我有一个javascript，它从mysql中获取数据，并在不刷新页面的情况下以html形式显示。考虑到我对javascript的不了解，我得到了以下脚本，StackOverflow社区，即@Brant-Olsen的大力帮助。
该脚本可以很好地获取3个mysql结果字段。我想再添加2个字段，并将它们显示在我的html表单中
工作脚本是：
<script type="text/javascript">
function showUser(userNumber, str)
{
if (str=="")
{
document.getElementById("txtHint" + userNumber).innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
//document.getElementById("txtHint" + userNumber).innerHTML=xmlhttp.responseText;
var responseText = xmlhttp.responseText;
var description = responseText;
var warehouse = "";
var sellingUnits = "";
if (responseText.indexOf("NOT A VALID") == -1)
{
description = responseText.substring(12, responseText.indexOf(",Warehouse:"));
warehouse = responseText.substring(responseText.indexOf(",Warehouse:")+11, responseText.indexOf(",SellingUnits:"));
sellingUnits = responseText.substring(responseText.indexOf(",SellingUnits:")+14);
}
document.getElementById("whse" + userNumber).innerHTML = warehouse;
document.getElementById("txtHint" + userNumber).innerHTML = description;
document.getElementById("su" + userNumber).innerHTML = sellingUnits;
}
}
xmlhttp.open("GET","getdata1.php?q="+str,true);
xmlhttp.send();
}
</script>
getdata1.php是：
<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'dbuser', 'dbpass');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("dbname", $con);
$sql="SELECT Category, Description,SellingUnits,Grouping,CasesPerPallet,ShrinksPerPallet FROM skudata WHERE packcode = '".$q."'";
$result = mysql_query($sql);
$rows=mysql_num_rows($result);
if($rows==0){echo "<font color=red><b>NOT A VALID PRODUCT CODE</b></font>";} else {
while($row = mysql_fetch_array($result))
{
echo "Description:" . $row['Description'] . ",Warehouse:" . $row['Grouping'] . ",SellingUnits:" . $row['SellingUnits'];
}
}
mysql_close($con);
?>
我已经修改了getdata1.php，并添加了要传递给html表单的额外字段：
$sql="SELECT Category, Description,SellingUnits,Grouping,CasesPerPallet,ShrinksPerPallet,if(SellingUnits='cs', CasesPerPallet,ShrinksPerPallet) as SUQTY FROM skudata WHERE packcode = '".$q."'";
$result = mysql_query($sql);
$rows=mysql_num_rows($result);
if($rows==0){echo "<font color=red><b>NOT A VALID PRODUCT CODE</b></font>";} else {
while($row = mysql_fetch_array($result))
{
echo "Description:" . $row['Description'] . ",Warehouse:" . $row['Grouping'] . ",SellingUnits:" . $row['SellingUnits'] . ",SUQTY:" . $row['SUQTY'] . ",Category:" . $row['Category']; ;
}
}
从这里开始，我正在努力对另外两个字段进行正确的编码。表格行的示例如下：
<tr id="r1">
<td>
<input type=checkbox name=kvi1 id=kvi1 value=1>
</td>
<td>
<input size=10 type=number id=sku1 name=sku1 onchange="showUser(1, this.value)"><a href="sku.php" target="_blank"><img src=q.png border=0></a>
</td>
<td>
<div align="left" id="txtHint1"> </div>
</td>
<td>
<div align="left" id="whse1"> </div>
</td>
<td>
<div align="left" id="su1"> </div>
</td>
<td>
<div align="left" id="suqty1"> </div>
</td>
<td>
<div align="left" id="category1"> </div>
</td>
</tr>
我编辑了如下javascript，但没有成功，有人能帮忙吗？
<script type="text/javascript">
function showUser(userNumber, str)
{
if (str=="")
{
document.getElementById("txtHint" + userNumber).innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
//document.getElementById("txtHint" + userNumber).innerHTML=xmlhttp.responseText;
var responseText = xmlhttp.responseText;
var description = responseText;
var warehouse = "";
var sellingUnits = "";
var SUQTY = "";
var Category = "";
if (responseText.indexOf("NOT A VALID") == -1)
{
description = responseText.substring(12, responseText.indexOf(",Warehouse:"));
warehouse = responseText.substring(responseText.indexOf(",Warehouse:")+11, responseText.indexOf(",SellingUnits:"));
sellingUnits = responseText.substring(responseText.indexOf(",SellingUnits:")+11, responseText.indexOf(",SUQTY:"));
suqty = responseText.substring(responseText.indexOf(",SUQTY:")+11, responseText.indexOf(",Category:"));
category = responseText.substring(responseText.indexOf(",Category:")+14);
}
document.getElementById("whse" + userNumber).innerHTML = warehouse;
document.getElementById("txtHint" + userNumber).innerHTML = description;
document.getElementById("su" + userNumber).innerHTML = sellingUnits;
document.getElementById("suqty" + userNumber).innerHTML = SUQTY;
document.getElementById("category" + userNumber).innerHTML = Category;
}
}
xmlhttp.open("GET","getdata1.php?q="+str,true);
xmlhttp.send();
}
</script>
谢谢，Ryan
我知道你不会喜欢这个答案，因为它需要你重新编写代码，但我会尽力提供帮助。
实际上，通过ajax调用在PHP和JS之间发送的任何类型的这种性质的数据都应该使用JSON，它在>=PHP5.3中是原生的。这样做的好处是Ajax和PHP可以来回传递数据，而无需在应用程序级别进行任何真正的解析。而且在路上换衣服也非常容易。如果你需要再次更改，这是完美的。
注意：如果您的PHP版本低于5.3，并且您还没有努力安装JSON插件，请不要使用下面的代码。此外，JSON对象在较旧的浏览器中不是本机的。http://caniuse.com/json
PHP端
使while循环内的回声看起来像这样：
echo json_encode(array(
"description" => $row['Description'],
"warehouse" => $row['Grouping'],
"sellingunits" => $row['SellingUnits'],
"suqty" => $row['SUQTY'],
"category" => $row['Category']
));
JS侧
var data = {};
if (responseText.indexOf("NOT A VALID") == -1) {
data = JSON.parse(responseText);
}
// Check the values or just place an empty string if undefined
document.getElementById("whse" + userNumber).innerHTML = data.warehouse || '';
document.getElementById("txtHint" + userNumber).innerHTML = data.description || '';
document.getElementById("su" + userNumber).innerHTML = data.sellingunits || '';
document.getElementById("suqty" + userNumber).innerHTML = data.suqty || '';
document.getElementById("category" + userNumber).innerHTML = data.category || '';