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);
最佳答案
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.
}
关于php - 无法获取 Mysqli_result,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12237683/