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php AJAX实例根据邮编自动完成地址信息

2019年05月09日  | 移动技术网IT编程  | 我要评论

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<script>
function createrequestobject() {
var ro;
var browser = navigator.appname;
if(browser == "microsoft internet explorer"){
ro = new activexobject("microsoft.xmlhttp");
}else{
ro = new xmlhttprequest();
}
return ro;
}
var http = createrequestobject();
function sndreq(zip) {
http.open('get', 'zipcode.php?zip='+zip);
http.onreadystatechange = handleresponse;
http.send(null);
}
function handleresponse() {
if(http.readystate == 4){
var response = http.responsetext;
var update = new array();
if(response.indexof('|' != -1)) {
update = response.split('|');
document.getelementbyid("city").value = update[0];
document.getelementbyid("state").value = update[1];
}
}
}
</script>
<h3>enter a united states zipcode, then tab</h3>
<table align="center">
<tr>
<td>enter zipcode:</td>
<td><input type="text" id="zipcode" name="zipcode" onblur="sndreq(this.value);"/></td>
</tr>
<tr>
<td>city:</td>
<td><input type="text" id="city" name="city"/></td>
</tr>
<tr>
<td>state:</td>
<td><input type="text" id="state" name="state"/></td>
</tr>
</table>
以上是客户输入页面,下面是服务端的处理页面'zipcode.php
<?php
$dbuser = 'root';
$dbpass = '111111';
$cn = mysql_connect("localhost", $dbuser, $dbpass);
$db = mysql_select_db("ajax");
$sql = "select city, state from zipcodes where zipcode = " . $_request['zip'];
$rs = mysql_query($sql);
$row = mysql_fetch_array($rs);
echo $row['city'] . "|" . $row['state'];
mysql_close($cn);
?>
当客户输入一个postcode后,zipcode.php就接收到它,然后进行从数据表中取出对应的资料,再按一定的格式返回给客户端(此处是以 | 分隔)。最后客户端接收返回的资料,显示在页面上。
if(response.indexof('|' != -1)) {
update = response.split('|');
document.getelementbyid("city").value = update[0];
document.getelementbyid("state").value = update[1]; 

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