hi i am getting an error during my execution of the code : PHP Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in C:apache2triadhtdocsimagedisplay.php on line 28 <?php $dir= "C:apache2triadhtdocsphppgadminimagesphpimages"; $file_display= array('jpg', 'jpeg', 'png', 'gif'); if(file_exists($dir)== false) { echo "directory x not found
您好，我在执行代码时出现错误：PHP解析错误：语法错误，意外的T_CONSTANT_ENCAPSED_STRING，期待'，'或';' 在第28行的C： apache2triad htdocs imagedisplay.php中 <?php $dir= "C:apache2triadhtdocsphppgadminimagesphpimages"; $file_display= array('jpg', 'jpeg', 'png', 'gif'); if(file_exists($dir)== false) { echo "directory x not found"; } else { $dir_content= scandir($dir);
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. <?php include_once('assets/libs/posicloud/posicloud.php'); $cloud=new posicloud(); $out=''; foreach ($cloud->list_templates() as $key=>$template) { $out.='<option value=''.$key.'">'.$value["name"].'; }
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。 <?php include_once('assets/libs/posicloud/posicloud.php'); $cloud=new posicloud(); $out=''; foreach ($cloud->list_templates() as $key=>$template) { $out.='<option value=''.$key.'">'.$value["name"].'; } return $out; ?> 感谢您的任何帮助！ 当你在PHP
code: public function isQuestion($query){ $questions = $this->getAllQuestions(); if (count($questions)){ foreach ($questions as $q){ if ($this->isQuestion$q($query)){ return $this->isQuestion$q($query); } } } return false; } error: Parse error: syntax error, unexpected T_VARIABLE in /App
码： public function isQuestion($query){ $questions = $this->getAllQuestions(); if (count($questions)){ foreach ($questions as $q){ if ($this->isQuestion$q($query)){ return $this->isQuestion$q($query); } } } return false; } 错误： Parse error: syntax error, unexpected T_VARIABLE in /Appl
I'm getting this error: "PHP Parse error: syntax error, unexpected T_VARIABLE in /var/www/vhosts/... on line 66" Here's my code: function combine($charArr, $k) { $currentsize = sizeof($charArr); static $combs = array(); static $originalsize = $currentsize; ###### <-- LINE 66 ###### static $firstcall = true; if ($originalsize >= $k) { # Get
我得到这个错误：“PHP解析错误：语法错误，意外的T_VARIABLE在/ var / www / vhosts / ...在线66” 这是我的代码： function combine($charArr, $k) { $currentsize = sizeof($charArr); static $combs = array(); static $originalsize = $currentsize; ###### <-- LINE 66 ###### static $firstcall = true; if ($originalsize >= $k) { # Get the First Combination $comb =
I'm trying to echo my information from my database in a simple blog. Now it just won't work. Whatever I try. I'm trying to figure it out myself but I am stuck behind a single error. php syntax error, unexpected T_VARIABLE, expecting ',' or ';' on line 29 I just can't find a solution for it.. Hope you guys can help me. I am getting pretty insane of being st
我试图在简单的博客中回复我的数据库中的信息。 现在它不会工作。 无论我尝试。 我试图弄清楚自己，但我被困在一个单一的错误后面。 PHP的语法错误，意想不到的T_VARIABLE，期待'，'或';' 在线29 我只是找不到解决方案..希望你们能帮助我。 我在这里呆了几个小时，真的很疯狂。 require('config.inc.php'); require('template.inc.php'); require('functions.inc.php'); $db_host = "***********"; $d
I can't seem to find where my code has went wrong. Here is my full error: Parse error: syntax error, unexpected T_VARIABLE in C:xampphtdocsGigaLoad.comregister.php on line 102 But I've been looking a that line for hours. I will give you the entire code cause my newbie brain cant find the error: $query = mysql_query("SELECT * FROM users WHERE username='$username'") Hope you can hel
我似乎无法找到我的代码出错的地方。 这是我的完整错误： 解析错误：语法错误，在线102上的C： xampp htdocs GigaLoad.com register.php中的意外T_VARIABLE 但我一直在寻找一条线，几个小时。 我会给你的整个代码，导致我的新手大脑找不到错误： $query = mysql_query("SELECT * FROM users WHERE username='$username'") 希望你能帮助，如果你需要更多的信息只是让我知道。 这是额外的代码 if (strstr($email, "@
While I was creating a class in php, I experienced this error: Parse error: syntax error, unexpected '[', expecting ',' or ';' on line 5 A simple example: <?php class MyClass { public $variable["attribute"] = "I'm a class property!"; } ?> I already had a look at Reference - What does this error mean in PHP? but this doesn't seem to fit to my case. The problem of all other exi
当我在php中创建一个类时，我遇到了这个错误： Parse error: syntax error, unexpected '[', expecting ',' or ';' on line 5 一个简单的例子： <?php class MyClass { public $variable["attribute"] = "I'm a class property!"; } ?> 我已经看过参考文献 - 这个错误在PHP中意味着什么？ 但这似乎不适合我的情况。 所有其他现有问题似乎都依赖于旧的PHP版本。 但我使用PHP 5.6.3！ 我能做什么？ 我只是没
I have a function that returns an array. I have another function that just returns the first row, but for some reason, it makes me use an intermediate variable, ie this fails: function f1(/*some args*/) { return /*an array*/; } function f2(/*some args*/) { return f1(/*some args*/)[0]; } . . . with: Parse error: syntax error, unexpected '[' in util.php on line 10 But, thi
我有一个函数返回一个数组。 我有另一个函数返回第一行，但由于某种原因，它使我使用一个中间变量，即失败： function f1(/*some args*/) { return /*an array*/; } function f2(/*some args*/) { return f1(/*some args*/)[0]; } 。 。 。 有： 解析错误：语法错误，在第10行的util.php中出现意外的'['' 但是，这工作： function f1(/*some args*/) { return /*an array*/; } function f2(/*
Is it possible that this PHP code line if ($this->greatestId()["num_rows"] > 0) works in PHP 5.5 and returns an error in 5.3?? PHP Parse error: syntax error, unexpected '[' in /var/www/app/AppDAO.php on line 43 How can I change it to work under PHP 5.3? Array dereferencing became available in PHP 5.4 That's why this doesn't work in PHP 5.3. So you have an extra step where yo
这可能是PHP代码行吗？ if ($this->greatestId()["num_rows"] > 0) 在PHP 5.5中工作，并在5.3返回一个错误？ PHP Parse error: syntax error, unexpected '[' in /var/www/app/AppDAO.php on line 43 我怎样才能改变它在PHP 5.3下工作？ 数组解引用在PHP 5.4中可用。这就是为什么这在PHP 5.3中不起作用。 所以你有一个额外的步骤，你需要从你的函数调用中获取数组值，然后你可以使用它： $variable = $this->gr
I am executing some queries and am getting the following the error: Parse error: syntax error, unexpected '$section2' (T_VARIABLE) on line 22 Line 22 is: $section2 = $db->prepare("INSERT INTO learning_style_scores VALUES (5,12,4)"); I don't have a clue why I am getting this, I have checked my syntax and all seems to be correct. It basically doesn't like anything after th
我正在执行一些查询，并得到以下错误： 解析错误：语法错误，第22行出现意外的'$ section2'（T_VARIABLE） 第22行是： $section2 = $db->prepare("INSERT INTO learning_style_scores VALUES (5,12,4)"); 我不知道为什么我得到这个，我检查了我的语法，似乎都是正确的。 它基本上不会在执行$section1查询后的任何事情 编辑： 我知道这很容易SQL注入，但我只是为了测试目的而这样做。 <?php session