mysqli准备好的语句num_rows返回0,而查询返回大于0 [英] mysqli prepared statement num_rows returns 0 while query returns greater than 0
问题描述
对于实际存在的电子邮件,我有一个简单的准备好的声明:
I have a simple prepared statement for an email that actually exists:
$mysqli = new mysqli("localhost", "root", "", "test");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sql = 'SELECT `email` FROM `users` WHERE `email` = ?';
$email = 'example@hotmail.com';
if ($stmt = $mysqli->prepare($sql)) {
$stmt->bind_param('s', $email);
$stmt->execute();
if ($stmt->num_rows) {
echo 'hello';
}
echo 'No user';
}
结果:在应回显hello
我在控制台中运行了相同的查询,并使用与上述相同的电子邮件获得了结果.
I ran the same query in the console and got a result using same email as above.
我也使用一个简单的mysqli查询进行了测试:
I tested using a simple mysqli query as well:
if ($result = $mysqli->query("SELECT email FROM users WHERE email = 'example@hotmail.com'")) {
echo 'hello';
}
结果:我所期望的hello
$result
的num_rows
也为1.
为什么准备的语句num_row
不大于0?
Why is the prepared statment's num_row
not greater than 0?
推荐答案
当您通过mysqli执行一条语句时,直到您获取它们,结果才真正在PHP中出现-结果由数据库引擎保存.因此mysqli_stmt
对象无法知道执行后立即有多少结果.
When you execute a statement through mysqli, the results are not actually in PHP until you fetch them -- the results are held by the DB engine. So the mysqli_stmt
object has no way to know how many results there are immediately after execution.
像这样修改您的代码:
$stmt->execute();
$stmt->store_result(); // pull results into PHP memory
// now you can check $stmt->num_rows;
这不适用于您的特定示例,但是如果您的结果集很大,则$stmt->store_result()
将占用大量内存.在这种情况下,如果您关心的只是弄清楚是否至少返回了一个结果,请不要存储结果.相反,只需检查结果元数据是否不为空:
This doesn't apply to your particular example, but if your result set is large, $stmt->store_result()
will consume a lot of memory. In this case, if all you care about is figuring out whether at least one result was returned, don't store results; instead, just check whether the result metadata is not null:
$stmt->execute();
$hasResult = $stmt->result_metadata ? true : false;
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