I'm currently trying to write a query that will return all users that have not logged time for a given date in our timesheet system. We currently have 2 tables, timesheets and users. I am trying to make a query that will return a list of users that do not have an entry in the timesheets table for a date range. There is only one record in the timesheets table per-day, this should be simple
我目前正在尝试编写一个查询，它将返回在我们的时间表系统中没有记录给定日期的所有用户。 我们目前有2个表格，时间表和用户。 我正在尝试进行查询，以返回日期范围的时间表中没有条目的用户列表。 每天在时间表中只有一条记录，这应该很简单，但我无法弄清楚如何在我的生活中接近这一点。 任何帮助，将不胜感激 ：）。 +-----------------------+------------------+------+-----+---------+----------------+ | Field
I have a testcode.php file need to encode: <?php $hello = "Hello World!"; ?> And I created file encode.php to encrypt and test that file: <?php /* read the PHP source code */ $source_code = file_get_contents("testcode.php"); /* create the encrypted version */ $redistributable_key = blenc_encrypt($source_code, "encrypt.php"); /* read which is the key_file */
我有一个需要编码的testcode.php文件： <?php $hello = "Hello World!"; ?> 我创建了文件encode.php来加密和测试该文件： <?php /* read the PHP source code */ $source_code = file_get_contents("testcode.php"); /* create the encrypted version */ $redistributable_key = blenc_encrypt($source_code, "encrypt.php"); /* read which is the key_file */ $key_file = ini_get('
The IEEE-754 floating point standard says: Four mutually exclusive relations are possible: less than, equal, greater than, and unordered. The last case arises when at least one operand is NaN. Every NaN shall compare unordered with everything, including itself. And yet (codepad here): <?php echo phpversion() . " " . zend_version() . " " . php_uname() . "n"; // 5.2.5 2.2.0 Linux 2cf38fb
IEEE-754浮点标准说： 四种相互排斥的关系是可能的：小于，等于，大于和无序。 最后一种情况出现在至少一个操作数是NaN的情况下。 每个NaN都会将无序与包括它自己在内的所有事物进行比较。 然后（键盘在这里）： <?php echo phpversion() . " " . zend_version() . " " . php_uname() . "n"; // 5.2.5 2.2.0 Linux 2cf38fbc9b9e 3.11.0-15-generic #25-Ubuntu SMP // Thu Jan 30 17:22:01 UTC 2014 x86_64 NAN < N
I am using Google Chart in one my Program, The program will run locally and have no internet access to use Google Charts. Can you please tell me where can i download the Library, I searched on net but failed to find out any possible way to download it? Thanks You can download the google chart API from the following link: https://www.google.com/jsapi Just hit the URL on your browser and s
我在一个程序中使用Google图表，该程序将在本地运行，并且无法通过互联网访问Google图表。 你能告诉我在哪里可以下载图书馆，我在网上搜索，但没有找到任何可能的方式来下载它？ 谢谢 您可以从以下链接下载谷歌图表API： https://www.google.com/jsapi 只需点击浏览器中的URL并使用Ctrl + S进行保存即可。 并将其包含在您的项目中。 有关更多信息，请参阅此处了解离线图表API。 编辑： 这个库使用了两个在线库，
When an array is passed as an argument to a method or function is it passed by reference? What about doing this: $a = array(1,2,3); $b = $a; Is $b a reference to $a ? For the second part of your question, see the array page of the manual, which states (quoting) : Array assignment always involves value copying. Use the reference operator to copy an array by reference. And the given exam
当数组作为参数传递给方法或函数时，它是通过引用传递的？ 那么做什么： $a = array(1,2,3); $b = $a; 为$b的参考$a ？ 对于问题的第二部分，请参阅手册的数组页面，其中指出（引用）： 数组分配始终涉及值复制。 使用引用运算符通过引用复制数组。 举个例子： <?php $arr1 = array(2, 3); $arr2 = $arr1; $arr2[] = 4; // $arr2 is changed, // $arr1 is still array(2, 3) $arr3 = &$arr1; $a
