如何使用jQuery ajax post传递多个复选框

Question

如何使用jQuery ajax post传递多个复选框

这是ajax功能

 function submit_form(){
 $.post("ajax.php", {
 selectedcheckboxes:user_ids,
 confirm:"true"
 },
 function(data){
  $("#lightbox").html(data);
  });
 }

这是我的表格

<form>
<input type='checkbox' name='user_ids[]' value='1'id='checkbox_1' />
<input type='checkbox' name='user_ids[]' value='2'id='checkbox_2' />
<input type='checkbox' name='user_ids[]' value='3'id='checkbox_3' />
<input name="confirm" type="button" value="confirm" onclick="submit_form();" />
</form>

Answer 1

从POST的jquery文档(第3个例子):

$.post("test.php", { 'choices[]': ["Jon", "Susan"] });

所以我只是遍历复选框并构建数组.就像是

var data = { 'user_ids[]' : []};
$(":checked").each(function() {
  data['user_ids[]'].push($(this).val());
});
$.post("ajax.php", data);

Answer 2

刚刚遇到这个尝试找到同样问题的解决方案.实现Paul的解决方案我做了一些调整以使这个功能正常.

var data = { 'venue[]' : []};

$("input:checked").each(function() {
   data['venue[]'].push($(this).val());
});

简而言之,添加输入:checked而不是:checked将输入到数组中的字段限制为仅在表单上的复选框.保罗确实是正确的,this需要被封闭$(this)

Answer 3

可以使用以下内容然后展开post结果explode(",", $_POST['data']);以提供一系列结果.

var data = new Array();
$("input[name='checkBoxesName']:checked").each(function(i) {
    data.push($(this).val());
});

Answer 4

这是一种更灵活的方式.

让我们说这是你的形式.

<form>
<input type='checkbox' name='user_ids[]' value='1'id='checkbox_1' />
<input type='checkbox' name='user_ids[]' value='2'id='checkbox_2' />
<input type='checkbox' name='user_ids[]' value='3'id='checkbox_3' />
<input name="confirm" type="button" value="confirm" onclick="submit_form();" />
</form>

这是你下面的jquery ajax ......

                // Don't get confused at this portion right here
                // cuz "var data" will get all the values that the form
                // has submitted in the $_POST. It doesn't matter if you 
                // try to pass a text or password or select form element.
                // Remember that the "form" is not a name attribute
                // of the form, but the "form element" itself that submitted
                // the current post method
                var data = $("form").serialize(); 

                $.ajax({
                    url: "link/of/your/ajax.php", // link of your "whatever" php
                    type: "POST",
                    async: true,
                    cache: false,
                    data: data, // all data will be passed here
                    success: function(data){ 
                        alert(data) // The data that is echoed from the ajax.php
                    }
                });

在你的ajax.php中,你试着回复或print_r你的帖子,看看里面发生了什么.这应该是这样的.仅返回您选中的复选框.如果您没有选中任何内容,则会返回错误.

<?php 
    print_r($_POST); // this will be echoed back to you upon success.
    echo "This one too, will be echoed back to you";

希望足够清楚.