请教个AJAX问题
- JScript code
<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> function nums(){ var n = document.natform.num.value; var url="index.php?c=nat&a=natid&id="+n+""; if(n == ''){ document.getElementById('num').style.display='block'; document.getElementById('num').innerHTML='编号不能为空'; document.getElementById('num').style.color='red'; return false; }else{ xmlHttpRequest = createXmlHttpRequest(); xmlHttpRequest.onreadystatechange = handle; xmlHttpRequest.open("GET",url,true); xmlHttpRequest.send(null); //我的问题就是 在这个else里怎么接收到 handle()的返回值(true false) } } function handle(){ if(xmlHttpRequest.readyState == 4){ var re = xmlHttpRequest.responseText; if(re){ document.getElementById('num').style.display='block'; document.getElementById('num').innerHTML='编号已存在'; document.getElementById('num').style.color='red'; return false; }else{ document.getElementById('num').style.display='block'; document.getElementById('num').innerHTML='可以使用'; document.getElementById('num').style.color='green'; return true; } } }
以上是一个ajax 判断 id 是否存在的, 如果不存在,可以提交,如果存在提示已存在,不能提交
- JScript code
<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> function saves(){ if(nums()){ document.natform.action="index.php?c=nat&a=addnatpost"; document.natform.submit(); } }
但是现在问题是 num() 接收不到 handle() 的返回值, 所以已存在的时候 num() 状态是未定义的, 请教下这块怎么改
------解决方案--------------------
- JScript code
var re = xmlHttpRequest.responseText; //这里的re值是多少?alert看看。//如果没有值,说明是后端接收处理的问题<br><font color="#e78608">------解决方案--------------------</font><br>ajax 是异步通讯方式,所以你不可能从 num() 得到正确的返回值<br>你所需要的提交动作应在 handle 中完成<div class="clear"> </div>

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