Possible Duplicates: When to pass-by-reference in PHP What's the difference between passing by reference vs. passing by value? In PHP by default we have pass by value.I am wondering what would be good situations for passing values by reference. If you want to "return" more than one variable it's helpful. Functions like preg_match use it. Populating a Combobox, for exa
可能重复： 何时通过PHP引用 传递引用与价值传递之间有什么区别？ 在PHP默认情况下，我们通过value.I想知道什么是通过引用传递值的良好情况。 如果你想“返回”多个变量，这是有帮助的。 像preg_match这样的函数使用它。 例如，填充组合框。 使用控件或主范围中的对象执行某些操作，但是从另一个层执行某些操作。 从数据层获取数据，并通过引用填充ComboBox，而不是检索主范围（页面）中的值列表并在其中填充ComboB
I'm trying to create a randomized string in PHP, and I get absolutely no output with this: <?php function RandomString() { $characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'; $randstring = ''; for ($i = 0; $i < 10; $i++) { $randstring = $characters[rand(0, strlen($characters))]; } return $randstring; } RandomString(); echo $randst
我想在PHP中创建一个随机化的字符串，并且我完全没有输出： <?php function RandomString() { $characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'; $randstring = ''; for ($i = 0; $i < 10; $i++) { $randstring = $characters[rand(0, strlen($characters))]; } return $randstring; } RandomString(); echo $randstring; 我究竟做错了什么？ 具体回答这
Our client had given me a REST API to which I need to make a PHP call to. But as a matter of fact the documentation given with the API is very limited, so I don't really know how to call the service. I've tried to Google it, but the only thing that came up was an already expired Yahoo! tutorial on how to call the service. Not mentioning the headers or anything in depth information.
我们的客户给了我一个REST API，我需要给它打个PHP电话。 但事实上API提供的文档非常有限，所以我不知道如何调用该服务。 我试图谷歌它，但唯一出现的是一个已经过期的雅虎！ 如何调用服务的教程。 不提及标题或任何深度信息。 有没有关于如何调用REST API的一些体面的信息，或者有关它的一些文档？ 因为即使在W3schools，他们也只描述SOAP方法。 在PHP中制作其他API有什么不同的选择？ 您可以使用PHP cURL Extension
My application routinely connects to a third-party server to fetch data via SOAP/WSDL: $this->soap_client = new SoapClient("https://[the-domain]:443/[path]?wsdl", array( 'trace'=>1, 'login'=>$this->username, 'password'=>$this->password, 'exceptions' => true, 'cache_wsdl' => WSDL_CACHE_NONE ) Everything has been great for the last year, but they recent
我的应用程序通常连接到第三方服务器以通过SOAP / WSDL获取数据： $this->soap_client = new SoapClient("https://[the-domain]:443/[path]?wsdl", array( 'trace'=>1, 'login'=>$this->username, 'password'=>$this->password, 'exceptions' => true, 'cache_wsdl' => WSDL_CACHE_NONE ) 去年一切都很好，但最近他们更新了他们的WSDL文件，现在当应用程序尝试连接时， 我得到以
Possible Duplicate: How to extract a file extension in PHP? I wish to get the file extension of an image I am uploading, but I just get an array back. $userfile_name = $_FILES['image']['name']; $userfile_extn = explode(".", strtolower($_FILES['image']['name'])); Is there a way to just get the extension itself? No need to use string functions. You can use something that's actually des
可能重复： 如何在PHP中提取文件扩展名？ 我希望得到我正在上传的图像的文件扩展名，但我只是得到一个数组。 $userfile_name = $_FILES['image']['name']; $userfile_extn = explode(".", strtolower($_FILES['image']['name'])); 有没有办法只是得到扩展本身？ 不需要使用字符串函数。 你可以使用实际设计的东西： pathinfo() ： $path = $_FILES['image']['name']; $ext = pathinfo($path, PATHINFO_EXTENSION); 这也