Question

所以，我有一份准备好的陈述，我已成功准备，装订和执行：

401 Unauthorized

看起来像预期的那样返回了一行。但是，当我调用@ SELECT * FROM users WHERE email = ? AND pass = SHA1(?)时，为什么$ result变量为空？提前谢谢。

$stmt->get_result()

Answer 1

将这两条评论放在一起并详细说明......

<?php
$result = $stmt->get_result();
$num_rows = mysqli_num_rows($result);

if ($num_rows == 1) {
    // already fetched the mysqli_result
    // now lets fetch the one record from that result set
    $row = $result->fetch_assoc();
    // ....do something with row
}
else {
    // either 0   ...or more than 1 row
    foo();
}

但你甚至可以摆脱对mysqli_num_rows（）的调用（所以它也适用于unbuffered queries）

$result = $stmt->get_result();
$row = $result->fetch_assoc();
if ( !$row ) {
    // no such record
    foo();
}
else {
    // ....do something with row

    // might want to check whether there are more matching records
    // given the context there shouldn't, but ...
}

php get_result（）返回一个空结果集

1 个答案: