小编Gia*_*yez的帖子

我正在尝试向phpmyadmin中的表添加值,我收到错误:'字段列表'中的未知列'...'.

这是我的代码:

    <?php   

    //preparing the patch to copy the uploaded file
    $target_path = "images/";

    //adding the name of the file, finishing the path
    $target_path = $target_path . basename( $_FILES['image']['name']); 

   //moving the file to the folder
   if(move_uploaded_file($_FILES['image']['tmp_name'], $target_path)) {
       echo "The file ".  basename( $_FILES['image']['name']). 
       " has been uploaded";
   } else{
      echo "There was an error uploading the file, please try again!";
   }

   //getting input from the form
   $name = $_POST['game'];
   $description = $_POST['beschrijving'];

   //preparing the query to insert the values …

我想在提交表单之前检查是否所有文本框都已填满,我正在执行此操作的方式是这样的:

If strGebruikersnaam <> String.Empty And strVoornaam <> String.Empty And strFamilienaam <> String.Empty And strEmail <> String.Empty And strBevestigEmail <> String.Empty And strWachtwoord <> String.Empty And strBevestigWachtwoord <> String.Empty And strAntispam <> String.Empty Then

    End If

我想知道是否有更有效的方法来做同样的事情.

我有这段代码,应该获取网站的源代码.

$homepage = file_get_contents('http://homepage.com');
echo $homepage;

而不是实际给我的源代码..它显示了我正在尝试从中获取源代码的页面.

所以我去了一个存储Feeds对象的列表,这些对象是用链接和类别构建的.现在我想删除具有特定链接和类别的给定Feeds对象.

我的清单声明:

public void addFeed(String link, String cat) {
    linkAcategory.Add(new feed(link, cat));
}

添加功能:

public void addFeed(String link, String cat) {
    linkAcategory.Add(new feed(link, cat));
}

删除功能不起作用,但显示我正在尝试做的事情:

public void removeFeed(String link, String cat) {
    linkAcategory.Remove(new feed(link,cat));
}

我希望我能得到任何有价值的帮助.提前致谢.