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我正在尝试整理我正在制作的网络应用程序上的查询，该查询一直失败，我不知道为什么！代码如下：

$update = $_GET['update'];
if($update == "true"){
    $setDetails="UPDATE users SET email='{$_POST['email']}', api_key='{$_POST['api_key']}', api_secret='{$_POST['api_secret']}' WHERE username={$_POST['username']}";
    if(mysql_query($setDetails)){
        $updatemsg = '<div class="alert alert-success"><a href="#" class="close" data-dismiss="alert">×</a><strong>Success!</strong> Your details have been updated in our database.</div>';
    }else{
        $updatemsg = '<div class="alert alert-error"><a href="#" class="close" data-dismiss="alert">×</a><strong>Failure!</strong> Your details could not be updated in our database. Please try again later or contact us if this keeps happening.</div>';
    }
}else if($update == "false"){
    $updatemsg = '<div class="alert alert-success"><a href="#" class="close" data-dismiss="alert">×</a><strong>Success!</strong> Your changed were discarded.</div>';
}

任何想法，帮助或提示？请注意，在我的网络应用程序的下方，我的网络应用程序SELECT * FROM users WHERE username='$username'运行良好，因此数据库连接没有问题。

4

2 回答 2

3 
$update = $_GET['update'];
if($update == "true"){
    $setDetails="UPDATE users SET email='{$_POST['email']}', api_key='{$_POST['api_key']}', api_secret='{$_POST['api_secret']}' WHERE username='{$_POST['username']}'";
    if(mysql_query($setDetails)){
        $updatemsg = '<div class="alert alert-success"><a href="#" class="close" data-dismiss="alert">×</a><strong>Success!</strong> Your details have been updated in our database.</div>';
    }else{
        $updatemsg = '<div class="alert alert-error"><a href="#" class="close" data-dismiss="alert">×</a><strong>Failure!</strong> Your details could not be updated in our database. Please try again later or contact us if this keeps happening.</div>';
    }
}else if($update == "false"){
    $updatemsg = '<div class="alert alert-success"><a href="#" class="close" data-dismiss="alert">×</a><strong>Success!</strong> Your changed were discarded.</div>';
}

详细信息： 您的代码：

$setDetails="UPDATE users SET email='{$_POST['email']}', api_key='{$_POST['api_key']}', api_secret='{$_POST['api_secret']}' WHERE username={$_POST['username']}";

正确代码：

 $setDetails="UPDATE users SET email='{$_POST['email']}', api_key='{$_POST['api_key']}', api_secret='{$_POST['api_secret']}' WHERE username='{$_POST['username']}'";

用户名字符串周围缺少 '。

于 2013-07-04T06:22:17.420 回答
1

像这样尝试。为username

$setDetails="UPDATE users 
            SET email='{$_POST['email']}', 
                api_key='{$_POST['api_key']}', 
                api_secret='{$_POST['api_secret']}' 
            WHERE username='{$_POST['username']}' ";

并尽量避免mysql_*声明，因为它们已被弃用。而是使用mysqli_*statements 或PDOstatements

于 2013-07-04T06:23:49.193 回答