html - Notice: Undefined index: picCat in E:\xampp\htdocs\Evako\admin.php on line 313 -
this question has answer here:
if not in form choose category returns error :
notice: undefined index: piccat in e:\xampp\htdocs\evako\admin.php on line 313
code: 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> and php:
if (isset($_get["showfilter"])) { $picname = $_get['nmpic']; $pricefrom = $_get['pricefrom']; $priceto = $_get['priceto']; $piccat = $_get['piccat']; $numpic = $_get['numpic']; $filter = " select * images status = '1'"; if ($numpic && !empty($numpic)) { $filter .= " , id='$numpic'"; } if ($picname && !empty($picname)) { $filter .= " , img_content='$picname'"; } if ($piccat && !empty($piccat)) { $filter .= " , category='$piccat'"; } if ($priceto && !empty($priceto)) { $filter .= " , price < '$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>"; } } i can not understand wrong let me return error. , if choose category works without problem ...? line 313 is:
$piccat = $_get['piccat']; thank you.
your select box option need value tag be
<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> read html select tag
youe script open sql injection read how can prevent sql injection in php? prevent it
Comments
Post a Comment