我无法获取我希望它们插入我想要的数据库中的图像文件 像我上传的那样的照片:
insert into('pic1.jpg');
insert into('pic2.jpg');
insert into('pic3.jpg');
不喜欢这样:
insert into(Array[0] => pic1.jpg, Array[1] => pic2.jpg, Array[2] => pic3.jpg);
所以我只能得到他们的名字并插入我的数据库 我需要使用 foreach循环
if(isset($_POST['upload'])){
$upload[] = ($_FILES['images']['name']);
print_r($upload);
}
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title></title>
</head>
<body>
<form method="POST" enctype="multipart/form-data">
<input type="file" name="images[]" multiple="multiple">
<input type="submit" name="upload" value="upload">
</form>
</body>
</html>
答案 0 :(得分:3)
这也是一种方法:
foreach ($_FILES['images']['name'] as $f => $name) {
$filename = $_FILES['images']['name'];
$mysql_query1 = mysql_query("insert into table values('".$filename."')");
}
答案 1 :(得分:2)
试试这个......
foreach ($_FILES['image']['tmp_name'] as $key => $val ) {
$filename = $_FILES['image']['name'][$key];
$filesize = $_FILES['image']['size'][$key];
$filetempname = $_FILES['image']['tmp_name'][$key];
$fileext = pathinfo($fileName, PATHINFO_EXTENSION);
$fileext = strtolower($fileext);
// here your insert query
}
答案 2 :(得分:1)
$images = $_FILES['images']['name'];
$sql = "insert into table (image_name) VALUES ('". implode("') ('", $a)."')";
结果如下:
insert into table (image_name) VALUES ('1.jpg') ('2.jpg') ('3.jpg') ('4.jpg')
答案 3 :(得分:0)
用户foreach。
示例代码:
foreach ($image as $image_path)
{
mysql_query("insert into `table` (column) values ('".$image_path."')");
}
答案 4 :(得分:0)
试试这个......
if(isset($_POST['upload'])){
$total = count($_FILES['images']['name']);
for($i=0; $i<$total; $i++){
$tmpFilePath=$_FILES['images']['tmp_name'][$i];
if($tmpFilePath != ""){
$img_name = $_FILES['images']['name'][$i];
mysqli_query($connection, "INSERT INTO table_name (column_name) value ('".$img_name."')");
}
}
}