我需要帮助,每次插入多张图像时,它仅显示最后一张已上传的图像,你们可以帮助我解决此问题。
这里是我用于上传和插入数据的代码
<?php
require_once('dbConfig.php');
$upload_dir = '../uploads/';
if(isset($_POST['btnSave'])){
$judul = $_POST['judul'];
$deskripsi = $_POST['deskripsi'];
$status = $_POST['status'];
$harga = $_POST['harga'];
$imgName = $_FILES['myfile']['name'];
$imgTmp = $_FILES['myfile']['tmp_name'];
$imgSize = $_FILES['myfile']['size'];
if(empty($judul)){
$errorMsg = 'Please input judul';
}elseif(empty($deskripsi)){
$errorMsg = 'Please input deskripsi';
}elseif(empty($status)){
$errorMsg = ' Input status';
}elseif(empty($harga)){
$errorMsg = 'Input Harga';
}elseif(empty($imgName)){
$errorMsg = 'Please select gambar';
}else{
$imgExt = strtolower(pathinfo($imgName, PATHINFO_EXTENSION));
$allowExt = array('jpeg', 'jpg', 'png', 'gif');
$userPic = time().'_'.rand(1000,9999).'.'.$imgExt;
if(in_array($imgExt, $allowExt)){
if($imgSize < 5000000){
move_uploaded_file($imgTmp ,$upload_dir.$userPic);
}else{
$errorMsg = 'Image too large';
}
}else{
$errorMsg = 'Please select a valid image';
}
}
if(!isset($errorMsg)){
$sql = "insert into rumah(judul, deskripsi, status, harga, gambar1)
values('".$judul."', '".$deskripsi."', '".$status."', '".$harga."', '".$userPic."')";
$result = mysqli_query($conn, $sql);
if($result){
$successMsg = 'New record added successfully';
header('refresh:5;index.php');
}else{
$errorMsg = 'Error '.mysqli_error($conn);
}
}
} ?>
上传按钮
<input type="file" name="myfile" multiple>
我已经在表单操作中使用了此方法-enctype =“ multipart / form-data”