我是Javascript和jquery的新手。我希望从一个php页面传递多个选定的dropdown值到另一个php页面的功能。
在我的下拉列表中,我有两个按钮。' Set'并且'取消设置' .Set会将值插入数据库并从数据库中取消设置删除。
这是我的php:(php1.php)
printf( '<h1>Update Capabilities</h1>' );
printf( '<form action="%s" method="POST" id="capabilityform">', $_SERVER['PHP_SELF'] );
printf( '<span class="subHeading">Capabilities</span>' );
printf('<select name="selectedCapabilities[]" multiple>');
foreach ($capabilityList as $i)
{
printf('<option value="%s">%s</option>', $i, $i);
}
printf( '</select>' );
printf( '<input type="hidden" name="hosts" value="%s"/>', implode( ',', $hosts ) );
printf( '<input id="capabilityHosts" type="hidden" name="sethosts" value="" />' );
printf( '<input id="setCapabilitiesButton" name="setcapabilities" type="summit" value="Set Capabilities" disabled="disabled" />' );
printf( '<input id="unsetCapabilitiesButton" name="unsetcapabilities" type="submit" value="Unset Capabilities" disabled="disabled" />' );
printf( '</form>' );
另一个php页面有两个set和unset函数。(php2.php)
function setCap($host , $capablities)
{
//....sql query here
}
function unsetCap($host , $capablities)
{
//....sql query here
}
如果我们按setCapabilities按钮,所有选定的值将转到setcap(),并且与unsetCapabilities相同。
请帮助您使用JS将选定的值传递给上述函数?
答案 0 :(得分:0)
首先,您必须添加类名以选择选项。
printf( '<h1>Update Capabilities</h1>' );
printf( '<form action="%s" method="POST" id="capabilityform">', $_SERVER['PHP_SELF'] );
printf( '<span class="subHeading">Capabilities</span>' );
printf('<select name="selectedCapabilities[]" class="selectedCapabilities" multiple>');
foreach ($capabilityList as $i)
{
printf('<option value="%s">%s</option>', $i, $i);
}
printf( '</select>' );
printf( '<input type="hidden" name="hosts" value="%s"/>', implode( ',', $hosts ) );
printf( '<input id="capabilityHosts" type="hidden" name="sethosts" value="" />' );
printf( '<input id="setCapabilitiesButton" name="setcapabilities" type="summit" value="Set Capabilities" disabled="disabled" />' );
printf( '<input id="unsetCapabilitiesButton" name="unsetcapabilities" type="submit" value="Unset Capabilities" disabled="disabled" />' );
printf( '</form>' );
然后添加此javascript代码。
<script type="text/javascript">
$("select.selectedCapabilities").change(function(){
var selectedCapabilities = $(".selectedCapabilities option:selected").val();
//uncomment alert you can check if you getting right data
//alert(selectedCapabilities);
$.ajax({
url: "php2.php",
type: "post",
data: {option: $(this).find("option:selected").val()},
success: function(data){
//this one you need to show some responce from php2 file
$("#getservicesdatas").html(data);
} });
});
</script>