javascript – 当选择值更改时,ajax调用从数据库查询填充表单字段

我一直在查看这里的问题,无法找到我所追求的确切答案:(但我已经设法得到了一些东西.

<select style="width:100%;" class="quform-tooltip chosen-select" id="company_select" name="company_select" title="Company Select" onChange="showUser(this.value)">
<option value="">Please select</option>
<?PHP
$userID = $user->getUserID();
$query = $user->database->query("SELECT * FROM tbl_businesses as business LEFT JOIN tbl_user_businesses as biz_user ON business.businessID = biz_user.businessID WHERE biz_user.userID ='$userID'");

while($row=$user->database->fetchArray($query))
{
    $bizID = $row['businessID'];
    $bizName = $row['businessName'];
    echo "<option value='$bizID'>$bizName</option>";
}?>
</select>

然后目前有2个其他文本框(最终可能会增加),我想在更改/选择上面的选择框值时填充

<input id="company_name" type="text" name="company_name" value="" />
<input id="company_email" type="text" name="company_email" value="" />

所以我在我的选择框上有一个onchange功能就是这个

<script>
function showUser(str)
{
if (str=="")
{
    document.getElementById("company_name").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)
    {
        var data = JSON.parse(xmlhttp.responseText);
        for(var i=0;i<data.length;i++) 
        {
          document.getElementById("company_name").innerHTML += data[i].id + ' - ' + data[i].name + ' - ' + data[i].web;
        }
    }
}
xmlhttp.open("GET","formdata.PHP?q="+str,true);
xmlhttp.send();
}
</script>

和我的formdata.PHP 文件是这样的

    <?PHP
include("include/user.PHP");

$q = intval($_GET['q']);

$sql="SELECT * FROM tbl_businesses WHERE businessID = '".$q."'";

$result = $user->database->query($sql);
$info = array();
while($row=$user->database->fetchArray($result))
{
    $cID = $row['bussinessID'];
    $cName = $row['businessName'];
    $cWeb = $row['businessWebsite'];
    $info[] = array( 'id' => $cID, 'name' => $cName, 'web' => $cWeb );
}
echo json_encode($info);?>

这正在使ajax调用正确并返回预期的数据但我现在需要帮助填充文本框值？

任何人都可以请帮助我这一点,已经花了很长时间试图搞清楚,我不熟悉javascript / json所以不知道从哪里开始

我希望company_name文本框值设置为$cName;和
company_email文本框值设置为$cWeb;

感谢任何帮助

卢克

解决方法:

好吧我使用的解决方案,对于任何想要知道我如何解决它的人来说

我的index.PHP包含javascript和表单代码

javascript代码

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js" type="text/javascript"></script>

<script>
function showUser(str)
{
if (str=="")
{
    document.getElementById("company_name").value="";
    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)
    {
        var data = JSON.parse(xmlhttp.responseText);
        for(var i=0;i<data.length;i++) 
        {
          document.getElementById("company_name").value = data[i].name;
          document.getElementById("company_email").value = data[i].web;
        }
    }
}
xmlhttp.open("GET","formdata.PHP?q="+str,true);
xmlhttp.send();
}
</script>

和表格代码

    <select style="width:100%;" class="quform-tooltip chosen-select" id="company_select" name="company_select" title="Company Select" onChange="showUser(this.value)">
<option value="">Please select</option>
<?PHP
$userID = $user->getUserID();
$query = $user->database->query("SELECT * FROM tbl_businesses as business LEFT JOIN tbl_user_businesses as biz_user ON business.businessID = biz_user.businessID WHERE biz_user.userID ='$userID'");

while($row=$user->database->fetchArray($query))
{
    $bizID = $row['businessID'];
    $bizName = $row['businessName'];
    echo "<option value='$bizID'>$bizName</option>";
}?>
</select>

<input id="company_name" type="text" name="company_name" value="" />
<input id="company_email" type="text" name="company_name" value="" />

然后我的formdata.PHP

    $q = intval($_GET['q']);

$sql="SELECT * FROM tbl_businesses WHERE businessID = '".$q."'";

$result = $user->database->query($sql);
$info = array();
while($row=$user->database->fetchArray($result))
{
    $cID = $row['businessID'];
    $cName = $row['businessName'];
    $cWeb = $row['businessWebsite'];
    $info[] = array( 'id' => $cID, 'name' => $cName, 'web' => $cWeb );
}
echo json_encode($info);?>

多数民众赞成,感谢charlietfl的帮助！

希望这有助于某人:)

javascript – 当选择值更改时,ajax调用从数据库查询填充表单字段

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