When I copy the words with double quotes, .php pages show an error. Double quotes don't get copied normally. How can I solve it? Which codes can I edit? (My script is Wordpress.) Example: $goster = getenv(“HTTP_USER_AGENT”); Before HTTP and after AGENT double quotes in example above. It is a problem with my users. I think I must edit in blockquotes function in Wordpress. Any idea?
当我用双引号复制单词时，.php页面显示错误。 双引号不会正常复制。 我该如何解决它？ 我可以编辑哪些代码？ （我的脚本是Wordpress。） 例如： $goster = getenv(“HTTP_USER_AGENT”); 在上面的例子中，在HTTP和AGENT之后加双引号。 这对我的用户来说是一个问题。 我想我必须在Wordpress的blockquotes函数中进行编辑。 任何想法？ 这些是聪明的引号（ e2 80 9c ），并不被PHP词法分析器视为字符串分隔符，这似乎你
I have a PHP application with me, which was done by myself and a few of us. I have not coded much, but it worked well in the localhost . When I tried to upload it in our university web server, I had got this error. Parse error unexpected : This happened on this line. So I believe that PHP has to do something with respect to the previous line too. So I am adding the previous and next lines
我有一个PHP应用程序，这是我自己和我们中的一些人完成的。 我没有编码太多，但在localhost运行良好。 当我试图在我们的大学Web服务器上传它时，我得到了这个错误。 意外解析错误: 这发生在这条线上。 所以我相信PHP也必须对前一行做些什么。 所以我添加了上一行和下一行： <?php session_start(); $page = $_GET["page"] ?: "index"; // Error in this line! 有趣的是，这在我的WAMP服务器本地工作，但它不适
In my PHP code I have a variable $message which includes the message to be sent to me which has post variables in it. It is supposed to have a semicolon at the end...but it gives me an error saying it is unexpected but I know I need it because it wont work without it. I am at a complete loss. Hopefully someone here can help me. Error Message: PHP Parse error: syntax error, unexpected ';'
在我的PHP代码中，我有一个变量$message ，其中包含要发送给我的消息，其中包含post变量。 它应该在最后有一个分号......但它给了我一个错误，说它是意想不到的，但我知道我需要它，因为如果没有它，它就无法工作。 我完全丧失了。 希望这里有人能帮助我。 错误信息： PHP Parse error: syntax error, unexpected ';' PHP代码 if(!empty($_POST["name"]) && !empty($_POST["address"]) && !empty($_POST
This thing is bugging me a lot. I'm getting Parse error: syntax error, unexpected '.', expecting ',' or ';' at this line public static $user_table = TABLE_PREFIX . 'users'; TABLE_PREFIX is a constant created by define function Static class properties are initialized at compile time. You cannot use a constant TABLE_PREFIX to concatenate with a string literal when
这东西让我感到很烦。 我得到Parse错误：语法错误，意外的'。'，期待'，'或';' 在这条线上 public static $user_table = TABLE_PREFIX . 'users'; TABLE_PREFIX是由定义函数创建的常量 静态类属性在编译时被初始化。 在初始化静态类属性时，不能使用常量TABLE_PREFIX与字符串文字进行连接，因为直到运行时才会知道该常量的值。 相反，在构造函数中初始化它： public static $user_table; // In
Well, basically I am working on a register and login tutorial on youtube. Which is using the old version of PHP, and I have attempted to update the code, however I get this error. Parse error: syntax error, unexpected ',' in C:Program Files (x86)EasyPHP-DevServer-14.1VC11datalocalwebprojectsForumforumcorefunctionsusers.php on line 23 users.php <?php function user_exists($username
那么，基本上我正在研究YouTube上的注册和登录教程。 这是使用旧版本的PHP，我试图更新代码，但是我得到这个错误。 解析错误：语法错误，在C： Program Files（x86） EasyPHP-DevServer-14.1VC11 data localweb projects Forum forum core functions users.php中出现意外的' users.php <?php function user_exists($username, $con) { $data = $username; $username = sanitize($data, $con);
im quite new on this php codings and really would like to see if is possible to get help here.. Below is a part of the coding since apparently the error is only on line 6 <?php class DBWrapper { function DBWrapper($server,$db,$user,$pass) { $this->Server = $server; $this->DB = $db; $this->User = $user; $this->Password = $pass;
即时通讯相当新的这个PHP编码，真的想看看是否有可能得到帮助在这里.. 以下是编码的一部分，因为显然错误仅在第6行 <?php class DBWrapper { function DBWrapper($server,$db,$user,$pass) { $this->Server = $server; $this->DB = $db; $this->User = $user; $this->Password = $pass; mysql_connect($this->Server, $this->User, $this-&
class Employee { public static $favSport = "Football"; public static function watchTV() { echo "Watching ".static::$favSport; } } class Executive extends Employee { public static $favSport = "Polo"; } echo Executive::watchTV(); Parse error: syntax error, unexpected T_STATIC on line 7 Why do I get parse error & and how to fix it? Thanks! The parse error he
class Employee { public static $favSport = "Football"; public static function watchTV() { echo "Watching ".static::$favSport; } } class Executive extends Employee { public static $favSport = "Polo"; } echo Executive::watchTV(); 解析错误：语法错误，第7行出现意外的T_STATIC 为什么我会得到解析错误？以及如何解决它？ 谢谢！ 解析错误在这里： echo "Watching ".static
My client says he is getting this error using my script: Warning: Unexpected character in input: '' (ASCII=92) state=1 in /path/to//header.php on line 34 Parse error: syntax error, unexpected T_STRING, expecting T_CONSTANT_ENCAPSED_STRING or '(' in/path/to/header.php on line 34 The line 34 in header.php is just use MainClass; Now, I told him he has to have PHP >= 5.3.0 and he says his PH
我的客户说他使用我的脚本得到这个错误： Warning: Unexpected character in input: '' (ASCII=92) state=1 in /path/to//header.php on line 34 Parse error: syntax error, unexpected T_STRING, expecting T_CONSTANT_ENCAPSED_STRING or '(' in/path/to/header.php on line 34 header.php的第34行只是use MainClass; 现在，我告诉他他必须有PHP >= 5.3.0 ，他说他的PHP版本是5.3.24 可能是什么问题呢？ 编辑：
I have this line in PHP: $bom != b"xEFxBBxBF" When I run it, I get the error: Parse error: syntax error, unexpected T_NS_SEPARATOR in C:xampphtdocsMediaAlbumWebUtilsUtils.php on line 218 What is the T_NS_SEPARATOR in php and why is it unexpected? You likely have an unclosed single or double quote above that line in your code. What is the b that's outside of the quotes? If it's
我在PHP中有这一行： $bom != b"xEFxBBxBF" 当我运行它时，出现错误： Parse error: syntax error, unexpected T_NS_SEPARATOR in C:xampphtdocsMediaAlbumWebUtilsUtils.php on line 218 什么是在PHP中的T_NS_SEPARATOR和它为什么是意外的？ 您的代码中可能有一个未封闭的单引号或双引号。 引号之外的b是什么？ 如果这是一个比较，它可能是这样的： if($bom != "bxEFxBBxBF") { //code } 在PHP中重现此错误的简单
$command = $_POST['cmd']; $args = $_POST['args']; if($args == !empty && $command != 'reload'){ } 谢谢你，让它工作！ 空是一个功能。 $command = $_POST['cmd']; $args = $_POST['args']; if(!empty($args) && $command != 'reload'){ } empty()是构造，你应该必须使用它，如： if(!empty($args) && $command!='reload'){ } 检查条件： if(!empty($args) && $command != 'reload')
$command = $_POST['cmd']; $args = $_POST['args']; if($args == !empty && $command != 'reload'){ } 谢谢你，让它工作！ 空是一个功能。 $command = $_POST['cmd']; $args = $_POST['args']; if(!empty($args) && $command != 'reload'){ } empty()是构造，你应该必须使用它，如： if(!empty($args) && $command!='reload'){ } 检查条件： if(!empty($args) && $command != 'reload')