我想要創建一個包含在每個選項多值的選擇「下拉菜單」 .. 是這樣的：如何創建一個選擇的多值
<select id="select" name="zone">
<option value="val1" value="val2" value="val3">zone1</option>
<option value="v1" value="v2" value="v3">zone2</option>
<option value="valu1" value="valu2" value="valu3">zone3</option>
</select>
請告訴我如何進行謝謝你
你能做到這一點
<select id="select" name="zone">
<option value="val1,val2,val3">zone1</option>
<option value="v1,v2,v3">zone2</option>
<option value="valu1,valu2,valu3">zone3</option>
</select>
，並在客戶端或類似上在腳本中使用拆分服務器上
感謝它非常有用 –
您可以輕鬆使用簡單的技巧，因爲您可以使用值屬性中的分隔符（例如:）連接所有值。
當你在php中獲得價值時，它會爆炸它，你將得到所有需要的值。
<select id="select" name="zone">
<option value="val1:val2:val3">zone1</option>
<option value="v1:v2:v3">zone2</option>
<option value="valu1:valu2:valu3">zone3</option>
</select>
現在在PHP中，你只需要做到這一點：
<?php
$value = $post['zone']; // assuming form method is post
$values = explode(':',$value);
$value_1 = $values[0];
$value_2 = $values[1];
$value_3 = $values[2];
?>
希望它是有用的:)
謝謝Rahul :)請告訴我，如果我使用php變量，它是一樣的嗎？ –
是的，你也可以使用php變量。如果我的回答是正確的，並且有幫助，我應該得到一個upvote .. :) – Rahul
，所以如果我明白你都告訴我，我可以這樣做：？
<form>
<select class="target">
<?php // loading rectangles from database
$dbname ='zone'; //Name of the database
$dbuser ='root'; //Username for the db
$dbpass =''; //Password for the db
$dbserver ='localhost'; //Name of the mysql server
$dbcnx = mysql_connect ("$dbserver", "$dbuser", "$dbpass");
mysql_select_db($dbname) or die(mysql_error());
$query = mysql_query("SELECT * FROM ville");
while ($row = mysql_fetch_array($query))
{
$name=$row['NOM_VILLE'];
$lat1=$row['LATITUDE1_V'];
$lon1=$row['LONGITUDE1_V'];
$lat2=$row['LATITUDE2_V'];
$lon2=$row['LONGITUDE2_V'];
echo "<option value=".$lat1.",".$lon1.",".$lat2.",".$lon2.">".$name."</option>";
}
?>
</select>
</form>
您無法在選項中添加許多值。相反，您可以使用的數據集（HTML5），如：
<select id="myselect">
<option data-this="foo" data-that="bar">
</select>
的JavaScript來讀取這些值是：
var d = document.getElementById("myselect");
var _this = d.options[d.selectedIndex].dataset["this"];
var _that = d.options[d.selectedIndex].dataset["that"];
如果你不想與數據集的一塌糊塗，你可以存儲一個JSON對象：
...
<option value='<?php echo json_encode(array("foo"=>1,"bar"=>2)); ?>'>
...
並提取數據，如：
var d = document.getElementById("myselect");
var option_value = JSON.parse(d.options[d.selectedIndex].value);
我發現瞭如何做到這一點 ....這是我最後的代碼..謝謝:)大家..
<!DOCTYPE html>
<html>
<?php
include_once("config.php");
?>
<head>
<title>Simple Map</title>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no">
<meta charset="utf-8">
<style>
html, body, #map-canvas {
margin: 0;
padding: 0;
width: 500px;
height: 500px;
left;
}
</style>
<script src="https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false"></script>
<script>
var map;
//var bounds;
function addRectangle(lat1, lon1, lat2, lon2)
{
var Bounds = new google.maps.LatLngBounds(
new google.maps.LatLng(lat1, lon1),
new google.maps.LatLng(lat2, lon2)
);
var rectangle = new google.maps.Rectangle({
bounds: Bounds,
//draggable: true,
//editable: true,
map : map,
});
//rectangle.setMap(map);
}
function initialize() {
var mapOptions = {
zoom: 8,
center: new google.maps.LatLng(44.490, -78.649),
mapTypeId: google.maps.MapTypeId.ROADMAP
};
map = new google.maps.Map(document.getElementById('map-canvas'),
mapOptions);
<?php // loading rectangles from database
$dbname ='zone'; //Name of the database
$dbuser ='root'; //Username for the db
$dbpass =''; //Password for the db
$dbserver ='localhost'; //Name of the mysql server
$dbcnx = mysql_connect ("$dbserver", "$dbuser", "$dbpass");
mysql_select_db($dbname) or die(mysql_error());
$query = mysql_query("SELECT * FROM ville");
while ($row = mysql_fetch_array($query))
{
$name=$row['NOM_VILLE'];
$lat1=$row['LATITUDE1_V'];
$lon1=$row['LONGITUDE1_V'];
$lat2=$row['LATITUDE2_V'];
$lon2=$row['LONGITUDE2_V'];
echo "addRectangle($lat1, $lon1, $lat2, $lon2);";
}
?>
}
google.maps.event.addDomListener(window, 'load', initialize);
</script>
</head>
<body>
<div id="map-canvas"></div>
<?php // loading makers from database
$dbname ='zone'; //Name of the database
$dbuser ='root'; //Username for the db
$dbpass =''; //Password for the db
$dbserver ='localhost'; //Name of the mysql server
$dbcnx = mysql_connect ("$dbserver", "$dbuser", "$dbpass");
mysql_select_db($dbname) or die(mysql_error());
$query = mysql_query("SELECT * FROM ville");
while ($row = mysql_fetch_array($query))
{
$name=$row['NOM_VILLE'];
$lat1=$row['LATITUDE1_V'];
$lon1=$row['LONGITUDE1_V'];
$lat2=$row['LATITUDE2_V'];
$lon2=$row['LONGITUDE2_V'];
//$desc=$row['zone'];
echo " ".$lat1.", ".$lon1.",".$lat2.", ".$lon2."";
//echo "addMarker($lat, $lon,'<b>$name</b><br/>$desc');";
}
?>
</body>
</html>
也許[這]（HTTP：// stackoverflow.com/questions/3245967/can-an-option-in-select-tag-carry-multiple-values）將有所幫助。每個選項有 – geekchic
單個值。如果你想擁有多個值，你可以給逗號分隔值並在提交時只用逗號分隔它們。 – swan
您不能多次設置相同的屬性。所以這裏的value屬性在你的選項標籤中不止一次是錯誤的。 –