Question

即使找不到'success'，以下内容始终会回显mname。如果找不到，则应显示'failed'。当我输入未找到的'success'时，为什么显示mname？

<?php 
$conn = mysqli_connect("localhost","root","","lpdb"); 

//update 
if (isset($_POST['forgot'])){ 
    $password = 'password'; 
    $mname = $_POST['mname']; 
    $query = "UPDATE logindb SET password ='$mname' WHERE mname = '$mname'"; 
    $result = mysqli_query($conn, $query); 
    if ($result) { 
        echo 'success'; 
    } else { 
        echo 'failed'; 
    } 
}
?>

Answer 1

来自the documentation：

失败时返回FALSE。对于成功的SELECT，SHOW，DESCRIBE或EXPLAIN个查询mysqli_query()将返回mysqli_result个对象。对于其他成功的查询，mysqli_query()将返回TRUE。

不更新任何内容的更新不是失败，因此结果为TRUE。使用mysqli_affected_rows查看UPDATE：

更新了多少行

<?php 
$conn = mysqli_connect("localhost","root","","lpdb"); 

//update 
if (isset($_POST['forgot'])){ 
    $password = 'password'; 
    $mname = $_POST['mname']; 
    $query = "UPDATE logindb SET password ='$mname' WHERE mname = '$mname'"; 
    mysqli_query($conn, $query);           // ***
    $result = mysqli_affected_rows($conn); // ***
    if ($result > 0) {                     // *** Or perhaps == 1, up to you
        echo 'success'; 
    } else { 
        echo 'failed'; 
    } 
}
?>

Answer 2

首先检查mname：

<?php

//check if user submitted first
if (isset($_POST['forgot'])){ 
$password = 'password'; 
$mname = $_POST['mname']; 

$conn = mysqli_connect("localhost","root","","lpdb"); 
//update 
$query = "UPDATE logindb SET password ='$mname' WHERE mname = '$mname'"; 
$result = mysqli_query($conn, $query); 
if($result){
    echo 'success'; 
} else { 
    echo 'failed to update for some reason';
//or maybe echo the db err
echo mysqli_error(); 
} 
?

更新行但在更新之前，输入框应该符合记录

2 个答案: