Question

我正在尝试调用函数来更新我的mysqli数据库，并在发生错误时报告。到目前为止，我的测试文件中有这个：

<?php

$success = test("cat",17);
echo $success;
echo "<br />";


function test($name, $number){
    $db = new mysqli("localhost","root","","test");
    if($db->connect_errno){
        return "Failed to connect to MYSQL Database: [" . $db->connect_errorno . "]" . $db->connect_error;
    }
    if(!($stmt = $db->prepare("UPDATE `test` SET `name` = ? WHERE `number` = ?"))){
        return "prepare failed: [" . $db->errno . "]" . $db->error;
    }
    if(!($stmt->bind_param('si', $name, $number))){
        return "bind failed: [" . $db->errno . "]" . $db->error;
    }
    if(!($stmt->execute())){
        return "execute failed: [" . $db->errno . "]" . $db->error;
    }
    $stmt->close();
    return "Successfully updated database!";
}

?>

并且我在那里（cat，17）的测试它返回成功，即使没有数量为17的列，我看到数据库中没有任何变化。如果无法在WHERE子句中找到数字，我怎样才能正确返回失败？

Answer 1

execute()完成后，检查$stmt->affected_rows是否大于零：

if(!($stmt->execute())){
    return "execute failed: [" . $db->errno . "]" . $db->error;
}

if(!$stmt->affected_rows) {
    return "no rows updated";
}

在MYSQLi中报告未知索引的错误

1 个答案: