Javascript 需要将select tag选项值传递给ajax函数

Javascript 需要将select tag选项值传递给ajax函数,javascript,php,html,ajax,Javascript,Php,Html,Ajax,当我按下add按钮时,我需要获取select tag value的值 这是我的ajax文件 function insertDep(){ var ajaxRequest; try{ ajaxRequest = new XMLHttpRequest(); }catch (e){ try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); }catch (e) {

当我按下add按钮时,我需要获取select tag value的值

这是我的ajax文件

function insertDep(){
   var ajaxRequest;
   try{
      ajaxRequest = new XMLHttpRequest();
   }catch (e){
      try{
         ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
      }catch (e) {

         try{
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
         }catch (e){
            alert("Your browser broke!");
            return false;
         }
      }
   }
   ajaxRequest.onreadystatechange = function(){   
      if(ajaxRequest.readyState == 4){
         var ajaxDisplay = document.getElementById('addmegp');
         ajaxDisplay.innerHTML = ajaxRequest.responseText;
      }
   }
   var unitnamedept = document.getElementById('unitnamedept').value;


   var queryString =  "?unitnamedept=" + unitnamedept;
   ajaxRequest.open("GET", "insertDept.php" + queryString, true);
   ajaxRequest.send(null); 
}
<select> <option name="unitnamedept" id="unitnamedept" value="0" >--</option> <option value="D01">D01</option> <option value="D02">D02</option> <option value="D03">D03</option> <option value="D04">D04</option> </select>

<input type="button" name="ang" align="right" value="Add" onclick="insertDep()"/>
这是我的html文件

function insertDep(){
   var ajaxRequest;
   try{
      ajaxRequest = new XMLHttpRequest();
   }catch (e){
      try{
         ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
      }catch (e) {

         try{
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
         }catch (e){
            alert("Your browser broke!");
            return false;
         }
      }
   }
   ajaxRequest.onreadystatechange = function(){   
      if(ajaxRequest.readyState == 4){
         var ajaxDisplay = document.getElementById('addmegp');
         ajaxDisplay.innerHTML = ajaxRequest.responseText;
      }
   }
   var unitnamedept = document.getElementById('unitnamedept').value;


   var queryString =  "?unitnamedept=" + unitnamedept;
   ajaxRequest.open("GET", "insertDept.php" + queryString, true);
   ajaxRequest.send(null); 
}
<select> <option name="unitnamedept" id="unitnamedept" value="0" >--</option> <option value="D01">D01</option> <option value="D02">D02</option> <option value="D03">D03</option> <option value="D04">D04</option> </select>

<input type="button" name="ang" align="right" value="Add" onclick="insertDep()"/>
——D01 D02 D03 D04
**我有另一个页面叫做insertDept.php** 我使用此页面显示


当我按下只接收0值的按钮时,任何帮助请……
--D01 D02 D03 D04

将id和name属性更改为select not to the option,如我所给出的

需要将name和id属性放置在select not in option中@NarendraSisodia

使用选择器获取选项的值-jQuery

使用
e.options[e.selectedIndex].value


其中
var e=document.getElementById('SelectID')
不是选项

您需要在select not in选项中放置name和id属性,这将对您有所帮助。。您可以使用JavaScript获取值。这里有一个例子@Narendra。。。。。信息技术works@sreetharraja我需要把它作为答案发布吗