Can't seem to get this to work. What Im trying to achieve is check if the variable isn't longer than 68, but also not shorter than 64. However, I can't seem to achieve even that...can someone please point out what I'm doing wrong? Also, is there a way to simplify this procedure? if (!strlen($pre_session_key_1) > 68) { if (!strlen($pre_session_key_1) < 64) {
似乎无法得到这个工作。 我试图实现的是检查变量是否不超过68，但也不少于64。 然而，我似乎无法达到甚至......有人能指出我做错了什么吗？ 另外，有没有办法简化这个程序？ if (!strlen($pre_session_key_1) > 68) { if (!strlen($pre_session_key_1) < 64) { ...... } } 谢谢，请在这里和蔼可亲， 尝试： if (!(strlen($pre_session_key_1) > 68)) { if (!(strlen($pre_session_key
From the question Type-juggling and (strict) greater/lesser-than comparisons in PHP I know PHP interpret strings as numbers whenever it can. "10" < "1a" => 10 less than 1 expecting false "1a" < "2" => 1 less than 2 expecting true "10" > "2" => 10 greater than 2 expecting true But in the case of "10" < "1a" php returns true. I am
从PHP中的类型杂耍和（严格）更大/更小比较的问题 我知道PHP会在任何时候将字符串解释为数字。 "10" < "1a" => 10 less than 1 expecting false "1a" < "2" => 1 less than 2 expecting true "10" > "2" => 10 greater than 2 expecting true 但在"10" < "1a"的情况下，php返回true。 我不理解这个概念，请帮我澄清一下。 编辑： 但是当我添加"1
I have a value as a number. For instance, 502. I want to write a php if statement that will display some text if the value is lesser or greater than certain numbers, or between a range. Eg number is 502, text will say: "Between 500-600" number is 56, text will say: "Between 0-60" etc. So far I have this: <?php $count=0;?> <?php $board = getUserBoard($userDetails
我有一个数值。 例如，502.我想编写一个php if语句，如果该值小于或大于某个数字，或者在一个范围内，将显示一些文本。 例如数字是502，文字会说：“在500-600之间”的数字是56，文字会说：“在0-60之间”等等 到目前为止，我有这样的： <?php $count=0;?> <?php $board = getUserBoard($userDetails['userId']);?> <?php if(is_array($board)):?> <?php $boardCount = count($board);?> <?php for
I encountered a strange and unexpected behavior in PHP while comparing some string values. The first two statements below return true when I would expect them to return false. The last statement returns false as expected. I'm aware of PHP's Type Juggling, but what I understand from the docs is that type juggling happens when you are comparing two different data types like a string and
我在PHP中比较了一些字符串值时遇到了一个奇怪且意外的行为。 当我期望它们返回false时，下面的前两个语句返回true。 最后一条语句按预期返回错误。 我知道PHP的类型杂耍，但我从文档中了解到，类型杂耍发生在比较两种不同的数据类型（如字符串和整数）时。 在下面的例子中，虽然这两个文字都是字符串。 这是否意味着，当你在PHP中进行字符串比较时，它会检查这两个字符串，看看它们是否看起来像整数，如果是，则将这两个字
如何在PHP中检测使用哪种请求类型（GET，POST，PUT或DELETE）？ By using $_SERVER['REQUEST_METHOD'] Example if ($_SERVER['REQUEST_METHOD'] === 'POST') { // The request is using the POST method } For more details please see the documentation for the $_SERVER variable. REST in PHP can be done pretty simple. Create http://example.com/test.php (outlined below). Use this for REST calls, eg ht
如何在PHP中检测使用哪种请求类型（GET，POST，PUT或DELETE）？ 通过使用 $_SERVER['REQUEST_METHOD'] 例 if ($_SERVER['REQUEST_METHOD'] === 'POST') { // The request is using the POST method } 有关更多详细信息，请参阅$ _SERVER变量的文档。 PHP中的REST可以非常简单地完成。 创建http://example.com/test.php（如下所述）。 将其用于REST调用，例如http://example.com/test.php/testing/123/hello。 这适
