选择值更改时，用ajax调用从数据库查询中填充表单字段

Question

我一直在看这里的问题，找不到我想要的确切答案:(但是我设法得到了一些东西。

我有一个从数据库查询中填充的表单选择字段

<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>

然后当前有2个其他文本框（最终可能会增加），当上面的选择框值被更改/选择时，我想填充该文本框

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

所以我的选择框上有一个onchange函数，这是

<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);?>

哪个使ajax调用正确并返回预期的数据，但是我现在需要帮助来填充文本框值？

谁能帮我解决这个问题，花了很多年的时间试图弄清楚，我对javascript / json不熟悉，所以不确定从哪里开始

我想将company_name文本框值设置为$ cName; 并将company_email文本框值设置为$ cWeb；

感谢任何帮助

路加

Answer 1

确定我使用的解决方案，对于任何想知道我如何解决的人

我的index.php，其中包含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>

和表单代码

    <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);?>

就是这样，感谢charlietfl的帮助！

希望这可以帮助某人:)