如何使用萤火虫进行调试并显示单选按钮的值 [英] How to debug using firebug and show the value of radio button

问题描述

 < div id =lensType> 
 
< input type =radioname =designstyle =vertical-align：middlevalue =1/> 
< label for =design> Single Vision< / label>< br /> 
 
< input type =radioname =designstyle =vertical-align：middlevalue =2/> 
< label for =material>调节支持< / label>< br /> 
 
< input type =radioname =designstyle =vertical-align：middlevalue =3/> 
< label for =design> Bifocal< / label> <峰; br /> 
 
< input type =radioname =designstyle =vertical-align：middlevalue =4/> 
< label for =material> Varifocal（Intermediate / Near）< / label>< br /> 
 
< input type =radioname =designstyle =vertical-align：middlevalue =5/> 
< label for =material> Varifocal（Distance / Near）< / label> 
 
< / div> 
  

我正在做一个动态选择。我有我的JavaScript代码发布的价值。现在看来供应商的价值已经发布了。下面是我的脚本的代码。

  $（document）.ready（function（）{
 
函数populate（）{
 fetch.doPost（'getSupplier.php'）; 
} 
 
 $（'＃lensType'）。change（populate）; 
 
 var fetch = function（）{
 
 var counties = $（'＃county'）; 
 
 return {
 doPost：function（src ）{
 $ b $（'＃loading_county_drop_down'）。show（）; //显示Loading ... 
 $（'＃county_drop_down'）。hide（）; //隐藏（隐藏）隐藏无县消息（如果是这种情况）
 
 $（'＃no_county_drop_down'）。hide（）; .post（src，{supplier：$（'＃lensType'）.val（）}，this.getSupplier）; 
 
 else new error（'No source was passed！'）; 
 $ 
 $ b getSupplier：function（results）{
 if（！results）return; 
 counties.html（results）; 
 
 $ （'#loading_county_drop_down'）。hide（）; //隐藏加载... 
 $（'＃county_drop_down'）。s how（）; //显示下拉列表
} 
} 
}（）; 
 
 populate（）; 
 
}）; 
  

PHP代码：

 <？php 
 if（isSet（$ _ POST ['supplier']））{
 
 include'db.php'; 
 $ b $ stmt = $ mysql-> prepare（SELECT DISTINCT SupplierBrand FROM plastic WHERE HeadingNo ='。$ _ POST ['supplier']。'）; 
 $ stmt-> execute（）; 
 $ stmt-> bind_result（$ supplierBrand）; 
 
 while（$ row = $ stmt-> fetch（））：？> 
 
< option value =<？php echo $ supplierBrand;？> width =100px><？php echo $ supplierBrand; ？>< /选项> 
  

我的问题是当我调试我注意到没有值传递给PHP脚本，这使得选择空。我试图通过萤火虫输出console.log跟踪或调试，并在这方面的失败。
请协助使用这个代码来显示一个单选按钮的动态列表。

解决方案

你得到的是div的值，而不是单选按钮：

$ $ $''code $（'＃lensType'）。val（） < ---更改

类似于这样的内容：

  $（＃lensType input：radio [name ='design']：checked）。val（）
  

I have radio buttons as shown below.

        <div id="lensType">

                        <input type="radio"  name="design" style="vertical-align: middle"  value="1"/>
                        <label for="design">Single Vision</label><br/>

                        <input type="radio" name="design" style="vertical-align: middle" value="2" />
                        <label for="material" >Accommodative Support</label><br/>

                        <input type="radio"  name="design" style="vertical-align: middle"  value="3"/>
                        <label for="design">Bifocal</label> <br/>

                        <input type="radio"   name="design" style="vertical-align: middle" value="4" />
                        <label for="material" >Varifocal (Intermediate/Near)</label><br/>

                        <input type="radio"   name="design" style="vertical-align: middle" value="5"/>
                        <label for="material" >Varifocal (Distance/Near)</label>

                    </div>

I am making a dynamic select. I have my javascript code that post the value . It seems the supplier value is now posted. Below is the code for my script.

  $(document).ready(function(){

     function populate() {
      fetch.doPost('getSupplier.php');
   }

 $('#lensType').change(populate);

  var fetch = function() {

 var counties = $('#county');

return {
doPost: function(src) {

$('#loading_county_drop_down').show(); // Show the Loading...
$('#county_drop_down').hide(); // Hide the drop down
$('#no_county_drop_down').hide(); // Hide the "no counties" message (if it's the case)


    if (src) $.post(src, { supplier: $('#lensType').val() }, this.getSupplier);

    else throw new Error('No source was passed !');
},

getSupplier: function(results) {
    if (!results) return;
    counties.html(results);

$('#loading_county_drop_down').hide(); // Hide the Loading...
$('#county_drop_down').show(); // Show the drop down
}   
  }
 }();

 populate();

 });

Php code :

<?php
  if(isSet($_POST['supplier'])) {

   include 'db.php';

  $stmt = $mysql->prepare("SELECT DISTINCT SupplierBrand FROM plastic WHERE          HeadingNo='".$_POST['supplier']."'");
  $stmt->execute();
  $stmt->bind_result($supplierBrand);

  while ($row = $stmt->fetch()) : ?>

 <option value="<?php echo $supplierBrand; ?>" width="100px"><?php echo $supplierBrand; ?></option>

My problem is when I debug I notice there is no value passed to the php script and this makes the select empty. I have tried to trace or debug by having firebug output the console.log and failed in this regard. Please assist with this code which is meant to show a dynamic list from a radio button selection.

解决方案

In your javascript you are getting the value of the div, not the radio button:

$('#lensType').val() <--- change that

To something like this:

$("#lensType input:radio[name='design']:checked").val()

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