PHP $ stmt-> num_rows在准备好的语句中不起作用 [英] PHP $stmt->num_rows doesnt work by prepared statements

问题描述

我要检查if ($numRows >= 1)，然后它应该返回一些内容.

I want to check if ($numRows >= 1) then it should return something.

当我使用$con->mysql("{QUERY}")时，它可以工作.

When I use $con->mysql("{QUERY}"), it works.

但是当我使用$stmt = $con->prepare("{QUERY}")时，它不起作用.

But when I use $stmt = $con->prepare("{QUERY}"), it doesn't work.

有人知道吗?

<?php
if ($result = $con->query("SELECT username FROM users WHERE username = 'test'")) {
    $numRows = $result->num_rows;

    echo $numRows;
}
?>

结果:1​​

<?php
$name = 'test';

$stmt = $con->prepare("SELECT username FROM users WHERE username = ?");
$stmt->bind_param('s', $name);

$name = 'test';

$stmt->execute();

$numRows = $stmt->num_rows;

echo $numRows;
?>

结果:0

推荐答案

在调用->num_rows()方法之前，您需要从SELECT查询中转移结果集.

You need to transfer the result set from the SELECT query before calling the ->num_rows() method.

// your code    

$stmt->execute();
$stmt->store_result();  
$numRows = $stmt->num_rows;

echo $numRows;

---

这里是参考:

---

Here's the reference:

• http://php.net/manual/en/mysqli.store- result.php

这篇关于PHP $ stmt-> num_rows在准备好的语句中不起作用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！