PHP从mySQL到依赖于表中组的下拉列表 [英] PHP from mySQL into a drop down list dependant on group in table

问题描述

大家好，



我正在尝试使用从数据库中标记为productpr [number]和qty [number]的项目填充每个下拉列表，然而它不断为每个结果创建一个新的下拉菜单，可能是非常简单的东西，但是会很感激帮助。



问候，

詹姆斯





数据看起来像这样...



ID | PIDx | QTYx |价格| jPrice | CartDescription | ViewDescription

1 | productpr1 | qty1 | 10 | 10 | 1 | 1

2 | productpr1 | qty1 | 10 | 10 | 1a | 1a

3 | productpr2 | qty2 | 11 | 11 | 2 | 2

4 | productpr2 | qty2 | 11 | 11 | 2a | 2a



创建这样的html;

Hi Guys,

I'm trying to populate each drop down with items that are labled as productpr[number] and qty[number] from a database, however it keeps creating a new drop down for each result, probably something really simple, but would appreciate the help.

Regards,
James


data look like this...

ID | PIDx | QTYx | Price | jPrice | CartDescription | ViewDescription
1 | productpr1 | qty1 | 10 | 10 | 1 | 1
2 | productpr1 | qty1 | 10 | 10 | 1a | 1a
3 | productpr2 | qty2 | 11 | 11 | 2 | 2
4 | productpr2 | qty2 | 11 | 11 | 2a | 2a

creating a html like this;

<b>Choose an Operating System</b>
<input type="hidden" value="1" name="qty1">
<select name="productpr1">
<option>---</option>
<option id="10" value="1:10">1</option>
</select>
<br><br>

<?
$objConnect = mysql_connect("localhost","username","password") or die(mysql_error());
$objDB = mysql_select_db("database");
$strSQL = "SELECT * FROM puterz";
$objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]");
?>
<br><br><br><br><br><br><br>
<h1>Configure your own AMD Based PC and let us build it for you.</h1>

        <form action="http://wwx.aitsafe.com/cf/addmulti.cfm" method="post">
        <input name="userid" type="hidden" value="1234567">

<?
while($objResult = mysql_fetch_array($objQuery))
{
?>

<?
if($objResult["PIDx"] == "productpr1" && $objResult["QTYx"] == "qty1")
{
?>

Choose an Operating System
<input type="hidden" value="1" name="<?=$objResult["QTYx"];?>">
<select name="<?=$objResult["PIDx"];?>">
<option>---</option>
<option id="<?=$objResult["jPrice"];?>" value="<?=$objResult["CartDescription"];?>:<?=$objResult["Price"];?>"><?=$objResult["ViewDescription"];?></option>
</select>
<br><br>


<?
}
if($objResult["PIDx"] == "productpr2" && $objResult["QTYx"] == "qty2")
{
?>


Choose a Case
<input type="hidden" value="1" name="<?=$objResult["QTYx"];?>">
<select name="<?=$objResult["PIDx"];?>">
<option>---</option>
<option id="<?=$objResult["jPrice"];?>" value="<?=$objResult["CartDescription"];?>:<?=$objResult["Price"];?>"><?=$objResult["ViewDescription"];?></option>
</select>

<?
}
}
?>
			

			<!-- Send Button -->
			<input name="return" type="hidden" value="#">
			<input type="submit" value="Buy Now!">
			
			
        </form>

<?
mysql_close($objConnect);
?>

推荐答案

objConnect = mysql_connect（ localhost， username， password）< span class =code-keyword>或 die（mysql_error（））;

objConnect = mysql_connect("localhost","username","password") or die(mysql_error());

objDB = mysql_select_db（ 数据库）;

objDB = mysql_select_db("database");

strSQL = SELECT * FROM puterz;

strSQL = "SELECT * FROM puterz";

这篇关于PHP从mySQL到依赖于表中组的下拉列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！