将jquery ajax中的许多变量提交给php时遇到了麻烦

This JS is giving me some problems. Specifically FF3.5 is saying there is Error: missing ; before statement and its pointing at the 'comsn=' + comsn section. I know enough js to get myself into trouble, but specifics, im not there yet. Any ideas?

这个JS给我带来了一些问题。具体来说，FF3.5说有错误:缺失;在语句之前，它指向'comsn=' + comsn部分。我知道有足够多的js让我陷入麻烦，但具体来说，我还没到那一步。什么好主意吗?

<script type="text/javascript" >
$(function () {
    $(".submit").click(function () {
        var comsn = $("#comsn").val();
        var comrn = $("#comrn").val();
        var compic = $("#compic").val();
        var comment = $("#comment").val();
        var eventid = $("#eventid").val();
        var comuserid = $("#comuserid").val();
        var owner = $("#ownerid").val();
        var dataString = 'comsn=' + comsn '&comrn=' + comrn '&compic=' + compic '&comment=' + comment '&eventid=' + eventid '&comuserid=' + comuserid '&owner=' + owner;
            if (comment == '') {
                alert('Must Type Comment to Post Comment');
            }else{
                $("#flash").show();
                $("#flash").fadeIn(400).html('<img src="assets/uploading.gif" />Loading Comment...');
                $.ajax({
                    type: "POST",
                    url: "comments_post.php",
                    data: dataString,
                    cache: false,
                    success: function (html) {
                        $("ol#update").prepend(html);
                        $("ol#update li:last").fadeIn("slow");
                        $("#flash").hide();
                    }
                });
            }return false;
    });
});
</script>

3 个解决方案

#1

var dataString = 'comsn=' + comsn + '&comrn=' + comrn + '&compic=' + compic + '&comment=' + comment + '&eventid=' + eventid '&comuserid=' + comuserid + '&owner=' + owner;

You've got a lot of missing +.

你有很多东西不见了。

Also jquery have a method called serialize which do the same think you are doing by hand probably worth to give a try.

此外，jquery还有一个名为serialize的方法，它可以实现与手工操作相同的功能，可能值得一试。

#2

You have missing + signs in your concatenation:

您在连接中缺少+符号:

var dataString = 'comsn=' + comsn + '&comrn=' + comrn +
                 '&compic=' + compic + '&comment=' + comment +
                 '&eventid=' + eventid + '&comuserid=' + comuserid +
                 '&owner=' + owner;

However if you want to get all your form element values in a string of data, you could use the Ajax/serialize method:

但是，如果您想在一串数据中获取所有的表单元素值，您可以使用Ajax/serialize方法:

var dataString =  $("#formId").serialize();

#3

Your missing some +s

你缺少一些+ s

+ comrn '&compic='...

should be

应该是

+ comrn + '&compic='...

#1

var dataString = 'comsn=' + comsn + '&comrn=' + comrn + '&compic=' + compic + '&comment=' + comment + '&eventid=' + eventid '&comuserid=' + comuserid + '&owner=' + owner;