-1

I'm trying to replace a select element. However, instead of replacing the select with id "Base" a new select is generated within that select with the same ID. The relevant parts of my code are below. What part do you think is generating the duplicate select? The problem can be seen at this site when searching. http://bahcalculator.org/nano/

<select class="base" id="Base" onChange="codeAddress2();">
    <option>Select State First...</option>
</select>

function getbases2(){
    var S=$('#search').val();
    $.post("getbases2.php", {S:S},
        function(data) {
            $('#Base').html(data);
        });
} 
//getbases2.php

$S=$_POST['S'];
$re=mysql_query(" SELECT * FROM state_bases WHERE Base='$S' ");
$ro=mysql_fetch_assoc($re);
extract($ro);

echo "
     <select class='base' id='Base' onchange='codeAddress2();'>
         <option>Select base</option>
    ";

$re=mysql_query(" SELECT Base FROM state_bases WHERE State='$State' ORDER BY Base ASC");
while ($ro=mysql_fetch_assoc($re)){
    extract($ro);
    if ($Base!=""){
    echo "
        <option
        ";
        if ($Base=="$S"){ echo " selected='selected' ";}
            echo " 
            >$Base</option>
        ";
    }
}
echo "
    </select>
    <script type='text/javascript'>
        $('#State').val('$State');
            codeAddress2();
    </script>
    ";
4

2 回答 2

1

该$.html()调用将更新元素的内容：它将“数据”插入到选择中。您应该$.replaceWith()改用：http ://api.jquery.com/replaceWith/ 。

于 2013-08-18T21:07:54.727 回答
1

好吧，它正在按照您所说的做 -<select>在 #Base 中生成一个<select>. 如果您想更换顶部，请尝试<select>用 a包围顶部。<div id="wrap-around"></div>

于 2013-08-18T21:12:34.170 回答