Ajax不工作？

Question

function submitForm()
{
    $.ajax({
        type: 'GET', 
        url: 'iccrefresh.php', 
        data: $('this').serialize(),
        dataType:'script',
        error: function()
                { $( "#dialog_error" ).dialog( "open" ); },
        success: function()
                { $( "#dialog_success" ).dialog( "open" ); }
            });
    return false;
}

在Php中

echo "<form name='refresh' onsubmit='return submitForm();'>";
echo "<input type='hidden' name='team1' value=$teamx />";
echo "<input type='hidden' name='team2' value=$teamy />";
echo "<input type='submit' name='submit' align='middle' value='Refresh Match Score'/>";
echo "</form>";

有什么理由说我无法派出1队和2队？我收到了成功消息，但iccrefresh无法访问团队1和团队2的值？

Answer 1

不存在名称为this的标记。您必须写$(this)而不是$('this') ...

但这也是错误的，您必须将this（元素）传递给函数，然后将其用作“ selector ”：

function submitForm(form)
{
    $.ajax({
        type: 'GET', 
        url: 'iccrefresh.php', 
        data: $(form).serialize(),
        dataType:'script',
        error: function()
                { $( "#dialog_error" ).dialog( "open" ); },
        success: function()
                { $( "#dialog_success" ).dialog( "open" ); }
            });
    return false;
}

在PHP中：

echo "<form name='refresh' onsubmit='return submitForm(this);'>";
echo "<input type='hidden' name='team1' value='$teamx' />";
echo "<input type='hidden' name='team2' value='$teamy' />";
echo "<input type='submit' name='submit' align='middle' value='Refresh Match Score'/>";
echo "</form>";

Answer 2

您不应将其用引号括起来，$('this')会找到标记名称this，而不会引用当前的form对象。

更改

data: $('this').serialize(),

到的

data: $(this).serialize(),

此外，您使用javascript绑定事件，因此您需要明确传递源。

HTML 的

echo "<form name='refresh' onsubmit='return submitForm();'>";  

的Javascript

function submitForm(formobj)
{
    $.ajax({
        type: 'GET', 
        url: 'iccrefresh.php', 
        data: $(formobj).serialize(),
        dateType:'script',
        error: function()
                { $( "#dialog_error" ).dialog( "open" ); },
        success: function()
                { $( "#dialog_success" ).dialog( "open" ); }
            });
    return false;
}

Answer 3

删除数据类型'script'。您正从请求中获取html，因此jquery应该为您解析它

Answer 4

试试这个：

echo "<form name='refresh' onsubmit='return submitForm();'>";
echo "<input type='hidden' name='team1' value='".$teamx."' />";
echo "<input type='hidden' name='team2' value='".$teamy."' />";
echo "<input type='submit' name='submit' align='middle' value='Refresh Match Score'/>";
echo "</form>";