提问人:Paspartu 提问时间:1/27/2016 更新时间:1/27/2016 访问量:105
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]
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。
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>
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。
评论