Question

我正在尝试使用这些if / else if语句来显示这些php页面。 if / elseif语句允许显示php页面。数据存储在mysql中。我们如何得到它，如果它已经显示它只进入一次？谢谢。我希望你能提供帮助。对不起，这有点令人困惑。我最近刚学英语。谢谢。

if ($result_array[0] = Politics) {
        require 'news/political.php';
    } elseif ($result_array[0] = Gossip) {
        require 'news/celebgossib';
    }   elseif ($result_array[0] = Entertainment) {
            require 'news/entertainment.php';
        }   elseif ($result_array[0] = Finance) {
                require 'news/finance.php';

            }   elseif ($result_array[0] = Health) {
                    require 'news/health.php';
                }   elseif ($result_array[0] = Leisure) {
                        require 'news/leisure.php';
                    }   elseif ($result_array[0] = Sports) {
                            require 'news/sports.php';
                        }   elseif ($result_array[0] = Tech) {
                                require 'news/tech.php';
                            }   elseif ($result_array[0] = World) {
                                    require 'news/world.php';
                                } else {
                                    echo "There is no interests in your database";
                                }

                                if ($result_array[1] = Politics) {
                                    require 'news/political.php';
                                } elseif ($result_array[1] = Gossip) {
                                    require 'news/celebgossib';
                                }   elseif ($result_array[1] = Entertainment) {
                                        require 'news/entertainment.php';
                                    }   elseif ($result_array[1] = Finance) {
                                            require 'news/finance.php';

                                        }   elseif ($result_array[1] = Health) {
                                                require 'news/health.php';
                                            }   elseif ($result_array[1] = Leisure) {
                                                    require 'news/leisure.php';
                                                }   elseif ($result_array[1] = Sports) {
                                                        require 'news/sports.php';
                                                    }   elseif ($result_array[1] = Tech) {
                                                            require 'news/tech.php';
                                                        }   elseif ($result_array[1] = World) {
                                                                require 'news/world.php';
                                                            } else {
                                                                echo "There is no interests in your database";
                                                            }

Answer 1

类似的东西：

$pages = array(
  'Politics' => 'political',
  'Gossip' => 'celebgossib',
   ...
);

$used = array();

for ($i = 0; $i < 2; ++$i)
{
   if (array_key_exists($result_array[$i], $pages)
   {
      if (!array_key_exists($result_array[$i], $used))
      {
         # only display this section once
         include 'news/'.$pages[$result_array[$i]].'.php';
         $used[$result_array[$i]] = true;
      }
   }
   else
   {
      echo "Nothing to see here.";
   }
}

如果找不到页面，我不确定你想要做什么;或者后续记录是重复的。

Answer 2

您需要使用==而不是= 我不知道声明的右边是什么，例如政治。如果它不是变量，那么你应该把它放在引号中。

像这样的东西

if ($result_array[0] == "Politics") {
        require 'news/political.php';
    }else ...

Answer 3

看起来你正在迭代一系列选项，包括一堆设置文件？

我会尝试以下内容：

switch( TRUE )
{
    case in_array("Politics", $result_array):
        require 'news/political.php';
        break; 

    case in_array("Gossip", $result_array):
        require 'news/celebgossib';
        break;

    // etc.
}

Answer 4

如何使用switch语句？

switch $result_array[0] {
     case 'Politics':     include('politics.php'); break;
     case 'This': include(...); break;
     case 'That': include(....); break;
     default: include('default.php'); break;
}

对于使用简单“todo”部分的loooong if / then / else测试集，switch语句是理想的。

PHP问题：如何修复这些if / elseif语句

4 个答案: