如何使用php从mysql显示4个图像（blob） [英] How to display 4 Images (blob) from mysql using php

问题描述

我正在尝试使用PHP和MySQL数据库显示4个图像。我必须将4个图像显示为行。



我使用带有字段的表车（`id_car`，`car_image1`，`car_image2`，`car_image3`， `car_image4`），所有图像都是blob数据类型。





$ id = $ _GET ['id'];



$ link = mysql_connect（localhost，root，）;

mysql_select_db（cars_database）;

$ sql =SELECT car_image1，car_image2， car_image3，car_image4 FROM cars WHERE id_car ='$ id';

$ result = mysql_query（$ sql）;



mysql_close （$ link）;

while（$ row = mysql_fetch_array（$ result））

{

header（'Content-type：image / jpeg'）;

echo $ row ['car_image1'];

echo $ row ['car_image2'];

echo $ row [ 'car_image3'];

echo $ row ['car_image4'];

}





我只能显示1张图片而不能显示其他图片。由于我是这项技术的新手，我需要帮助。

I am trying to display 4 images using PHP and MySQL database. I have to display the 4 images as rows.

I use the table cars with fields (`id_car`, `car_image1`, `car_image2`, `car_image3`, `car_image4`), with all the images being blob datatype.


$id = $_GET['id'];

$link = mysql_connect("localhost", "root", "");
mysql_select_db("cars_database");
$sql = "SELECT car_image1, car_image2, car_image3, car_image4 FROM cars WHERE id_car='$id'";
$result = mysql_query("$sql");

mysql_close($link);
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
echo $row['car_image1'];
echo $row['car_image2'];
echo $row['car_image3'];
echo $row['car_image4'];
}


I can only display 1 image and not the other images. Since I am a newbie to this technology I need help.

推荐答案

id =

_GET ['id'];

_GET['id'];

link = mysql_connect（localhost，root，）;

mysql_select_db（cars_database ）;

link = mysql_connect("localhost", "root", "");
mysql_select_db("cars_database");

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