'UnsafePointer<UnsafePointer<GLchar>>'
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如何正确传递String?(xCode版本6.1(6A1052d))
我有一些实现帕累托规则的通用代码.它似乎是格式良好的代码.
关于newResult.set<Criterion>( criterion() );表达式错误的GCC 4.4编译器消息.但我找不到问题.
完整错误日志:
trunk$ g++ -std=c++0x -o test test.cpp
t6.cpp: In member function ‘bool Pareto<Minimize<T>, Types ...>::operator()(Map&, Map&)’:
t6.cpp:24: error: expected primary-expression before ‘>’ token
t6.cpp:26: error: expected primary-expression before ‘>’ token
t6.cpp:26: error: expected primary-expression before ‘)’ token
t6.cpp:26: error: expected primary-expression before ‘>’ token
t6.cpp:26: error: expected primary-expression before ‘)’ token
t6.cpp: In member function ‘bool Pareto<Maximize<T>, Types ...>::operator()(Map&, Map&)’:
t6.cpp:43: error: expected primary-expression before ‘>’ token
t6.cpp:45: error: expected primary-expression before ‘>’ …Run Code Online (Sandbox Code Playgroud) 我用 C++ 编写了一个程序,使用以下命令模拟非活动窗口上的点击:
PostMessage (z, WM_LBUTTONDOWN, 0,MAKELONG(t.left+x,t.top+y));
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但每当单击时,它就会激活窗口,并且窗口会移动到顶部。
有没有办法让窗口保持非活动状态或通过其他方式单击它?
我曾经SetWindowPos(z , HWND_BOTTOM,....)使该窗口位于 z 顺序列表的底部,但它仍然激活。
编辑:窗口是游戏机