Question

我有一个代码，其中我想通过ajax调用将检查的checkbpxes值发送到下一个php页面。但是我无法正确发送它。以前我曾经问过类似的问题。

以下是代码

<html>
<head>
<title>Insert title here</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
</head>
<body>

<script type="text/javascript">
    function get_check_value() {
        var c_value = [];
        $('input[name="services"]:checked').each(function () {
            c_value.push(this.value);
        });
        //return c_value.join(',');
    }
    $(document).ready(function(){
        $('#btnSubmit').on('click', function () {
            alert("hi");
            //var os = $('#originState').val();
            //var c = $('#commodity').val();
            //var ds = $('#destState').val();
            var ser = get_check_value();
            //var queryString = "os=" + os;
            var queryString = "&ser=" + ser;
            $.ajax({
           type: "GET",
       url:'sortingajax.php',
       data:"queryString="+querystring,
               success:function(data){
            alert(data);
                console.log(data);
                    $('#results').html(data);
               }
            });

        });
    });
</script>
<form name="searchForm">
    <input type="checkbox" name="services" value="twic" />TWIC
    <br/>
    <input type="checkbox" name="services" value="enclosedTrucking" />Enclosed Trucking
    <br/>
    <input type="checkbox" name="services" value="flatBedTrucking" />Flat Bed Trucking
    <br/>
    <input type="submit" id="btnSubmit" value="Submit" />
</form>
<div id="results">
</div>
</body>
</html>

ajaxcall.php

    <?php

        include('connection.php');
$query=$_GET['querystring'];
echo $query;
        $countsql='SELECT * FROM table1 ';
        $countsql1=mysql_query($countsql);
        $numrows = mysql_num_rows($countsql1);  
        $countArray2=array();

        while($row = mysql_fetch_array($countsql1)) {
            // Append to the array
            $Array2[] = $row;
        }
    ?>
    <?php
        foreach($Array2 as $array)
        {
    ?>
    <div class="search">hi</div>

    <?php 
            $i++; 
        }
    ?>

我现在无法使用ajaxcall html内容获取内容。但我无法做到这一点.. 请指导我guyzzz

Answer 1

如果我正确理解了您要实现的目标，可以在表单中添加隐藏的输入并传递值，您的脚本会将其设置为隐藏控件。

<form name="searchForm">
<input type="hidden" id="checkboxCollection">
<input type="checkbox" name="services" value="twic" />TWIC
<br/>
<input type="checkbox" name="services" value="enclosedTrucking" />Enclosed Trucking
<br/>
<input type="checkbox" name="services" value="flatBedTrucking" />Flat Bed Trucking
<br/>
<input type="submit" id="btnSubmit" value="Submit" />
</form>

然后脚本将变为

 function get_check_value() {
    var c_value = [];
    $('input[name="services"]:checked').each(function () {
        c_value.push(this.value);
    });

$('#checkboxCollection').val(c_value.join(','));
    //return c_value.join(',');
}

然后在提交表单时阅读checkboxCollection。

无法通过ajax调用将已检查的复选框值传递给下一个php页面

1 个答案: