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Home Backend Development PHP Tutorial 怎么把php查询到的值显示到下拉框中

怎么把php查询到的值显示到下拉框中

Jun 23, 2016 pm 02:00 PM

想在把查询到的值显示在 下拉框 中
这个值只有上 中 下 三种 
我加个

但是这样做就会有两个相同的值
如果只显示一个不知道应该怎么写


$sql="select * from `xx` "; 
$query=mysql_query($sql);
$rs=mysql_fetch_array($query);
?>
 


回复讨论(解决方案)

<?php$sql="select * from `xx` "; $query=mysql_query($sql);?><select name="type">	<?php	while ($rs=mysql_fetch_array($query)) {		?>			<option  value="<?php echo $rs['type'];?>"><?php echo $rs['type'];?></option>		<?php	};	?></select> 
Copy after login
Copy after login

不知道是什么意思,

这里不是只有一个option吗,怎么会有两个相同的值。

$rs=mysql_fetch_array($query);
这个函数错误,他有ASSOC,NUM,BOTH方式返回,你的返回的就是BOTH的方式,就是键名和下标各一个,所以会有2个。
去看看mysql_fetch_arry,mysql_fetch_row的区别

不知道是什么意思,

这里不是只有一个option吗,怎么会有两个相同的值。
本来有3条数据,上中下,被他读成了,$rs['type'],$rs[0],各一个,所以读出来了6条数据

<?php$sql="select * from `xx` "; $query=mysql_query($sql);?><select name="type">	<?php	while ($rs=mysql_fetch_array($query)) {		?>			<option  value="<?php echo $rs['type'];?>"><?php echo $rs['type'];?></option>		<?php	};	?></select> 
Copy after login
Copy after login

这个代码应该和他自己的结果是一样的。你换成mysql_fetch_assoc($query)才是自己想要的结果吧


不知道是什么意思,

这里不是只有一个option吗,怎么会有两个相同的值。
本来有3条数据,上中下,被他读成了,$rs['type'],$rs[0],各一个,所以读出来了6条数据
你仔细看下他的代码,有$rs[0]么?

是我理解错了。



不知道是什么意思,

这里不是只有一个option吗,怎么会有两个相同的值。
本来有3条数据,上中下,被他读成了,$rs['type'],$rs[0],各一个,所以读出来了6条数据
你仔细看下他的代码,有$rs[0]么?

是这样 以我写代码 返回的值如果是 上
下拉框就会 形成




返回的值如果是 下
就是




会有重复

不知道你这样写有什么用意。也不知道你想干什么。


因为写了个表单 这个项是固定的 三个值 上 中 下
表单一INSERT下拉选择其中的一项
表单二先select 这项的值 然后可以用下拉选择其它 UPDATE更新

语言组织能力不好。。。

很简单啊,用条件判断一下$rs['type']是否等于上中下之一,用

<select name="type"><option value="上" <?php echo $rs['type']=='上'?'selected':'' ?>>上</option><option value="中" <?php echo $rs['type']=='中'?'selected':'' ?>>中</option><option value="下" <?php echo $rs['type']=='下'?'selected':'' ?>>下</option></select>
Copy after login
 

$rs=mysql_fetch_array($query);
打印出来,看看是什么玩意。

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