我有一个查询
$qdrop=mysqli_query($connect,"select TABLE_NAME,create_time
FROM INFORMATION_SCHEMA.TABLES
WHERE table_schema = 'purbalingga'
AND CREATE_TIME <='2016-10-07' and TABLE_NAME LIKE '2016%'");
一会儿展示数组
while ($datadrop=mysqli_fetch_array($qdrop)) {
echo $qdrop['TABLE_NAME'];echo"<br/>";
}
但是我有一个错误:
Fatal error: Cannot use object of type mysqli_result as array in E:\xampp\htdocs
最佳答案
更换:
while ($datadrop=mysqli_fetch_array($qdrop)) {
echo $qdrop['TABLE_NAME'];
echo"<br/>";
}
带有:
while ($datadrop=mysqli_fetch_array($qdrop)) {
echo $datadrop['TABLE_NAME'];
echo"<br/>";
}
$qdrop
仅保存查询,而$datadrop
是保存result in array
的查询。关于php - 如何在php中的显示表模式查询中显示数组,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/39908160/