如何输出object?

[Y2zz]

我想实现类似php里的var_dump, 但是遇到对象型参数时不知如何输出了...

[ruanwei1987]

[Y2zz]

[Quote=引用 5 楼 dingzongyinnihao 的回复:]

到底是C#还是Php！
[/Quote]

将php里的var_dump移值到C# 呵呵

[Snowdust]

没有直接使用的方法

[Ryan20082009]

你是使用 cakephp框架吗？
1、你先定义一个model 与表名对应：
class Hotel extends AppModel
{
var $name='Hotel';
}
2、写一个controllers ：
var $name='Hotels';
var $ext=".html";
var $uses = array('Homeroom','Hotel','Wineshop');
var $helpers=array('Html','Javascript','Ajax','Form');
function hotels_name()
{
$hotel_name=$this->data['select']['hotel_name'];
$this->set('hotel_name',$hotel_name);
$this->set('hotel_name',$this->Homeroom->find('all', array('homeroom_id', 'hometype', 'menprice','firstprice'), 'homeroom_id DESC'));
}
3、页面得到：
<?php
var_dump($hotel_name);
?>
<?php foreach($hotel_name as $Hotel_name){?>

<td class="fontSize">RMB
<del class="fontSize"><?php echo $Hotel_name['Homeroom']['menprice'];?></del>

</td>
<td class="fontSize"><?php echo $Hotel_name['Homeroom']['zaocan'];?>
</td>
<td class="fontSize"><?php echo $Hotel_name['Homeroom']['chuangxing'];?>
</td>
大概就是这样一个流程 你看看。。。

[我说我行就是行]

到底是C#还是Php！

[phil999]

///<summary>
/// equiv of PHP's var dump for an object’s properties because i cbf writing all the properties out.
///</summary>
///<param name="info"></param>
private static string var_dump(object info)
{
StringBuilder sb = new StringBuilder();

Type t = info.GetType();
PropertyInfo[] props = t.GetProperties();
sb.AppendFormat("{0,-18} {1}", "Name", "Value");

foreach (PropertyInfo prop in props)
{
sb.AppendFormat("{0,-18} {1}", prop.Name, prop.GetValue(info, null).ToString());
}

return sb.ToString();
}


这个问题不错，经常有这样的需求，但是懒于整理。在网上搜了一下，复制给楼主，楼主测试一下，如果能用告诉我

[wuyq11]

贴出var_dump,的使用看看

[龍过鸡年]

反射

var customObject = new { ID = 9527, Name = "王二麻子" };
Type type = customObject.GetType();

foreach (System.Reflection.PropertyInfo pi in type.GetProperties())
{
    this.tbEmployeeID.Text += pi.GetValue(customObject, null);
}