Question

我正在向php脚本发送一个值，并希望在我的主页上的div中显示该脚本的结果，但不知何故没有显示。当我检查网络设置时，我可以看到值正确地发布到php脚本。

我的代码：

在我的索引上：

<li class="active lidcat">
    <input type="radio" name="bedno" '.$checked.'>
    <a class="zoekvalue" href="'.$void.'">'.$ledeninfo['title'].'</a>
</li>

<div id="result">
</div>

我的js文件：

<script>
    $(document).ready(function () {
        $(".lidcat").click(function () {
            var zoekvalue = $(this).children(".zoekvalue").text();
            var posting = $.post("includes/leden.php", {zoekwaarde: zoekvalue});

            posting.done(function (data) {
                var content = $(data).find("#content");
// Gooi $result eerst leeg voordat de inhoud wordt getoond vanwege vorige ajax requests
                $("#result").empty().append(content);
                console.log(zoekvalue);
            });
        });
    });
</script>

最后我的php文件：

<div class="blog-content" id="content">
    <div class="home-wrapper">
        <div class="row">
            <div class="col-xs-12 col-sm-12 col-md-12 col-lg-12">
                <h2 class="osLight">Onze leden</h2>
                <div class="row">
                    <?php
                    $zoekwaarde = $_POST['zoekwaarde'];
                    echo $zoekwaarde;

                    // De leden zelf worden hiermee opgehaald
                    $ledeninfo1 = "
                      SELECT ct.id, ct.catid, ct.title as content_title, ct.alias, ct.introtext, ct.state, ct.images, ct.ordering , cn.id, cn.title as category_title
                      FROM snm_content ct
                      INNER JOIN snm_categories cn
                      ON ct.catid = cn.id
                      WHERE ct.catid IN (" . $useableids . ")
                      AND ct.state = 1
                      AND ct.title LIKE %" . $zoekwaarde . "%
                      ORDER BY ct.ordering";
                    $ledeninfocon1 = $conn->query($ledeninfo1);
                    while ($ledeninfo1 = $ledeninfocon1->fetch_assoc()) {

                        $ledenimage = $ledeninfo1['images'];
                        $ledenimg = json_decode($ledenimage);

                        if ($ledenimg->image_intro != '') {
                            $image = '<img class="ledenimg" src="cms/' . $ledenimg->image_intro . '" alt="image">';
                        } else {
                            $image = '<img class="ledenimg" src="cms/images/afbeeldingen/logo.jpg" alt="image">';
                        }

                        if (strlen($ledeninfo1['introtext']) > 150) {
                            $shortstr = substr($ledeninfo1['introtext'], 0, 150) . '...';
                        } else {
                            $shortstr = $ledeninfo1['introtext'];
                        }

                        $ledenoverzicht .= '
                        <div class="col-xs-12 col-sm-4 col-md-3 col-lg-3">
                            <div class="article">
                                <a href="' . $ledeninfo1['alias'] . '.html" class="image">' . $image . '</a>
                                <div class="article-category"><a href="#" class="text-green">' . $ledeninfo1['category_title'] . '</a></div>
                                <h3 class="osLight"><a href="' . $ledeninfo1['alias'] . '.html">' . $ledeninfo1['content_title'] . '</a></h3>
                                <p>' . strip_tags($shortstr) . '</p>
                            </div>
                        </div>';
                    }
                    echo $ledenoverzicht;
                    ?>
                </div>
            </div>
        </div>
    </div>
</div>

php的工作方式不是问题，即使没有，我也可能会在结果div或至少h2标记中看到错误。

正确发送zoekwaarde的值，它向leden.php显示200状态

我错过了什么吗？我从官方的jquery网站上获得了代码。

Ajax结果没有加载到div中

0 个答案: