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Javascript jquery不刷新div_Javascript_Php_Jquery_Html - Fatal编程技术网

Javascript jquery不刷新div

Javascript jquery不刷新div,javascript,php,jquery,html,Javascript,Php,Jquery,Html,我有这个密码 HTML 我的javascript代码正在将值传递给php文件(php正在很好地更新这些值),但是,当它返回到html页面时,会显示警报,当按ok时,会刷新整个页面。我希望它只刷新div而不是整个页面。 我不知道我做错了什么,任何建议都将受到重视。您错过了函数中的返回false: function updateValues(thisForm) { // your code... xmlhttp.send(); return false; } 你能试试这个吗

我有这个密码 HTML

我的javascript代码正在将值传递给php文件(php正在很好地更新这些值),但是,当它返回到html页面时,会显示警报,当按ok时,会刷新整个页面。我希望它只刷新div而不是整个页面。
我不知道我做错了什么,任何建议都将受到重视。

您错过了函数中的
返回false

function updateValues(thisForm)
{
   // your code...
   xmlhttp.send();
   return false;
}
你能试试这个吗

        <form id="myForm" method="post" onsubmit="return updateValues(this)">
            <input id="id" type="hidden" value="<?php echo $row[0]; ?>" />
            <input type="text" id="dd" value="<?php echo $row[1];?>" />
            <input type="text" id="da" value="<?php echo $row[2];?>" />       
            <input type="text" id="db" value="<?php echo $row[3];?>" />
            <input type="submit" id="submit" value="Save Changes" />
        </form>

        <div id="successNote"></div>

        <script>

        function updateValues(thisForm)
        {

           if (thisForm == "")
           {
              document.getElementById("successNote").innerHTML = "please fill form.";
              return false;
           } 
           var id = document.getElementById('id').value;
           var dd = document.getElementById('dd').value;
           var da = document.getElementById('da').value;
           var db = document.getElementById('db').value;
           var str = 'id='+id+'&dd='+dd+'&da='+da+'&db='+db;

           if (window.XMLHttpRequest)
           {
              // code for IE7+, Firefox, Chrome, Opera, Safari
              xmlhttp = new XMLHttpRequest();
           }
           else
           {
               // code for IE6, IE5
               xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
           }
           xmlhttp.onreadystatechange=function()
           {
              if (xmlhttp.readyState==4 && xmlhttp.status==200)
              {
                document.getElementById("successNote").innerHTML = "Form has been updated";
                alert('Form has been updated successfully!');
              }
            }   
           xmlhttp.open("GET","update_myForm.php?"+str,true);
           xmlhttp.send();

           return false;
        }


        </script>


您需要防止表单提交的默认操作。此外,您根本没有使用jQuery,因为它目前的状况是这样的。
function updateValues(thisForm)
{
   // your code...
   xmlhttp.send();
   return false;
}
        <form id="myForm" method="post" onsubmit="return updateValues(this)">
            <input id="id" type="hidden" value="<?php echo $row[0]; ?>" />
            <input type="text" id="dd" value="<?php echo $row[1];?>" />
            <input type="text" id="da" value="<?php echo $row[2];?>" />       
            <input type="text" id="db" value="<?php echo $row[3];?>" />
            <input type="submit" id="submit" value="Save Changes" />
        </form>

        <div id="successNote"></div>

        <script>

        function updateValues(thisForm)
        {

           if (thisForm == "")
           {
              document.getElementById("successNote").innerHTML = "please fill form.";
              return false;
           } 
           var id = document.getElementById('id').value;
           var dd = document.getElementById('dd').value;
           var da = document.getElementById('da').value;
           var db = document.getElementById('db').value;
           var str = 'id='+id+'&dd='+dd+'&da='+da+'&db='+db;

           if (window.XMLHttpRequest)
           {
              // code for IE7+, Firefox, Chrome, Opera, Safari
              xmlhttp = new XMLHttpRequest();
           }
           else
           {
               // code for IE6, IE5
               xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
           }
           xmlhttp.onreadystatechange=function()
           {
              if (xmlhttp.readyState==4 && xmlhttp.status==200)
              {
                document.getElementById("successNote").innerHTML = "Form has been updated";
                alert('Form has been updated successfully!');
              }
            }   
           xmlhttp.open("GET","update_myForm.php?"+str,true);
           xmlhttp.send();

           return false;
        }


        </script>