执行成功，但num_rows返回0 [PHP-MySQL] [英] Execute Success but num_rows return 0 [PHP-MySQL]

问题描述

我刚遇到的问题是

$update_stmt->execute()正常，并且数据库中的数据已更新

但是，$update_resultrow = $update_stmt->num_rows; 返回0吗?

我试图复制MySQL命令以在查询中运行，它也运行良好，如下所示:

I tried to copy MySQL command to run in query and it also worked well, like this:

UPDATE ACCOUNT_EMPLOYEE SET NAME = 'cccccc' WHERE ID = 1

---

问题代码在这里:

---

Problem's Code here:

$update_sql = "UPDATE ACCOUNT_EMPLOYEE SET NAME = ? WHERE ID = ?";
if ($update_stmt = $conn -> prepare($update_sql)) {
    if ($update_stmt->bind_param("si",
        $newname,
        $acc_id
    )
    ) {
        if ($update_stmt->execute()) {
            // must declare here to be able to get num_rows
            $update_stmt->store_result();
            $update_resultrow = $update_stmt->num_rows;
            if ($update_resultrow == 0) {
                echo $error_forgot_noresult . '????' . $acc_id ;
                $update_stmt->close();
                $conn->close();
                exit();
            }
        }
    }
}

推荐答案

是的，Fred -ii-，我从没注意到它有-> affected_rows.请张贴为答案，我会在这里标记出来

根据OP的要求.

看到这里的目标是测试查询是否确实成功，您需要使用affected_rows.

Seeing that the goal here is to test if the query was indeed successful, you need to use affected_rows.

按照手册 http://php.net/manual/en/mysqli.affected-rows .php

printf(受影响的行(UPDATE):％d \ n"，$ mysqli-> affected_rows);

printf("Affected rows (UPDATE): %d\n", $mysqli->affected_rows);

• 面向对象的样式

int $ mysqli-> affected_rows;

int $mysqli->affected_rows;

---

旁注:

---

Sidenote:

使用

$update_resultrow = $update_stmt->num_rows;

并检查错误，将抛出一个错误，而不是返回0".

and checking for errors, would have thrown an error, rather than "return 0".

• http://php.net/manual/en/mysqli.error.php

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