ajax详解

虽然ajax资料泛如烟海，但是许多都对新人不友好。有的缺少一些关键文件，给新人造成了困扰；有的直接就是抄袭，百度到的许多资料都是雷同。。。这里我特地写一篇关于ajax的博客，既是给自己总结经验，也是给新手提供帮助。

jquery实现

首先是html的表单，用了bootstrap

<div class="form-group has-feedback" id="username-div">
        <label class="control-label" for="username"></label>
        <input type="text" class="form-control" id="username" placeholder="用户名" >
        <span class="help-block" id="username-hint"></span>
        <span class="glyphicon form-control-feedback" id="username-iconhint"></span>
</div>

<script>
    $("#username").focus('on',function() {
        $("#username-hint").html('请输入用户名');
    });
    
    $("#username").keyup('on',function() {

        $.get("../controller/CheckLogin.Action.php?username=" + $("#username").val(),function(data,status){
            if(data == '1') {
                $("#username-hint").html('有效账号');
                $("#username-iconhint").attr('class','glyphicon form-control-feedback glyphicon-ok');
                $("#username-div").attr('class','form-group has-feedback has-success');
            }
            else {
                $("#username-hint").html('非法用户');
                $("#username-iconhint").attr('class','glyphicon form-control-feedback glyphicon-remove');
                $("#username-div").attr('class','form-group has-feedback has-error');
            }
        });


    });
</script>

function checkUserName($username) {
  $sql = "SELECT * FROM User WHERE user_name='{$username}'";
  $flag = null;


  if( !checkUserNameFormat($username) )
    $flag = 0;
  else if(strlen($username)<5 || strlen($username)>25)
    $flag = 0;
  else if( !fetchOne($sql) )
    $flag = 0;
  else
    $flag = 1;
  return $flag;
}
if( $_GET['username'] ) {
  $res = checkUserName( $_GET['username'] );
  if( $res ) $_SESSION['username'] = $_GET['username'];
  echo $res;
}

（编辑：李大同）

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