我有一个 javascript，可以从 mysql 获取数据并将其显示在我的 html 表单中，而无需刷新页面。鉴于我对 javascript 不了解，我得到了以下脚本，并得到了 StackOverflow 社区@Brant Olsen 的大力帮助。
该脚本可以完美地获取 3 个 mysql 结果字段。我想添加另外 2 个字段并将其显示在我的 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);
?>
我已经用我想要传递到 html 表单的附加字段更改了 getdata1.php:
$sql="SELECT Category, Description,SellingUnits,Grouping,CasesPerPallet,ShrinksPerPallet,if(SellingUnits='cs', CasesPerPallet,ShrinksPerPallet) as SUQTY FROM skudata WHERE packcode = '".$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，我正在努力正确编码另外两个字段。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>
我按如下方式编辑了 JavaScript，但没有成功，有人可以帮忙吗？
<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， 瑞安
最佳答案
我知道您不会喜欢这个答案，因为它需要您重新编写代码，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数据都应该使用 >= PHP5.3 中原生的 JSON。这样做的优点是 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。而且以后改变也非常容易。如果您需要再次更改它，这是完美的。
注意:如果您的 PHP 版本低于 5.3 并且您尚未安装 JSON 插件，请不要使用以下代码。此外，旧版浏览器中的 JSON 对象并不是原生的。 http://caniuse.com/json
PHP 端
使 while 循环内的 echo 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 端
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 || '';
关于php - javascript - 从 mysql 获取附加字段并显示在表中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10000526/