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这条线怎么了?Php连接错误_Php_String_Codeigniter_Concatenation - Fatal编程技术网

这条线怎么了?Php连接错误

这条线怎么了?Php连接错误,php,string,codeigniter,concatenation,Php,String,Codeigniter,Concatenation,我在这行有错误: $previous_next = '<div class="previous">POPRZEDNI WPIS:<h4><a href="#">'.<?= $previous_row->title ?>.'</a></h4><br></div>'; $previous_next='POPRZEDNI wpi:'; 错误是 An uncaught Exception was e

我在这行有错误:

$previous_next = '<div class="previous">POPRZEDNI WPIS:<h4><a href="#">'.<?= $previous_row->title ?>.'</a></h4><br></div>';
$previous_next='POPRZEDNI wpi:
';
错误是

An uncaught Exception was encountered

Type: ParseError

Message: syntax error, unexpected '<'

Filename: E:\XAMPP\htdocs\projekt2\application\models\Devloger.php

Line Number: 68

Backtrace:

File: E:\XAMPP\htdocs\projekt2\application\controllers\Main.php
Line: 9
Function: model

File: E:\XAMPP\htdocs\projekt2\index.php
Line: 315
Function: require_once
遇到未捕获的异常
类型:ParseError
消息:语法错误,意外“请尝试此操作

$previous_next = '<div class="previous">POPRZEDNI WPIS:<h4><a href="#">'.$previous_row->title.'</a></h4><br></div>';
$previous_next='POPRZEDNI wpi:
';
应该是
$previous\u row->title
您没有使用
添加到上面的注释中;抛出错误是因为您已经在php标记中,当您已经在一对内时,不需要再添加一组
。只需删除php标记即可