php - 无法获取 Mysqli_result

Question

我有这个错误

Warning: mysqli_fetch_array() [function.mysqli-fetch-array]: Couldn't fetch mysqli_result in /home/fights7/public_html/include/load_more_home_posts.php on line 12

想知道我在下面的代码中做错了什么？

$articles_data = mysqli_query($mysqli,"SELECT * FROM streamdata WHERE streamitem_id < '$lastID' ORDER BY streamitem_id DESC LIMIT 10") or die(mysql_error());
while($articles_info = mysqli_fetch_array($articles_data)) {
$json = array();
$json['streamitem_id'] = $articles_info['streamitem_id'];
$json['streamitem_content'] = $articles_info['streamitem_content'];
$json['streamitem_timestamp'] = $articles_info['streamitem_timestamp'];
mysqli_free_result($articles_data);

Answer 1

马上，您似乎在获取循环中调用了 mysqli_free_result()，因此在第一次循环迭代之后，您的结果资源已关闭并释放，并且不会有更多结果可用.

while($articles_info = mysqli_fetch_array($articles_data)) {
  $json = array();
  $json['streamitem_id'] = $articles_info['streamitem_id'];
  $json['streamitem_content'] = $articles_info['streamitem_content'];
  $json['streamitem_timestamp'] = $articles_info['streamitem_timestamp'];
  // Don't do this!
  //mysqli_free_result($articles_data);
}
// If you need to, free it outside the loop
mysqli_free_result($articles_data);

我注意到您在调用 mysqli_fetch_array() 时未指定 MYSQLI_ASSOC，因此您将返回数字键和关联键。如果您使用 JSON 中的所有内容，并且使用 MYSQLI_ASSOC 或 mysqli_fetch_assoc()，则无需执行所有这些分配:

while($articles_info = mysqli_fetch_assoc($articles_data)) {
  // No need for the $json array. Just use $articles_info directly
  // if you were going to json_encode() it.
}