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  • jQuery ajax的前台代码编写

    jQuery ajax的前台代码:

    <script type="text/javascript" src="/include/jquery/jquery-1.1.3.1.pack.js"></script>
    <form name="form1" id="form1" action="ajax.php?action=1" method="post">
      <input type="hidden" id="comid" name="comid" value="111">
      <input type="hidden" id="jobid" name="jobid" value="222">
      <input type="hidden" id="userid" name="userid" value="333">
      <input type="submit" /></form>
    <script language="javascript">
    $("#form1").submit
        (
           function()
            {
            login();
            return false;
            }
        );
    function login()
    {
    var comid = $("#comid").val();
    var jobid = $("#jobid").val();
    var userid = $("#userid").val();
    $.ajax({
        type: "POST",
        url: "../include/ajax.php",
        data: "action=1" + "&comid=" + comid + "&jobid=" + jobid + "&userid=" + userid,
        beforeSend: function(2881064151){
        },
        success: function(msg){
      if(msg=="success"){
    alert("求职信息,已经提交!请等候通知!")
       }else {
           alert("信息已发送,请不要重复提交!");
        }
        }
    });
    }
    </script>

    PHP的后台代码:

    <?
    header("Cache-Control: no-cache"); 
    require_once("function.php");
    switch ($_POST["action"]){
    case 1:
    $comid=verify_id($_POST["comid"]);
    $jobid=verify_id($_POST["jobid"]);
    $userid=verify_id($_POST["userid"]);
    $appdate=time();
    $hasapp=getValue("select * from csj_appjob where jobid=$jobid and userid=$userid and comid=$comid","id");
    if($hasapp==""){
    $sql="insert into csj_appjob(comid,jobid,userid,appdate) values($comid,$jobid,$userid,$appdate)";
    query($sql);
    echo "success";
    }else{
    echo "wrong";
    }
    break;
    default:
    break;

    }
    ?>
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  • 原文地址:https://www.cnblogs.com/cbryge/p/6062952.html
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