在PHP中使用Heredoc有什么好处，并且你能展示一个例子吗？ The here doc syntax is much cleaner to me and it is really useful for multi-line strings and avoiding quoting issues. Back in the day I used to use them to construct SQL queries: $sql = <<<SQL select * from $tablename where id in [$order_ids_list] and product_name = "widgets" SQL; To me this has a lower probability of introd
在PHP中使用Heredoc有什么好处，并且你能展示一个例子吗？ 这里的doc语法对我来说比较干净，对于多行字符串和避免引用问题非常有用。 早在我使用它们来构建SQL查询的那一天： $sql = <<<SQL select * from $tablename where id in [$order_ids_list] and product_name = "widgets" SQL; 对我来说，引入语法错误的概率比使用引号的要低： $sql = " select * from $tablename where id in [$order_ids_list]
I would think the following piece of code should work, but it doesn't (Edited: Now works in PHP 5.5+) : if (!empty($r->getError())) Where getError() is simply: public function getError() { return $this->error; } Yet I end up with this error: can't use method return value in write context What does this mean? Isn't this just a read? empty() needs to access the valu
我会认为下面的一段代码应该可以工作，但它不会（编辑：现在在PHP 5.5+中工作） ： if (!empty($r->getError())) getError()简单来说就是： public function getError() { return $this->error; } 然而我最终发现了这个错误： 在写入上下文中不能使用方法返回值 这是什么意思？ 这不仅仅是一个阅读？ empty()需要通过引用访问值（为了检查该引用是否指向存在的东西），并且5.5之前的PHP不支持对从函数返回
Possible Duplicate: PHP - Remove last character if it's a period? Which is fastest way to remove last character from string? I have a string like a,b,c,d,e, I would like to remove last ',' and get the remaining string back OUTPUT: a,b,c,d,e What the fastest way to do this? First I try without space rtrim($arraynama,","); and get error result. Then I add a spac
可能重复： PHP - 删除最后一个字符，如果它是一个时期？ 从字符串中删除最后一个字符的最快方法是哪种？ 我有一个字符串 a,b,c,d,e, 我想删除最后一个'，'并获取剩余的字符串 OUTPUT: a,b,c,d,e 什么最快的方式来做到这一点？ 首先，我尝试没有空间rtrim($arraynama,","); 并得到错误结果。 然后我添加一个空格并获得好结果： $newarraynama=rtrim($arraynama,", "); 你可以使用su
What does PHP syntax $var1[] = $var2 mean? Note the [] after $var1 varable. It means that $var1 is an array, and this syntax means you are inserting $var2 into a new element at the end of the array. So if your array had 2 elements before, like this: $var1=( 1 => 3, 2 => 4) and the value of $var2 was 5, it would not look like this: $var1=( 1 => 3, 2 => 4, 3 => 5) It also m
PHP语法$var1[] = $var2是什么意思？ 注意$var1 varable之后的[] 。 这意味着$var1是一个数组，并且这个语法意味着您要将$var2插入到数组末尾的新元素中。 所以如果你的数组之前有2个元素，就像这样： $var1=( 1 => 3, 2 => 4) $ var2的值是5，它看起来不是这样的： $var1=( 1 => 3, 2 => 4, 3 => 5) 这也意味着，如果$ var2本身就是一个数组，那么您刚刚创建了一个二维数组。 假设如下： $var1=( 1 =
以下语法的解释是什么？ $var1->$var2 // Note the second $ You are calling a property on $var1 that is named the same as the value of $var2 . For example: $var2 = "name"; // The following are equivalent $var1->name; $var1->$var2; $var1 is an object. $var2 is (possibly) the name of a variable inside $var1 . If $var2="test"; this is evaluated to: $var1->test; You can
以下语法的解释是什么？ $var1->$var2 // Note the second $ 您正在调用与$var2的值相同的$var1属性。 例如： $var2 = "name"; // The following are equivalent $var1->name; $var1->$var2; $var1是一个对象。 $var2 （可能）是$var1的一个变量的名称。 如果$var2="test"; 这被评估为： $var1->test; 你可以用各种各样的东西做到这一点： $test = array(); $name="test"; print_r($$name); /