如何将mysqli_query结果转换为int?

时间:2018-02-09 04:16:16

标签: php mysql database mysqli

目前,我正在开发一个移动图书馆应用,需要查看图书。我希望能够检查与可用总数相比检出的书籍数量。我正在努力将mysqli_query结果转换为整数,以便与可用的总书数进行比较。如果这是一个简单的答案我很抱歉(我非常新),因为我做了很多挖掘,无法找到一个我能适应我的情景的答案。

<?php
require "conn.php";

$userID = "2";
$book_id = "1";

$mysql_qry = "select COUNT(*) from books_checked_out where bookID=$book_id";
$result = mysqli_query($conn, $mysql_qry);

//Somewhere here convert the $result into $quantity, which would be an int

if($quantity >= 2) {
    echo "more books checked out than in stock";
}
else if($quantity < 2) {
    echo "less books checked out than in stock";
}
?>

感谢任何帮助,谢谢!

4 个答案:

答案 0 :(得分:1)

这将有效

$result = $result->fetch_array();
$quantity = intval($result[0]);

答案 1 :(得分:0)

你可以这样做:
    

$userID = "2";
$book_id = "1";

$mysql_qry = "select COUNT(*) as quantity from books_checked_out where bookID=$book_id";
$result = mysqli_query($conn, $mysql_qry);

$row = mysqli_fetch_array($result, 'MYSQLI_ASSOC'); // Use something like this to get the result
$quantity = $row['quantity'];

if($quantity >= 2) {
    echo "more books checked out than in stock";
}
else if($quantity < 2) {
    echo "less books checked out than in stock";
}
?>

答案 2 :(得分:0)

这应该有效。请注意SQL注入。 此外,我对数据库中的库存书进行了疯狂猜测,需要进行更改。

require "conn.php";

$userID = "2";
$book_id = "1";

$mysql_qry = "SELECT 
(SELECT COUNT(*) FROM books_checked_out WHERE bookID='" . $book_id . "') as num1,
(SELECT COUNT(*) FROM books_in_stock WHERE bookID='" . $book_id . "') as num2";
$result = mysqli_query($conn, $mysql_qry);
$quantity = mysqli_fetch_assoc($result);

if($quantity['num1'] >= 2) {
    echo "more books checked out than in stock";
}
else {
    echo "less books checked out than in stock";
}
//If you want to see the difference you can do something like this.

echo "Checked Out" . $quantity['num1'] . '<br/>';
echo "In Stock" . $quantity['num2'] . '<br/>';

答案 3 :(得分:0)

尝试使用 intval("0".$row['quantity']) 这会将您的数量转换为整数。