$mysqli = new mysqli("localhost","root","", "mydb") or die("Error " . mysqli_error($con));
$books = "select * from books";
$book_results = $mysqli->query($booked_pen_query);
$booked_arr = array();
foreach($book_results as $b){
print_r($b);
array_push($booked_arr, $b);
}
预期结果:Array ([book_id] => 1 [book_name] => "abc") Array ([book_id] => 2 [book_name] => "xyz")
它适用于PHP 5.4.19,但较低版本似乎显示为空白。
问题是:如何在PHP 5.3中执行此代码工作?
答案 0 :(得分:0)
在PHP 5.4(http://us1.php.net/manual/en/class.mysqli-result.php)
中更改了此行为5.4.0添加了迭代器支持,因为mysqli_result现在实现了Traversable。
使其在PHP 5.3中运行:
while($b = $book_results->fetch_assoc()){
print_r($b);
array_push($booked_arr, $b);
}