我收到以下错误消息:
mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given
使用以下代码:
$postTags="NO Tags";
if(!empty($id)){
$query = mysql_query($con,"SELECT tags.* FROM blog_post_tags LEFT JOIN (tags) ON (blog_post_tags.tag_id = tags.id) WHERE blog_post_tags.blog_post_id = " . $id);
$tagsArray=array();
$tagsIdArray=array();
while($row = mysqli_fetch_assoc($query)){
array_push($tagsArray, $row["name"]);
array_push($tagsIdArray, $row["id"]);
}
发布于 2018-06-04 17:07:59
尝试使用[mysqli_query]在line:3
$query = mysqli_query($con,"SELECT tags.* FROM blog_post_tags LEFT JOIN (tags) ON (blog_post_tags.tag_id = tags.id) WHERE blog_post_tags.blog_post_id = " . $id);
https://stackoverflow.com/questions/-100004736
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