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如何把一个列表中选项的值传到下个php界面中

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Release: 2016-06-13 12:48:01
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怎么把一个列表中选项的值传到下个php界面中
if($roomID!=null)
{
$result=mysql_query("SELECT roomID,roomtype,roomsize FROM room WHERE roomID NOT IN (SELECT roomID FROM bookingroom WHERE bookingdate='$bookingdate' AND bookingtime='$bookingtime') AND roomID='$roomID'");
    $num_rows = mysql_num_rows($result);
if ($num_rows==0) {
echo "

Sorry, No result found. Please search again.

";
}
else{

while($row = mysql_fetch_array($result))
{
echo "";
echo "".$row['roomID']."";
echo "".$row['roomtype']."";
echo "".$row['roomsize']."";
echo "";
echo "";
}
}
}
怎么把我网页界面选择的那行数据传到insert_data.php去呢
PHP MySQL 搜索
------解决方案--------------------
<br />
echo "<td><input type=\"submit\" name=\"submit\"         value=\"Book\"onClick=\"if(confirm('Are you sure you want to book this room?'))location='insert_data.php?id=".$row['roomID']."'\")/></td>";<br />
Copy after login

在文件名后面增加ID参数,参数值为对应行数据的ID
------解决方案--------------------
get方法试试
------解决方案--------------------
通过js获取每行属性,你可以通过查找table tr td,定位到每一行,再通过url传值即可.
------解决方案--------------------
js onSubmit的时候存入cookies
在另一边php得到就行了
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