PHP 语法错误，“意外的 T_VARIABLE"- isQuestion($query) [英] PHP syntax error, "unexpected T_VARIABLE" - isQuestion($query)

问题描述

代码:

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:

解析错误:语法错误，第 7 行/Applications/XAMPP/xamppfiles/htdocs/ai/application/models/question_model.php 中的意外 T_VARIABLE

Parse error: syntax error, unexpected T_VARIABLE in /Applications/XAMPP/xamppfiles/htdocs/ai/application/models/question_model.php on line 7

问题出现在:

if ($this->isQuestion$q($query)){

return $this->isQuestion$q($query);

我有一些函数，例如 isQuestion1、isQuestion2、isQuestion3 等...我调用另一个函数 getAllQuestions 将返回数组中所有问题的数字，例如 1,2,3,4,5....

I have some functions like isQuestion1, isQuestion2, isQuestion3, etc... and I call another function getAllQuestions that will return me all the numbers of the questions in an array like 1,2,3,4,5....

然后我使用上面的代码来检查每个函数是否是基于查询的问题.

Then I use the above code to check if each function is a question based on a query.

推荐答案

好吧，以下是无效语法:

Well, the following is invalid syntax:

if ($this->isQuestion$q($query)){

试试这个:

foreach ($questions as $q) {
    if ($result = $this->{'isQuestion' . $q}()) {
        return $result;
    }
}
return false;

这篇关于PHP 语法错误，“意外的 T_VARIABLE"- isQuestion($query)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！