注意:未定义的索引:313 行 E:\xampp\htdocs\Evako\admin.php 中的 picCat [重复]

Notice: Undefined index: picCat in E:\xampp\htdocs\Evako\admin.php on line 313 [duplicate]

提问人:Paspartu 提问时间:1/27/2016 更新时间:1/27/2016 访问量:105

问:

如果不在我选择我的类别的表单中,则返回错误:

注意:未定义的索引:313 行 E:\xampp\htdocs\Evako\admin.php 中的 picCat

代码:form

<form class="form" method="GET">
            <input type="text" id="nmPic" name="nmPic" placeholder="име на снимката" onfocus="this.placeholder = ''" onblur="this.placeholder = 'име на снимката'"></br>
                <input type="text" id="price" class="priceFrom" name="priceFrom" placeholder="цена от" onfocus="this.placeholder = ''" onblur="this.placeholder = 'цена от'">
                <input type="text" id="price" class="priceTo" name="priceTo" placeholder="цена до" onfocus="this.placeholder = ''" onblur="this.placeholder = 'цена до'"></br>
                <select name="picCat" id="picCat">
                    <option value="" selected disabled>Изберете категория</option>
                    <option>Детски</option>
                    <option>Сватби</option>
                    <option>Рожден ден</option>
                    <option>18+</option>
                    <option>Други</option>
                </select></br>
                <input type="text" id="numPic" name="numPic" placeholder="номер на снимката" onfocus="this.placeholder = ''" onblur="this.placeholder = 'номер на снимката'"></br>
                <input type="submit" name="showFilter" value="покажи" />
            </form>

还有 php:

if (isset($_GET["showFilter"])) 
            {
                $picName = $_GET['nmPic'];
                $priceFrom = $_GET['priceFrom'];
                $priceTo = $_GET['priceTo'];
                $picCat = $_GET['picCat'];
                $numPic = $_GET['numPic'];

                $filter = " SELECT * FROM images WHERE status = '1'";
                if ($numPic && !empty($numPic)) {
                    $filter .= " AND id='$numPic'";
                }
                if ($picName && !empty($picName)) {
                    $filter .= " AND img_content='$picName'";
                }
                if ($picCat && !empty($picCat)) {
                    $filter .= " AND category='$picCat'";
                }
                if ($priceTo && !empty($priceTo)) {
                    $filter .= " AND price < '$priceTo'+1";
                }
                $resFilter = $connect->query($filter);

                if ($resFilter->num_rows > 0) 
                {

                    while($row = mysqli_fetch_array($resFilter))
                    {

                        echo    "<div class='col-md-4 picture'>
                                    <img class='child-img' src='".$row["picture"]." '/></br>
                                        <div class='number'><span class='id'>№ ".$row['id']." | име: ".$row['img_content']." | категория: ".$row['category']." | цена: ".$row['price']."лв. | дата: ".$row['time']."ч.</span></div>

                                </div>";

                    }

                }

我不明白哪里错了让我返回这个错误。如果您选择一个类别,一切正常......? 313 号线是:

$picCat = $_GET['picCat'];

谢谢。

php html 选择 未定义索引

评论


答:

1赞 Saty 1/27/2016 #1

您的选择框选项需要它value tag

<select name="picCat" id="picCat">
                    <option value="" selected disabled>Изберете категория</option>
                    <option value="Детски">Детски</option>
                    <option value="Сватби">Сватби</option>
                    <option value="Рожден ден">Рожден ден</option>
                    <option value="18+">18+</option>
                    <option value="Други">Други</option>
                </select>

读取 html select 标签

Youe 脚本已打开以进行 sql 注入阅读 如何在 PHP 中防止 SQL 注入? 防止它

评论

0赞 Paspartu 1/27/2016
感谢您选择标签和注射的信息,我忘记了。将尝试使用选项的值。谢谢。
0赞 Paspartu 1/27/2016
不,问题不在于选择标记或重复。