Question

这是来自另一页的ajax数据：

<? $cat=strval($_GET['cat']);
//Connect to mysql server
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
    die('Failed to connect to server: ' . mysql_error());
}
//Select database
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
    die("Unable to select database");
}
$query="SELECT * FROM cats WHERE cat_id='$cat'";
$result=mysql_query($query);

?>
<select name="sub_cat">
<option>Select Sub-Category</option>
<? while($row2=mysql_fetch_array($result)) { ?>
<option value="<?=($row2['sub_cat_id'])?>"><?=$row2['sub_cat']?></option>
<? } ?></select>

这是显示在其上的页面：

<select name="cat" onChange="getSub_cat(this.value)">
    <option>Select Category</option>
    <?php
    //Open database connection
    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    if(!$link) {
    die('Failed to connect to server: ' . mysql_error());
                                                }
    //Select database
    $db = mysql_select_db(DB_DATABASE);
    if(!$db) {
    die("Unable to select database");
    }
    //Select categories
    $sql = "SELECT DISTINCT cat, cat_id FROM cats ";
    $rs = mysql_query($sql);
    while($row = mysql_fetch_array($rs))
    {
    echo "<option value=\"".$row['cat_id']."\">".$row['cat']."</option>\n  ";
    }
    ?>
    </select>

    <b>Sub-Category</b></td>
    <div id="sub_catdiv"><select name="sub_cat" id="sub_cat"></div>
    <option>Select Category First</option>
    </select>
    </div>

这是它传递的网址：

https://mysitewontworkcom/somepagephp?state=state&cat=cat&submit=submit

从ajax拉入表单的sub_cat根本没有显示在URL中！

“状态”（表格中较早）和“猫”拉得好吗？有任何想法吗？请帮忙！

function getXMLHTTP() { //function to return the xml http object
var xmlhttp=false;  
    try{
        xmlhttp=new XMLHttpRequest();
            }
            catch(e)    {       
            try{            
        xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
        }
        catch(e){
        try{
        xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
                }
                    catch(e1){
                        xmlhttp=false;
                        }
                            }
                               }

                    return xmlhttp;
                                }

                function getSub_cat(cat_id) {       

            var strURL="findCat.php?cat="+cat_id;
            var req = getXMLHTTP();

            if (req) {

                 req.onreadystatechange = function() {
            if (req.readyState == 4) {
                // only if "OK"
            if (req.status == 200) {
        document.getElementById('sub_catdiv').innerHTML=req.responseText;                       
                        } else {    
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
        }
                    }               
                            }           
    req.open("GET", strURL, true);
                        req.send(null);
                                }       
                                    }

Answer 1

在显示数据的页面上，html结构出错：

<div id="sub_catdiv"><select name="sub_cat" id="sub_cat"></div>
<option>Select Category First</option>
</select>

更改为：

<div id="sub_catdiv">
</div>

正如您所看到的那样，您首先关闭div而不是添加option，而不是关闭似乎是主要问题的select。

URL不会从ajax输入传递第二个“select”值

1 个答案: