小编Mur*_*san的帖子

我已将图像上传到文件夹中并将路径存储到 MySQL 数据库中。路径已存储，图像已成功插入文件夹。但我的问题是当我显示存储在数据库中的路径中的图像时。它没有显示。当我回显图像路径时，它会显示图像路径。我检查了浏览器设置，一切正常。这是我的代码。请任何人帮忙。

$up=move_uploaded_file($_FILES['profile']['tmp_name'],dirname($_SERVER['DOCUMENT_ROOT']).'/htdocs/upload/image/'.$name);
$path = dirname($_SERVER['DOCUMENT_ROOT']).'/htdocs/upload/image/';
$location = $path . $_FILES['profile']['name'];
$ins=mysqli_query($con, "INSERT into image (url) values ('$location')");
echo 'image uploaded and stored';
echo "$location"; //It displays D:/xampp/htdocs/upload/image/Chrysanthemum.jpg
echo '<img width="250" height="250" src= "'.$location.'"/>';//It doesn't display anything.

Laravel提供了各种内置标签,可以轻松安全地处理HTML表单.HTML的所有主要元素都是使用Laravel生成的.所以我将采取这种设施,但它不起作用.

我使用composer require laravelcollective/html.

命令提示符说:

laravel>composer require laravelcollective/html
Using version ^5.2 for laravelcollective/html
./composer.json has been updated
Loading composer repositories with package information
Updating dependencies (including require-dev)
Nothing to install or update

包装照明/ html被放弃,你应该避免使用它.请改用laravelcollective/html.

Generating autoload files
> Illuminate\Foundation\ComposerScripts::postUpdate
> php artisan optimize
Generating optimized class loader

我正在尝试将一些数据存储到数组中.但是,我创建的公共数组在内部变为null each().怎么纠正？

public $questions = array();

public function crawl(){
    $questions = array();
    $client = new Client();
    $crawler = $client->request('GET', 'https://indiabusinessquiz.com/');
    $crawler->filter('p > strong')->each(function ($node) {       
      $ques = $node->text();
      echo $ques;
      var_dump($ques);
      array_push($questions, $ques);
    });
}

错误: Crawler.php第20行中的ErrorException:array_push()期望参数1为数组,给定null

我需要帮助在 PHP 中减去 2 次。

例子：

$date1 = 02:40;
$date2 = 00:00;

$finaldate = $date1 - $date2;

正确答案是 21:20。