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我是 php 新手。我正在阅读 RSS 提要并存储在我的数据库表中。为此我正在使用

$num = count($movies->channel->item);
for ($i=0; $i<=$num-1; $i++){
    $tittle= $movies->channel->item[$i]->title."<br/>";
    $link=$movies->channel->item[$i]->link."<br/>";
    $image=$movies->channel->item[$i]->medium_image_url."<br/>";
    $des=$movies->channel->item[$i]->description."<br/>";
    $city=$movies->channel->item[$i]->city;
}

如何使用 foreach 循环显示所有数据？

4

4 回答 4

1

您可以直接echo输出您的陈述结果（我也foreach根据要求将其更改为 a ）：

foreach ($movies->channel->item as $item) {
  $tittle= $item->title."<br/>";
  $link=$item->link."<br/>";
  $image=$item->medium_image_url."<br/>";
  $des=$item->description."<br/>";
  $city=$item->city;

  echo $tittle.$link.$image.$des.$city;
}
于 2011-11-03T12:01:54.237 回答
1 
 foreach($movies->channel->item as $opt){
     echo $tittle= $opt->title."<br/>";
     echo $link=$opt->link."<br/>";
     echo $image=$opt->medium_image_url."<br/>";
     echo $des=$opt->description."<br/>";
     echo $city=$opt->city;
 }
于 2011-11-03T12:01:59.220 回答
1 
foreach ($movies->channel->item as $item) {
  echo $item->title.'<br />';
  echo $item->link.'<br />';
  echo $item->medium_image_url.'<br />';
  echo $item->description.'<br />';
  echo $item->city;
}

我回应了它们，因为我认为这就是您想要做的 - 您的代码在每次迭代时重新分配变量，但对它们不做任何事情。它们每次都会被覆盖，因此它们将只保留最后一次迭代的值。

编辑

您可以通过执行以下操作轻松修改它以将行插入数据库：

foreach ($movies->channel->item as $item) {
  echo $item->title.'<br />';
  echo $item->link.'<br />';
  echo $item->medium_image_url.'<br />';
  echo $item->description.'<br />';
  echo $item->city.'<br />';

  // Build the query - change the names of your table and columns as appropriate
  $query = "INSERT INTO `tablename` 
              (`title`,`link`,`medium_image_url`,`description`,`city`)
            VALUES
              ('$item->title','$item->link','$item->medium_image_url','$item->description','$item->city')";
  // Do the query - do NOT show the output of mysql_error() in a production environment!
  if (!mysql_query($query)) echo 'Oh no! Something went wrong with the query: '.mysql_error();
}

或者，您可以将其全部转换为一个查询，以最大程度地减少数据库流量：

$rows = array();
foreach ($movies->channel->item as $item) {
  echo $item->title.'<br />';
  echo $item->link.'<br />';
  echo $item->medium_image_url.'<br />';
  echo $item->description.'<br />';
  echo $item->city.'<br />';
  // Add this entry to $rows
  $rows[] = "('$item->title','$item->link','$item->medium_image_url','$item->description','$item->city')";
}

// Build the query - change the names of your table and columns as appropriate
$query = "INSERT INTO `tablename` 
            (`title`,`link`,`medium_image_url`,`description`,`city`)
          VALUES ".implode(', ',$rows);
// Do the query - do NOT show the output of mysql_error() in a production environment!
if (!mysql_query($query)) echo 'Oh no! Something went wrong with the query: '.mysql_error();
于 2011-11-03T12:02:43.220 回答
1

这应该有效：

foreach ($movies as $movie)
    $tittle = $movie->title."<br/>";
    $link = $movies->link."<br/>";
    $image = $movie->medium_image_url."<br/>";
    $des = $movie->description."<br/>";
    $city = $movie->city;
}
于 2011-11-03T12:02:57.613 回答