如何在Php中将多个图像存储在数据库中 [英] How To I Store Multiple Images In Database In Php

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问题描述

我希望使用php在一列中将多个图像存储在数据库中。我试过这段代码

在这里输入代码

< html>

< head>

< title> ;商店详细信息注册表中的更多选项< / title>

< / head>

< body>

< br $> b $ b

< form action =method =postenctype =multipart / form-data>

I want o store multiple images in database in one column using php. i tried this code
enter code here
<html>
<head>
<title>more option in shop detail register form </title>
</head>
<body>



<form action="" method="post" enctype="multipart/form-data">




Image1 < input type =filename =image >


Image2 < input type =filename =image>


Image3 < input type =filename =image>


购物 < input type =textname =shop>


商店详情 < input type =textname =shopdetails>


电话 < inpu t type =textname =phone>


可用在 < input type =textname =availableon>












< / form> ;

< / body>

< / html>`在这里输入代码`$ msg =;

if(isset( $ _POST ['submit']))

{

$ filename = $ _ FILES ['image'] ['name'];

$ temppath = $ _ FILES ['image'] ['tmp_name'];

$ move = move_uploaded_file($ temppath,'upload /'.$ filename);

$ shop = $ _ POST ['shop'];

$ shopdetails = $ _ POST ['shopdetails'];

$ phone = $ _ POST ['phone'];

$ availableon = $ _ POST ['availableon'];

$ query =插入细节(图片,商店,店铺细节,电话,可用)值('$ filename', $店,'$ shopdetails','$ phone','$ availableon');

$ sql = mysql_query($ query);

if($ sql)

{

$ msg。=感谢您在这里注册;

}

其他

{

$ msg。=再试一次;

}

}

< input type =submitname =submitvalue =Submit>
Image1<input type="file" name="image">

Image2<input type="file" name="image">

Image3<input type="file" name="image">

Shop<input type="text" name="shop">

Shop Details<input type="text" name="shopdetails">

Phone<input type="text" name="phone">

Available On<input type="text" name="availableon">

<input type="submit" name="submit" value="Submit">





</form>
</body>
</html>`enter code here`$msg="";
if(isset($_POST['submit']))
{
$filename =$_FILES['image']['name'];
$temppath =$_FILES['image']['tmp_name'];
$move =move_uploaded_file($temppath,'upload/'.$filename);
$shop=$_POST['shop'];
$shopdetails=$_POST['shopdetails'];
$phone=$_POST['phone'];
$availableon=$_POST['availableon'];
$query="insert into details(image,shop,shopdetails,phone,availableon)values('$filename','$shop','$shopdetails','$phone','$availableon')";
$sql=mysql_query($query);
if($sql)
{
$msg.="Thank you for registering here";
}
else
{
$msg.="try again";
}
}

推荐答案

msg =;

if(isset(
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if(isset(


_POST ['submit']))

{
_POST['submit']))
{


filename =
filename =


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