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我有一个功能可以获取选择菜单的值，而且效果很好。但我正在尝试为该函数添加另一个值。所以我想我会使用 title 属性作为选项（请参阅下面的代码）。问题是我的 JavaScript 函数中的用户名参数是undefined.

有人对我做错了什么有任何想法吗？

形式

<form action="">
    <select id="acyear" name="acyear" onchange="showyearlogdays(this.value, this.title)">
    <option value="" label="">- Year -</option>
<?php

$is_business_result = mysql_query('SELECT DISTINCT(academic_year)FROM holiday_entitlement_business_manual WHERE employee = \'' . $username . '\''); 


    while($acyear_filter = mysql_fetch_array($is_business_result)) {
    echo '<option value="'.$acyear_filter['academic_year'].'" title="'.$username.'"';

    $datestr = $acyear_filter['academic_year'];
    $currentyear = substr($datestr, 0, 4);

    if(intval(substr($datestr,4,2)) < 8){$ayear = ($currentyear - 1).'/'.$currentyear;}
    else{$ayear = ($currentyear).'/'.($currentyear + 1);}       
        echo '>';

    echo $ayear;

    echo '</option>';
    }

?>    
    </select>
</form>

Javascript

   function showyearlogdays(str, username)
 {
 if (str=="")
   {
   document.getElementById("txtHint").innerHTML="";
   return;
   } 
 if (window.XMLHttpRequest)
   {// code for IE7+, Firefox, Chrome, Opera, Safari
   xmlhttp=new XMLHttpRequest();
   }
 else
   {// code for IE6, IE5
   xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
   }
 xmlhttp.onreadystatechange=function()
   {
   if (xmlhttp.readyState==4 && xmlhttp.status==200)
     {
     document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
     }
   }
 xmlhttp.open("GET","days_yearlog.php?username="+username+"&q="+str,true);
 xmlhttp.send();
 }
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1 回答 1

2

您需要获取title所选选项的属性。您的代码指向标签的title属性select。进行以下更改：

showyearlogdays(this.value, this.options[this.selectedIndex].title)

You should also address the security concern mentioned in the comments. The way your query is setup would make for a really simple SQL Injection attack. If you don't want to rearchitect it the way the commenter suggested, I would at least escape $username so that SQL can't be injected.

于 2012-08-16T16:05:52.760 回答