Ajax校验是否重复的实现代码
- 作者: 那晚越女说我?
- 来源: 51数据库
- 2021-07-28
直接贴代码!
一、controller层:
@responsebody
@requestmapping("entityfindbycode")
public string entityfindbycode(entity bean, httpsession httpsession, model m,httpservletresponse res) throws ioexception{
res.setcontenttype("text/plain; charset=utf-8");
//res.reset();
//printwriter out = res.getwriter();
string data=null;
@suppresswarnings("unchecked")
list<entity> list = (list<entity>) this.service.find("beanbycode", "entity", new object[]{bean.getcode()});
if(list.size()==0){
data="yes";
}else{
data="no";
}
res.getoutputstream().write(data.getbytes());
//out.print(data);
//out.flush();
//out.close();
return data;
}
二、html层:
$.ajax({
url :'entityfindbycode',
type :'post',
datatype :'text',
data :{code:$("#code").val()},//{key,value}
success :function(data) {
if (data == "yes"){
document.getelementbyid("msg-code").innerhtml = "(<fmt:message key='warn.field.required.unique' bundle='${bundle }' />)";
return true;
}
else {
$("#div-code").addclass("has-error");
document.getelementsbyname("code")[0].value = "";
//document.getelementsbyname("code")[0].setattribute("placeholder", "<fmt:message key='warn.duplicate' bundle='${bundle }' />");
document.getelementbyid("msg-code").innerhtml = "(<fmt:message key='warn.duplicate' bundle='${bundle }' /> !!!)";
return false;
}
}
});
以上所述是小编给大家介绍的ajax校验是否重复的实现代码,希望对大家有所帮助
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