小编lol*_*ola的帖子

为什么这不起作用？

该计划停止: this.textBox1.Text = "(New text)";

 Thread demoThread;
 private void Form1_Load(object sender, EventArgs e)
 {
     this.demoThread = new Thread(new ThreadStart(this.ThreadProcUnsafe));
     this.demoThread.Start();

     textBox1.Text = "Written by the main thread.";
 }

 private void ThreadProcUnsafe()
 {
     while (true)
     {
         Thread.Sleep(2000);
         this.textBox1.Text = "(New text)";           
     }
 }

if($ a == 4){echo'ok'; }

现在显示错误,因为$ $未定义变量.我的决定:

if(isset($a)){
    if($a == 4){
    echo 'ok';
    }
}

但也许有更好的解决方案？

我需要定义几个PHP常量,我需要在我的包中使用这个常量(控制器,自定义类,实体..)哪里有最好的位置来添加这个常量,这对他们来说会方便吗？

我有一个块(div),其中包含带链接的文本.当我将鼠标悬停在此块上时,我需要更改文本颜色(也链接颜色)." div:hover" - 使用此文本颜色更改,但链接颜色保持不变.

完整代码:

CSS:

a {
    color: #336699;
}
div {
    height: 50px;
    background-color: #FFF;
    color: red;
}
div a {
    color: red;
}
div:hover {
    background-color: #336699;
    color: #FFF;
}

HTML:

<div>
    text test <a href="#">URL</a> text
</div>

代码有什么问题？为什么第二次报告显示错误？

string level;
int key;

command.CommandText = "SELECT * FROM user WHERE name = 'admin'";

connection.Open();
Reader = command.ExecuteReader();

while (Reader.Read())
{
    level = Convert.ToString(Reader["level"]);
    key = Convert.ToInt32(Reader["key"]);

    MessageBox.Show(level); //Work fine
}

MessageBox.Show(level); //Show error:  Use of unassigned local variable 'level'

嗨,现在我得到所有的分类和子类别.怎么只获得子类别？

<?php foreach ($this->getStoreCategories() as $_category): ?>
    <?php $_menu .= $this->drawItem($_category) ?>
<?php endforeach ?>

为什么我得到错误:

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'IF EXISTS(SELECT id FROM mytable WHERE id = '1')' at line 1 

查询:

IF EXISTS(SELECT id FROM mytable WHERE id = '1')

谢谢

我有这个链接:

<a id = "link_1" href = "#">Cars</a>
<a id = "link_2" href = "#">Colors</a>
<a id = "link_3" href = "#">Users</a>
<a id = "link_4" href = "#">News</a>

如何获取我点击的ID号码？例如,我推动链接汽车,我希望获得1,推送用户,获得3号.

谢谢

我从POST方法得到了回复.在里面我看到了值,但是POST方法返回一个空值.为什么,热修复这个？

样品:

$.post('/news/add/', {parent: name, title: 'none'}, function(data){ 
 new_id = data;
 alert(new_id); //11
});
alert(new_id); //empty

是否可以将子查询中的两个值相加？

我需要选择三个值：total_view、total_comments 和 rating。

两个子查询都非常复杂，所以我不希望重复它。

我的查询示例：

SELECT p.id,
(
    FIRST subquery  
) AS total_view,
(
    SECOND subquery 
) AS total_comments,
(
    total_view * total_comments
) AS rating
FROM products p
WHERE p.status = "1"
ORDER BY rating DESC