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PHP AJAX example automatically completes address information based on zip code_PHP tutorial
- WBOYOriginal
- 2016-07-21 15:48:311068browse
<script> <br>function createRequestObject() { <br>var ro; <br>var browser = navigator.appName; <br>if(browser == "Microsoft Internet Explorer"){ <br>ro = new ActiveXObject("Microsoft.XMLHTTP"); <br>}else{ <br>ro = new XMLHttpRequest(); <br>} <br>return ro; <br>} <br>var http = createRequestObject(); <br>function sndReq(zip) { <br>http.open('get', 'zipcode.PHP?zip='+zip); <br>http.onreadystatechange = handleResponse; <br>http.send(null); <br>} <br>function handleResponse() { <br>if(http.readyState == 4){ <br>var response = http.responseText; <br>var update = new Array(); <br>if(response.indexOf('|' != -1)) { <br>update = response.split('|'); <br>document.getElementById("city").value = update[0]; <br>document.getElementById("state").value = update[1]; <br>} <br>} <br>} <br></script>
Enter A United States Zipcode, Then Tab
| Enter Zipcode: | |
| City: | |
| State: |
以上是客户输入页面，下面是服务端的处理页面'zipcode.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];