2014-04-16 72 views
0

我在链接(表单元素)选择时遇到问题。 我似乎无法用jquery触发第二个选择选项来填充选择使用$ examplevar作为筛选结果从第一个下拉列表中选择的选项。php,mysql,jquery,表单元素链选择

<form name="vlan" action="" method="post"> 
     <div class="row"> 
     <div class="col-md-6"> 
     <?php 
     $con_string= "host=example port=xxxx dbname=example user=example password=example"; 
     $db_con = pg_connect($con_string); 
     $dev = pg_query($db_con, "SELECT DISTINCT \"Type\" FROM \"End_Devices\""); 

          echo " 
          <a class=\"current_page_item\">Location NASID:</a> 
          <input type=\"text\" name=\"nasid\" />  
          <br> 
          <a class=\"current_page_item\">Device Name:</a> 
          <input type=\"text\" name=\"name\" /> 
          <br> 
          <input type=\"hidden\" name=\"dvtype\"/> 
          <a class=\"current_page_item\">Device Type:</a> 
          <select name=\"slType\" id=\"slType\">"; 

          while ($device = pg_fetch_array($dev)) 
          { 
           echo "<option id=\"" .$device['Type']. "\">" .$device['Type']. "</option>"; 

          } 
          echo "<script type=\"text/javascript\"> 

           $(\"#slType\").chained(\"#slModel\"); 
           $(\"#slType\").on(\"change\",function(){ 
            var slTypeID = $(this).val(); 
            $.post('inventoryTEST.php', {slTypeID:slTypeID}, function(data){ 
             $('#slModel').html(data); 
            }); 
           }); 

          </script>"; 


          echo " 
          </select> 
          <br> 
          <a class=\"current_page_item\">Device Model:</a> 
          <select name=\"slModel\" id=\"slModel\" >"; 

          if (isset($_POST['slTypeid'])) 
          { 
           $selection = ($_POST(['slTypeid'])); 

          } 

          $mod = pg_query($db_con, "SELECT \"Model\" FROM \"End_Devices\" WHERE \"Type\"=\"". $selection . "\""); 
          while ($model = pg_fetch_array($mod)) 
          { 
           echo $_POST[type]; 
           echo "<option id=\"" .$model['Model']. "\">" .$model['Model']. "</option>"; 
          } 
          echo " 
          </select> 
          <br> 
          <a class=\"current_page_item\">Physical Location:</a> 
          <input type=\"text\" name=\"location\" /> 
          <br> 
          <a class=\"current_page_item\">Serial:</a> 
          <input type=\"text\" name=\"serial\" /> 
          <br> 
          <a class=\"current_page_item\">IP Address:</a> 
          <input type=\"text\" name=\"ip\" /> 
          <br> 
          <a class=\"current_page_item\">MAC:</a> 
          <input type=\"text\" name=\"mac\" />  
          <br> 
          <a class=\"current_page_item\">SSID:</a> 
          <input type=\"text\" name=\"ssid\" /> 
          <br> 
          <a class=\"current_page_item\">WLAN Area:</a> 
          <input type=\"text\" name=\"wlan\" /> 
          <br> 
          <br> 
          <br> 
         "; 
         ?> 
          <!--SUBMIT INPUT--> 
          <input type="submit" value="Save" name="submit" /> 
         </div> 
        </div> 
       </form> 

任何人都可以请建议上面的代码? 我无法触发on change事件来填充用作sql查询过滤器的var,然后填充第二个链接下拉列表。 感谢 此致

回答

0

管理通过使用jquery和阿贾克斯通过使所选择的行的内容.change与柱插入其中,然后我查询使用它作为一个过滤器的分贝外部PHP和在填充第二组合框解决了它。