Question

搜索了许多选项以找到一些简单的解决方案，我可以对结果进行排序，但却找不到任何结果。我想按选定的下拉值排序显示的结果。我不想使用“表格”。想要另一种方式对其进行排序。

<div class="col-lg-2">
<label class="margin-bottom:25px;" style="margin-left:75px;"> Sort by: <label>
</div>
<div class="col-lg-2">
    <select class="form-control" id="sortby" name="sortby">
        <option selected value="ID">ID</option>
        <option value="Name">Name</option>
        <option value="Source">Source</option>
        <option value="Location">Location</option>
    </select>
</div>

以上是我的下拉列表。以下是我们的$ sql初始查询：

$condition = implode(' AND ', $query);
$sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source,candidate_contact.cand_location FROM candidate ".$join.' where '.$condition;

现在我们已经尝试过这个，但似乎有些不对劲。

if($_POST['sortby']=="ID")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source,candidate_contact.cand_location FROM candidate ".$join.' where '.$condition." order by candidate.cand_number asc";
}

if($_POST['sortby']=="Name")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source,candidate_contact.cand_location FROM candidate ".$join.' where '.$condition." order by candidate.cand_fname asc";
}

if($_POST['sortby']=="Source")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source,candidate_contact.cand_source FROM candidate ".$join.' where '.$condition." order by candidate.cand_source asc";
}

if($_POST['sortby']=="Location")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source,candidate_contact.cand_source FROM candidate_contact ".$join.' where '.$condition." order by candidate.cand_location asc";
}

asc-按升序排序;

我的jquery / ajax脚本 -

<script>
        $(document).ready(function(){
            // Each time you change your sort list, send AJAX request
            $("#sortby").change(function(){
                $.ajax({
                    method: "POST",
                    url: "viewcandidate.php",
                    data: { sortby:$("#sortby").val() }
                })
                // Copy the AJAX response in the table
                .done(function( msg ) {
                    $("#list").html(msg);
                });
            });
        });
    </script>
    //even tried with $(window).load(function){.....but no result.

这有什么不对，因为这不适合我？我不想使用表格。建议我一些简单的解决方案。

Answer 1

首先出现问题...

<select class="form-control" id="sortby" name="sortby">
    <option selected value="ID">ID</option>
    <option value="Relevance">Name</option><!-- Value ???-->
    <option value="Name">Source</option><!-- Value ???-->
</select>

就像

<select class="form-control" id="sortby" name="sortby">
    <option selected value="ID">ID</option>
    <option value="Name">Name</option>
    <option value="Source">Source</option>
</select>

当您比较$ _POST [＆＃39; sortby＆＃39;] ==＆＃34;姓名＆＃34;和$ _POST [＆＃39; sortby＆＃39;] ==＆＃34;来源＆＃34;

Answer 2

尝试在订单前添加空格。

$sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source 
FROM candidate ".$join.' where '.$condition." order by candidate.cand_number" ;

Answer 3

您的下拉列表是：

<select class="form-control" id="sortby" name="sortby">
    <option selected value="ID">ID</option>
    <option value="Relevance">Name</option>
    <option value="Name">Source</option>
</select>

当您发布下拉列表时，该值将为值=＆＃34; ...＆＃34;所选选项的属性，而不是显示的文本。

所以你的排序＆＃39;条件不好，必须是：

if($_POST['sortby']=="ID")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source FROM candidate ".$join.' where '.$condition." order by candidate.cand_number" ;
}

if($_POST['sortby']=="Relevance")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source FROM candidate ".$join.' where '.$condition." order by candidate.cand_fname" ;
}

if($_POST['sortby']=="Name")
{
 $sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_source FROM candidate ".$join.' where '.$condition." order by candidate.cand_Source" ;
}

Answer 4

@ CodeSniper..try检查一些空格。我相信你会得到解决方案。

Answer 5

我做了一个小例子，向您展示如何使用jQuery动态重新加载数据。当然，你必须根据你的情况进行调整。

对于这个例子，我使用的是localhost，数据库='test'和一个名为'candidate'的表，有三列：ID，Name和Source。

有两个文件：index.php和request.php（在同一文件夹中）。

index.php文件：

<?php
    // Database connection
    $db = mysqli_connect('localhost', 'root', '', 'test');
    // First request, on load
    $sql = "SELECT ID, Name, Source FROM candidate";
    $exe = $db->query($sql);
?>
<html>
    <head>
        <!-- Including jQuery -->
        <script src="//code.jquery.com/jquery-1.11.3.min.js"></script>
        <script src="//code.jquery.com/jquery-migrate-1.2.1.min.js"></script>
    </head>
    <body>
        <div>
            <!-- Generate table on load -->
            <table border="1">
                <thead>
                    <tr>
                        <th>ID</th>
                        <th>Name</th>
                        <th>Source</th>
                    </tr>
                </thead>
                <tbody id="list">
                    <?php
                        while($row = mysqli_fetch_assoc($exe)) {
                            echo "
                                <tr>
                                    <td>".$row['ID']."</td>
                                    <td>".$row['Name']."</td>
                                    <td>".$row['Source']."</td>
                                </tr>";
                        }
                    ?>
                </tbody>
            </table>
            Sort by :
            <select class="form-control" id="sortby" name="sortby">
                <option selected value="ID">ID</option>
                <option value="Name">Name</option>
                <option value="Source">Source</option>
            </select>
        </div>
        <script>
            $(document).ready(function(){
                // Each time you change your sort list, send AJAX request
                $("#sortby").change(function(){
                    $.ajax({
                        method: "POST",
                        url: "request.php",
                        data: { sortby:$("#sortby").val() }
                    })
                    // Copy the AJAX response in the table
                    .done(function( msg ) {
                        $("#list").html(msg);
                    });
                });
            });
        </script>
    </body>
</html>

request.php文件：

<?php

$db = mysqli_connect('localhost', 'root', '', 'test');

if(isset($_POST['sortby'])) {
    if($_POST['sortby'] == "ID") {
        $sql = "SELECT ID, Name, Source FROM candidate ORDER BY ID";
    } else if($_POST['sortby'] == "Name") {
        $sql = "SELECT ID, Name, Source FROM candidate ORDER BY Name";
    } else if($_POST['sortby'] == "Source") {
        $sql = "SELECT ID, Name, Source FROM candidate ORDER BY Source";
    }
} else {
    $sql = "SELECT ID, Name, Source FROM candidate";
}

$exe = $db->query($sql);

while($row = mysqli_fetch_assoc($exe)) {
    echo "
        <tr>
            <td>".$row['ID']."</td>
            <td>".$row['Name']."</td>
            <td>".$row['Source']."</td>
        </tr>";
}

?>

演示：http://maxime-mettey.com/projects/dropdownajax/

下载文件：http://maxime-mettey.com/projects/dropdownajax/dropdownajax.zip

通过使用下拉列表中的选定值而不使用表单进行排序

5 个答案: