This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers Review $googleadsensecode ="a""script Maybe $googleadsensecode ="script
这个问题在这里已经有了答案： PHP解析/语法错误; 以及如何解决它们？ 13个答案 评论 $googleadsensecode ="a""script 也许 $googleadsensecode ="script
As I try to load a website the following error shows up: unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' It says that the error is on line 5. <?php ob_start(); header("Location: /localhost/site1"); echo '<meta http-equiv="refresh" content="0;URL='/localhost/site1'">'; //line 5 die(); ?> Update: At this point my code looks like: <?php ob_
当我尝试加载网站时，出现以下错误： 意外的T_CONSTANT_ENCAPSED_STRING，期待'，'或';' 它说错误在第5行。 <?php ob_start(); header("Location: /localhost/site1"); echo '<meta http-equiv="refresh" content="0;URL='/localhost/site1'">'; //line 5 die(); ?> 更新： 在这一点上，我的代码如下所示： <?php ob_start(); header("Location: /localhost/site1"); echo '&l
Notice: Undefined index: picture in C:xampphtdocspage1.php on line 33 Warning: pathinfo() expects parameter 2 to be long, string given in C:xampphtdocspage1.php on line 35 Sorry, only JPG, JPEG, PNG & GIF files are allowed.Sorry, your file was not uploaded. User Logged in //<?PHP //session_start(); //if (!isset($_SESSION['login_user'])) { //header ("Location: page2.php");/ //} //?&
注意：未定义的索引：第33行的C： xampp htdocs page1.php中的图片 警告：pathinfo（）期望参数2很长，字符串在第35行的C： xampp htdocs page1.php中给出。对不起，只有JPG，JPEG，PNG和GIF文件是允许的。 用户登录 //<?PHP //session_start(); //if (!isset($_SESSION['login_user'])) { //header ("Location: page2.php");/ //} //?> <html> <head> <title>Basic Login Scr
I am building an web app using PHP.I am creating a file upload section in it but i am having problem. <form action="upload.php" method="post" enctype="multipart/form-data"> Select image to upload: <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> upload.php: <?php $target_dir = "images/"; $targe
我正在使用PHP构建一个Web应用程序。我正在创建一个文件上传部分，但我遇到了问题。 <form action="upload.php" method="post" enctype="multipart/form-data"> Select image to upload: <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> upload.php的： <?php $target_dir = "images/"; $target_file = $target_dir
I made a nice simple userscript: When I browse the web, I can "bookmark" any image in 1 click My userscript Grab the img src Grab the url of the webpage Copy the .jpg .png .gif to my server Everything works perfectly, BUT in some cases, the script cannot copy the file... Actually the file is created but do not contains the img data, it only contains the content of an error
我做了一个很好的简单的userscript： 当我浏览网页时，我可以通过单击“书签”任何图像 我的用户 抓住img src 抓取网页的网址 将.jpg .png .gif复制到我的服务器上 一切正常，但在某些情况下，脚本无法复制文件... 实际上文件已创建，但不包含img数据，它只包含错误网页的内容： <!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN"> <html><head> <title>403 Forbidden</title> <
I've taken a direct clone of the User Profile plug-in for my Joomla 2.5.9 install. I've renamed the plugin and the files accordingly to 'profiletest' similar to the old 1.6 tutorial. I've added a new input to the form and everything works on the backend and the new entry shows up as expected in the registration form on the front end. However when you register I never see
我已经为我的Joomla 2.5.9安装直接克隆了User Profile插件。 我已经将插件和文件重命名为'profiletest'，类似于旧版1.6 教程。 我为表单添加了新的输入，并且所有内容都在后端工作，并且新条目按预期在前端的注册表单中显示。 但是，当您注册时，我从未看到#__user_profiles表已更新。 这里有很多代码，但它是用户配置文件插件（/ plugins / user / profile /）的副本。 这里是onUserAfterSave函数的profiletest
I am very new to PHP and have no idea why this is happening, I have looked at other online items, however I just cannot seem to see why I am getting this error. if (get_home_page() != HOMEPAGE_SITE) { // Redirect logged-in users to My Moodle overview if required. $redirect = optional_param('redirect', 1, PARAM_BOOL); if (optional_param('setdefaulthome', false, PARAM_BOOL)) { set_user_prefere
我对PHP很陌生，不知道为什么会发生这种情况，我查看了其他在线项目，但是我似乎无法明白为什么我会收到此错误。 if (get_home_page() != HOMEPAGE_SITE) { // Redirect logged-in users to My Moodle overview if required. $redirect = optional_param('redirect', 1, PARAM_BOOL); if (optional_param('setdefaulthome', false, PARAM_BOOL)) { set_user_preference('user_home_page_preference', HOMEPAGE_SITE); } else
This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers You have a syntax error here : $sql = 'SELECT * FROM `student` WHERE `Index No.` = '$se''; change to $sql = 'SELECT * FROM `student` WHERE `Index No.` = "'.$se.'"'; Please have a look to : PHP Parse/Syntax Errors; and How to solve them? And, WARNING: Your code is vulnerable to SQL i
这个问题在这里已经有了答案： PHP解析/语法错误; 以及如何解决它们？ 13个答案 你在这里有一个语法错误： $sql = 'SELECT * FROM `student` WHERE `Index No.` = '$se''; 改成 $sql = 'SELECT * FROM `student` WHERE `Index No.` = "'.$se.'"'; 请看看：PHP解析/语法错误; 以及如何解决它们？ 和， 警告：您的代码容易受到SQL注入的攻击，可能会危及数据库和服务器的安全。 您应该使用PDO或mysqli API来保护您
This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers 请以这种方式尝试你的代码， <?php $con=mysqli_connect("localhost","my_user","my_password","my_db"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $sql="SELECT id, email FROM users WHERE id = 1"; $result=mysqli_query(
这个问题在这里已经有了答案： PHP解析/语法错误; 以及如何解决它们？ 13个答案 请以这种方式尝试你的代码， <?php $con=mysqli_connect("localhost","my_user","my_password","my_db"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $sql="SELECT id, email FROM users WHERE id = 1"; $result=mysqli_query($con,$sql); // Associati
Class Config{ public $levels = 10; public $points_difference = 100; public $diff_level = 3; public $timer_seconds = 60; public $maxBonus = 0; public $maxScore = 0; public $maxTotalScore = 0; public $pointsLevel = $this->points_difference * $this->diff_level; } I get Parse error: syntax error, unexpected T_VARIABLE error on the last line. An
Class Config{ public $levels = 10; public $points_difference = 100; public $diff_level = 3; public $timer_seconds = 60; public $maxBonus = 0; public $maxScore = 0; public $maxTotalScore = 0; public $pointsLevel = $this->points_difference * $this->diff_level; } 我得到Parse错误：语法错误，最后一行出现意外的T_VARIABLE错误。 有什么想法吗？