小编the*_*tao的帖子

每次我安装EasyPHP 13.1VC9它都会工作几天,然后在我开启它时随机点我得到这个错误,我无法启动MySQL.我尝试过以管理员身份运行,我尝试过编辑权限.此外,当我通过PHPadmin访问MYSQL时,我收到以下错误:

#2002 - No connection could be made because the target machine actively refused it.

这是日志文件所说的:

2013-06-12 23:47:18 4148 [ERROR] InnoDB: Attempted to open a previously opened     tablespace. Previous tablespace mysql/innodb_index_stats uses space ID: 2 at filepath: .\mysql\innodb_index_stats.ibd. Cannot open tablespace userauth/userinfo which uses space ID: 2 at filepath: .\userauth\userinfo.ibd

InnoDB: Error: could not open single-table tablespace file .\userauth\userinfo.ibd

InnoDB: We do not continue the crash recovery, because the table may become

InnoDB: corrupt if we cannot apply the …

我在这个查询中收到错误:

"SELECT username 
and password 
FROM users 
WHERE username = $email , password = $password"

它正在返回此错误:

错误1064 - 您的SQL语法中有错误; 查看与您的MySQL服务器版本对应的手册,以便在''用户'附近使用正确的语法WHERE username = $ email,password = $ password LIMIT 0,30'在第1行

如果我要做以下事情:

$pds= $pdo->prepare("SELECT * FROM userinfo  WHERE   username=:username AND   password=:password");

$pds->execute(array(':username' => $username, ':password' => $password));

$row = $pds->fetch(PDO::FETCH_ASSOC);

我是否需要为每个执行的命令设置try {},或者尝试使用单个catch块覆盖整个代码？

谢谢!

该节目只是在墙上输出了九十九瓶啤酒.它编译没有错误,但我的代码没有任何反应.我认为它与我如何设置最后两个方法的参数以及我的实例变量有关,但我在理解这一点时遇到了一些麻烦.

package beersong;

public class BeerSong

{
    private int numBeerBottles;
    private String numberInWords;
    private String secondNumberInWords;
    private int n;

    private String[] numbers = {"", "one", "two", "three", "four", "five", 
        "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen",
        "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"
    };

    private String[] tens = {"", "", "Twenty", "Thirty", "Forty", "Fifty",
        "Sixty", "Seventy", "Eighty", "Ninety" 
    };

    //constructor
    public BeerSong (int numBeerBottles)
    {
        this.numBeerBottles = numBeerBottles;
    }

    //method to return each line of song
    public String convertNumberToWord(int n) …

这是我在html页面底部的jQuery:

<script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript">
    $(document).ready(function () {
        $('#login').click(function() {
            var co_id = $('#companies').val();
            console.log(co_id); 
        });
    });
    </script>

当我在控制台中输入这个jQuery,然后键入co_id它有效,但是当我在浏览器中运行它并选择一个公司并单击正确的按钮时,我会假设当我输入co_idchrome浏览器控制台时,它会显示我的选项值？但它回来了:

>co_id
ReferenceError: co_id is not defined

这是html:

<select name="" id="companies">
    <option value="--">--</option>
    <option value="1">Company</option>
    <option value="2">Company2</option>
    <option value="3">Company3</option>
    <option value="68">Company4</option>
    <option value="69">Company5</option>
    <option value="70">Company6</option>
</select><input id="login" type="submit" value="Login">

我有 .replace(' ', ', ')

我想要的是:

Hanks, Tom

我得到的是:

Hanks,Tom

你怎么做到这一点？

输入字符串是Tom Hanks.

我想要汉克斯,汤姆