Question

我有两个下拉列表，其中的值是从MySQL填充的。第二个下拉值取决于第一个下拉选项。

无论如何，代码正在运行。现在使用我的代码我将hospital_id发布到另一个php。但我想在下拉列表中显示hospital_name作为文本，但截至目前我只能显示hospital_id。

请在下面看到我的代码并建议我一个解决方案：

$query = "SELECT bp_id,bp_name FROM mfb_billing";
  $result = $db->query($query);

  while($row = $result->fetch_assoc()){
    $categories[] = array("bp_id" => $row['bp_id'], "val" => $row['bp_name']);
  }

  $query = "SELECT bp_id, hospital_id, hospital_name FROM mfb_hospital";
  $result = $db->query($query);

  while($row = $result->fetch_assoc()){
    $subcats[$row['bp_id']][] = array("bp_id" => $row['bp_id'], "val" => $row['hospital_id']);
  }

  $jsonCats = json_encode($categories);
  $jsonSubCats = json_encode($subcats);

这是剧本：

<script type='text/javascript'>
      <?php
        echo "var categories = $jsonCats; \n";
        echo "var subcats = $jsonSubCats; \n";
      ?>
      function loadCategories(){
        var select = document.getElementById("categoriesSelect");
        select.onchange = updateSubCats;
        for(var i = 1; i < categories.length; i++){
          select.options[i] = new Option(categories[i].val,categories[i].bp_id);          
        }
      }
      function updateSubCats(){
        var catSelect = this;
        var catid = this.value;
        var subcatSelect = document.getElementById("subcatsSelect");
        subcatSelect.options.length = 0; //delete all options if any present
        for(var i = 0; i < subcats[catid].length; i++){
          subcatSelect.options[i] = new Option(subcats[catid][i].val,subcats[catid][i].hosp);
        }
      }
    </script>

这是我的表格：

<body onload='loadCategories()'>
   <form id="reportvalue" action="testpj2.php" method="post">

    <select id='categoriesSelect'>
        <option value="1">Select Billing Provider</option>
    </select>

    <select name='hospitalname[]' id='subcatsSelect' multiple='multiple'>
        <option value="all">Select Billing Provider</option>
    </select>
<label for="from">From</label>
<input type="text" id="from" name="from">
<label for="to">to</label>
<input type="text" id="to" name="to">
<?php
//$a = $_REQUEST['hospitalname[]'];
//echo $a;
?>
   <input type="submit" name="Submit" value="Submit">
  </form>

Answer 1

HTML

 <select id="someId"></select>

的Javascript

 document.getElementById('someId').innerHTML="
 <option value='value1'>"+option1+"</option>
 <option value='value2'>"+option2+"</option>
 <option value='value3'>"+option3+"</option>";

更新

来自SQL查询的动态数据，

尝试在执行查询后输入这些代码。用期望的动态值替换$ row ['value']和$ row ['option']。

 echo"<script type='text/javascript'> var str = '' </script>";
 while($row = $result->fetch_assoc())
 {
     echo"<script type='text/javascript'>
    str = str + '<option value='+".$row['value']."+'>'+".$row['option']."+'</option>   
    </script>";
 }
 echo"<script type='text/javascript'>
     document.getElementById('someId').innerHTML = str
    </script>";

Answer 2

如果您获得医院ID，那么通过从mysql获取它来填充所选值非常容易

 $query = "SELECT hospital_id, hospital_name FROM mfb_hospital where hospital_id = '$_post['hospital_id']'";
  $result = $db->query($query);
$res = mysql_fetch_array($result)

在您选择的选项值代码中输入此代码

<option value="your ids here will populate" <?php if($hospital_array['hospital_id']==$res['hospital_id']) echo "selected=selected" ?>>Your hospital name will display here which you selected</option>

如何使用Javascript在Dropdown中显示选项名称

2 个答案: