class book {
function __construct() {
$conn = mysql_connect('localhost', 'root', '') or die('can nit connect to DB');
mysql_select_db('atomic_project', $conn) or die('can not connect to db');
}
public function listView() {
$allData = array();
$query = "SELECT * FROM `book`";
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
print_r($row);
$allData [] = $row;
}
return $allData;
}
}
我的索引页
$listViewObj = new book();
$allData = $listViewObj->listView();
echo "<pre>";
print_r($allData);
echo "<pre>";
这是我的代码和表格 here is my table data ,我可以将数据插入表中,但没有找到行 我不明白为什么没有显示数据,请帮助我
最佳答案
<?php
class book {
public function listView() {
$conn = mysql_connect('localhost', 'root', '') or die('can nit connect to DB');
mysql_select_db('atomic_project', $conn) or die('can not connect to db');
$allData = array();
$query = "SELECT * FROM `book`";
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
//echo "<pre>";
//print_r($row);
$allData [] = $row;
}
return $allData;
}
}
book::listView(); //scope resolution operator
?>
或通过对象调用函数
$obj = new book();
$obj->listView();
关于php - 使用Php mysql查询不从php类获取数据,尽管有数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35957173/