0

我有一个表单，用户根据出现的州和城市选择框或文本框选择国家。

如果在国家/地区INDIA被选中，那么它应该使用州和城市选择填充选择框，否则城市和州的文本框将被写入。我能够实现这一点，但它有两个问题 -

  • 第一个是当用户选择任何一个国家时，隐藏部分都有空间
  • 第二个表格只提交给其他国家，而不是印度

脚本和html中可能有什么问题？下面是代码 -

<script type="text/javascript">
    function loadbox()
    {
    //var cnty=document.getElementById('country').slected;
    var x=document.getElementById("country1").selectedIndex;
    var y=document.getElementsByTagName("option")[x].value;
    if(y==22)
    {
    document.getElementById("selectbox").style.visibility = "visible";
    document.getElementById("textbox4cnty").style.visibility = "hidden";
    }
    else {
    document.getElementById("selectbox").style.visibility = "hidden";
    document.getElementById("textbox4cnty").style.visibility = "visible";
    }
    }
    </script>

html部分

    <label for="country">COUNTRY</label>
<select id="country1" name="country1" onChange="loadbox()">         
<option value="" selected="selected" />SELECT</option>
<?php
$sql="SELECT * FROM  `country` ";
$result = mysql_query($sql);
while($row=mysql_fetch_array($result))
{
 ?>
<option value="<?php echo $row['country_id']; ?>">
<?php echo $row['country_name']; ?></option>
<?php                                            
  }
 ?>                                  
<option value="others">Others</option>
</select>         
<div id="selectbox" style="visibility:hidden">
<label for="State">STATE </label>
<select id="state1" name="state1"
onchange="getCity('select_city.php?state_id='+this.value)">
<option value="">SELECT</option>
<?php
$sql1="SELECT * FROM  `state`  ";
$result1 = mysql_query($sql1);
 while($row1=mysql_fetch_array($result1))
{
?>
<option value="<?php echo $row1['state_id']; ?>">
<?php echo $row1['state_name']; ?></option>
<?php
}
?>
</select>
<label for="STREET">CITY </label>
<select id="city1" name="city1">
<option value="">select</option>
 </select> <br/>
<div>
<div id="textbox4cnty" style="visibility:hidden">
<label for="State">STATE </label>
<input type="text" placeholder="STATE" name="state1" required><br />
<label for="STREET">CITY </label>
<input type="text" placeholder="CITY" name="city1" required><br />
</div>
4

2 回答 2

0

你的代码有问题。您使用页面上所有选项的通用选择

以下是我如何在不深入挖掘并保留内联事件处理程序的情况下将其从头顶上编码出来。它应该不引人注目，但一次只做一件事

function loadbox(sel)     {
  var isIndia=sel.options[sel.selectedindex].text==="INDIA";
  document.getElementById("selectbox").style.display = (isIndia)?"":"none";
  document.getElementById("textbox4cnty").style.display = (isIndia)?"none":"";
  // and because you do not have ID on the second field names state1
  document.getElementsByName("state1")[0].disabled=!isIndia; 
  document.getElementsByName("state1")[1].disabled=isIndia;
  document.getElementsByName("city1")[0].disabled=isIndia;
}

使用

<select id="country1" name="country1" onChange="loadbox(this)">

和

<div id="selectbox" style="display:none">

和

<div id="textbox4cnty" style="display:none">
于 2013-06-10T07:56:11.120 回答
0

改变

if(y==22)

    {

    document.getElementById("selectbox").style.visibility = "visible";

    document.getElementById("textbox4cnty").style.visibility = "hidden";

    }

    else {

    document.getElementById("selectbox").style.visibility = "hidden";

    document.getElementById("textbox4cnty").style.visibility = "visible";

    }

至

if(y==22)

    {

    document.getElementById("selectbox").style.visibility = "visible";

    document.getElementById("textbox4cnty").style.visibility = "hidden";

    document.getElementById("selectbox").disabled=false;
    document.getElementById("textbox4cnty").disabled=true;

    }

    else {

    document.getElementById("selectbox").style.visibility = "hidden";

    document.getElementById("textbox4cnty").style.visibility = "visible";

    document.getElementById("selectbox").disabled=true;
    document.getElementById("textbox4cnty").disabled=false;

    }
于 2013-06-10T06:50:30.147 回答