Question

从一个表单中的下拉列表中获取所选值，然后单击表单外部的按钮将值发送到php页面

<div class="drpvendorname">
      <font style="color: white;">
        <label>Distribution Point:</label>        
    </font>
    </div>
        <select class="form-control" id="drpvendor" name="pointname" required="">
        <option selected disabled>Choose distribution point</option>
         <?php
          $strSQL = "SELECT * FROM point_tbl ORDER BY pointname ASC";
          $query = mysqli_query($db, $strSQL);
          while($result = mysqli_fetch_array($query))
          {          
            echo '<option onClick="distribution('.$result['pointshortname'].')" value="' .$result['pointshortname'].'">'. $result['pointname'].'</option>';
          }
          ?>
        </select>      
      </form><!--form1 ends here-->

      <form action="../customer/form.php"><!--form2 starts here-->
      <button class="btn pos7" name="abc" method="GET" style="margin-left:5%;">New Customer</button>
      </form><!--form2 ends here-->

      <div class="dailybreakupbtn">  
      <input class="btn" type="submit" id="dailybreakupbutn" name="dailybreakup" value="Enter Daily Breakup" onClick="distribution(<?=$pointname?>)"/>
      </div>
      <?php
      if(isset($pointname)){      
       ?>
      <script type="text/javascript">
        function distribution(pointname){
          var pointname;
          window.location.href="dailybreakup.php?query=" +pointname;
        }
      </script>
      <?php
      }
      ?>

我尝试使用功能名称分发
我结束了将undefined发送到下一页。
任何人都可以帮助我将所选值发送到下一页而不将该按钮放入<form>

Answer 1

愿这会对你有所帮助:) Html代码：

<form action= "" method= "post">
<select class="form-control drpvendor" id="drpvendor" name="pointname" required="">
        <option selected disabled>Choose distribution point</option>
         <?php
          $strSQL = "SELECT * FROM point_tbl ORDER BY pointname ASC";
          $query = mysqli_query($db, $strSQL);
          while($result = mysqli_fetch_array($query))
          {          
            echo '<option value="'.$result['pointshortname'].'">'. $result['pointname'].'</option>';
          }
          ?>
        </select> 
</form>

JQuery代码：

$(doucment).on('change','.drpvendor',function(){
    var data=$(this).attr('selected','selected');
      $.ajax({
      url: "dailybreakup.php",
      data:'query='+ data,
      type: "POST",
      success: function(data) {
         window.location.href='customer/form.php';
      }
      });
    });

Answer 2

使用代码

    <script   src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script> // please link the path or use cdn 
<script type="text/javascript">
function call_me(){
    var pointname=document.getElementById("drpvendor").value;
    alert(pointname); // comment it after testing 
    window.location.href="dailybreakup.php?query=" +pointname;
}
</script>
<form>
    <select class="form-control" id="drpvendor" name="pointname" required="">
    <option selected disabled>Choose distribution point</option>
     <?php
      $strSQL = "SELECT * FROM point_tbl ORDER BY pointname ASC";
      $query = mysqli_query($db, $strSQL);
      while($result = mysqli_fetch_array($query))
      {          
        echo '<option  value="' .$result['pointshortname'].'">'. $result['pointname'].'</option>';
      }
      ?>
    </select>  
   <input type="button" onclick="call_me()"/>       
  </form><!--form1 ends here-->

    Note : 1) Code is not tested 
           2) use your select id instead of your_select_id
           3)  check variable

使用表单外部的重定向按钮中的表单中的值

2 个答案: