我可以为多个选项制作一个值吗？ [英] can i make one value to many option ?

问题描述

1。假设我在MySQL中的表有一个列调用NUMBER。在这一栏中，它的值为1,2,3。现在我想让1,2,3作为选项供用户选择。

代码应该是什么？


2.这是代码是对吗？

<选择NAME =" field2">

< Option VALUE =''NUMBER''> 1< / Option>

< Option VALUE =''NUMBER> 2< / Option>

< / Select>

解决方案

如果如果要从数据库中填充这些字段，则必须以不同方式执行此操作。但基本上，您提供的代码应为（3个数字）

（值=传递给您的表单，>和<之间的值显示给用户）。

[HTML]

< Select NAME =" field2">

< Option VALUE =''1''> Number 1< /选项>

<选项VALUE =''2''>数字2< /选项>

<选项VALUE =''3''>数字3< / Option>

< /选择> [/ HTML]


Ronald：很酷：

< blockquote class =post_quotes>
如果要从数据库中填充这些字段，则必须采用不同的方式。但基本上，您提供的代码应为（3个数字）

（值=传递给您的表单，>和<之间的值显示给用户）。

[HTML]

< Select NAME =" field2">

< Option VALUE =''1''> Number 1< /选项>

<选项VALUE =''2''>数字2< /选项>

<选项VALUE =''3''>数字3< / Option>

< /选择> [/ HTML]


罗纳德：很酷：

谢谢你的回复。现在我有另一个问题是，我上次从这个论坛获得的代码生成了超链接。



while（

< blockquote> result = mysql_fetch_array（

1. let say my table in MySQL have a column call NUMBER. In this column it would be value 1,2,3. Now i want to make the 1,2,3 as option to user to choose.
what the code should be?

2. is it this code is right?
<Select NAME="field2">
<Option VALUE=''NUMBER''>1</Option>
<Option VALUE=''NUMBER>2</Option>
</Select>

解决方案

If you want to populate these fields from the database, you have to do that differently. But basically, the code you presented should be (for 3 numbers)
(the value= is passed to your form, the value between > and < is displayed to the user).
[HTML]
<Select NAME="field2">
<Option VALUE=''1''>Number 1</Option>
<Option VALUE=''2''>Number 2</Option>
<Option VALUE=''3''>Number 3</Option>
</Select>[/HTML]

Ronald :cool:

If you want to populate these fields from the database, you have to do that differently. But basically, the code you presented should be (for 3 numbers)
(the value= is passed to your form, the value between > and < is displayed to the user).
[HTML]
<Select NAME="field2">
<Option VALUE=''1''>Number 1</Option>
<Option VALUE=''2''>Number 2</Option>
<Option VALUE=''3''>Number 3</Option>
</Select>[/HTML]

Ronald :cool:

thank you for your reply. now i have another question is that i had generate the hyperlink as the code last time i get from this forum.


while(

result = mysql_fetch_array(

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