This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers I hate seeing people use array_push - I know it's legal. In this case, you can't push a key => value to your array, just do this instead: $datax['mem'] = $str; Manual: http://php.net/manual/en/function.array-push.php edit If you insist on using the array_push type method, y
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 我讨厌看到人们使用array_push - 我知道这是合法的。 在这种情况下，你不能将一个key => value传递给你的数组，只需做到这一点： $datax['mem'] = $str; 手册：http://php.net/manual/en/function.array-push.php 编辑 如果你坚持使用array_push类型的方法，你需要用你的新的键值对创建一个新的数组，然后使用array_merge来加入它
This question already has an answer here: PHP parse/syntax errors; and how to solve them? 13 answers Using class as a name of a constant is available in PHP 5.5 only. To get the class name you can replace VisiteMaisonInvestisseur::class with get_class(new VisiteMaisonInvestisseur) . Or change the name of the constant. For example: VisiteMaisonInvestisseur::class_name . The problem is
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 仅在PHP 5.5中提供使用class作为常量的名称。 要获得类名称，您可以用get_class(new VisiteMaisonInvestisseur)替换VisiteMaisonInvestisseur::class 。 或者改变常量的名字。 例如： VisiteMaisonInvestisseur::class_name 。 问题出在VisitMaisonInvestisseur::class 。 class是PHP中的一个保留字，因此您不能将其用作常量的名称。
How can I add a condition inside a php array? Here is the array $content['custom_fields'] = array( array( "key" => "_yoast_wpseo_focuskw", "value" => $_POST["title"] ), array( "key" => "_yoast_wpseo_metadesc", "value" => $_POST["titleenfa"] ), array( "key" => "_yoast_wpseo_metakeywords", "value" => $_POST["metakey"] ), if($_POST["link128"]){ array( "ke
我如何在php数组中添加一个条件？ 这是数组 $content['custom_fields'] = array( array( "key" => "_yoast_wpseo_focuskw", "value" => $_POST["title"] ), array( "key" => "_yoast_wpseo_metadesc", "value" => $_POST["titleenfa"] ), array( "key" => "_yoast_wpseo_metakeywords", "value" => $_POST["metakey"] ), if($_POST["link128"]){ array( "key" => "_link128", "v
Running this MySQL query in PHP yields a syntax error I can't solve. function queryTest(array $dataArray) $query = "Insert into users (email,fullName,password, accessLevel) values (($dataArray['formEmail']),($dataArray['formFullName']),($dataArray['formPassword']),($dataArray['formAccessLevel']))"; if( $this->mysqli->real_quer
在PHP中运行这个MySQL查询会产生一个我无法解决的语法错误。 function queryTest(array $dataArray) $query = "Insert into users (email,fullName,password, accessLevel) values (($dataArray['formEmail']),($dataArray['formFullName']),($dataArray['formPassword']),($dataArray['formAccessLevel']))"; if( $this->mysqli->real_query($query)) == FA
When I run this it says "PHP Syntax Check: Parse error: syntax error, unexpected '}' in your code on line 12 - if(mail) {$success == true}; PHP $to = "Admin@mywebsite.com"; $subject = $_POST["Subject"]; $message = 'Hello World!'; $headers = 'From: User@Yourwebsite.com' . "rn" . 'Reply-To: ' . $_POST["Email_From"] . "rn" . 'X-Mailer: PHP/' . phpversion(); mail($to, $subje
当我运行它时，它会在第12行的代码中显示“PHP Syntax Check：Parse error：syntax error，unexpected'}' if（mail）{$ success == true}; PHP $to = "Admin@mywebsite.com"; $subject = $_POST["Subject"]; $message = 'Hello World!'; $headers = 'From: User@Yourwebsite.com' . "rn" . 'Reply-To: ' . $_POST["Email_From"] . "rn" . 'X-Mailer: PHP/' . phpversion(); mail($to, $subject, $message, $h
I'am getting the following error: "PHP Syntax Check: Parse error: syntax error, unexpected 'qtip' (T_STRING), expecting ')' in your code on line 14" in this code: <?php // Always use wp_enqueue_scripts action hook to both enqueue and register scripts add_action( 'wp_enqueue_scripts', 'enqueue_and_register_qtip_scripts_and_style' ); function enqueue_and_register_
我得到以下错误：“PHP语法检查：解析错误：语法错误，意外的'qtip'（T_STRING），期待'）'在您的代码在第14行”在此代码中： <?php // Always use wp_enqueue_scripts action hook to both enqueue and register scripts add_action( 'wp_enqueue_scripts', 'enqueue_and_register_qtip_scripts_and_style' ); function enqueue_and_register_qtip_scripts_and_style(){ // Use `get_stylesheet_directo
i get this error when using my php code PHP Syntax Check: Parse error: syntax error, unexpected '}' in your code on line 7 please help me i dont understand what ive done wrong here php code : <?php if ( $ready == "false" ) { $rand = chr(rand(97, 122)). chr(rand(97, 122)). chr(rand(97, 122)); $folder = "uploads/".$rand } if ( !is_dir($folder) ) { $ready = "true" } if ( i
我使用我的PHP代码时出现此错误 PHP语法检查：解析错误：语法错误，在第7行代码中出现意外的'}' 请帮助我，我不明白什么是做错了这里的PHP代码： <?php if ( $ready == "false" ) { $rand = chr(rand(97, 122)). chr(rand(97, 122)). chr(rand(97, 122)); $folder = "uploads/".$rand } if ( !is_dir($folder) ) { $ready = "true" } if ( is_dir($folder) ) { $ready = "false" } if ( $ready =
For some reason, PHP is throwing the error on my web server but not locally: Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 16 in /home/a8576801/public_html/indev/core/functions/users.php on line 79 Now, the function in question at line 79 goes as such: function user_id_from_username($username) { $username = sanitize($username); return mys
出于某种原因，PHP将错误发布到我的Web服务器上，但不是本地的： Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 16 in /home/a8576801/public_html/indev/core/functions/users.php on line 79 现在，第79行中提到的函数如下所示： function user_id_from_username($username) { $username = sanitize($username); return mysql_result(mysql_query("SELECT `user_
This question already has an answer here: Coalesce function for PHP? 9 answers PHP parse/syntax errors; and how to solve them? 13 answers 你应该使用： $id = $_GET['id']?$_GET['id']:null;
这个问题在这里已经有了答案： 用于PHP的合并功能？ 9个答案 PHP解析/语法错误; 以及如何解决它们？ 13个答案 你应该使用： $id = $_GET['id']?$_GET['id']:null;
Hello im writing some PHP and HTML but im not sure whats going on an error keeps coming up saying: Parse error: syntax error, unexpected 'else' (T_ELSE) in C:wampwwwadminindex.php on line 19" here is my code <?php session_start(); if(!isset($_SESSION['user_id'])){ header('Location: login.php'); exit(); } include ('../includes/db_connect.php'); $post_count = $db->qu
你好，我写一些PHP和HTML，但我不知道什么是一个错误继续说： 解析错误：语法错误，在第19行的C： wamp www admin index.php中出现意外'else'（T_ELSE） 这是我的代码 <?php session_start(); if(!isset($_SESSION['user_id'])){ header('Location: login.php'); exit(); } include ('../includes/db_connect.php'); $post_count = $db->query("SELECT * FROM posts"); $comment_count = $db->q