我是编程新手，一直在研究一个可搜索的数据库，该数据库可以通过输入关键字来检索图像，按下提交后将显示结果和图片。
但到目前为止，我没有运气让图片显示（损坏的链接/图像），但我的搜索表单确实有效并且确实正确检索了名称或结果。
我在 phpmyadmin 名称中的表是shoes，我有 3 列，1 id (int15 PRI) ，2 品牌/型号 (varchar 50)，3 图片 (longblob)。
我的代码比较简单，希望你能帮助我=）
文件名：search.php
<form action="search.php" method="POST">
Name: <input type ="text" name="search_name"> <input type="submit" value="Search">
<?php
if (isset($_POST['search_name'])) {
$search_name = $_POST['search_name'];
if (!empty($search_name)){
if (strlen($search_name)>=3) {
$query = "SELECT * FROM `shoes` WHERE `brand/model` LIKE '%".mysql_real_escape_string($search_name)."%'";
$query_run = mysql_query($query);
$query_num_rows = mysql_num_rows($query_run);
if ($query_num_rows>=1) {
echo $query_num_rows.' Results found:<br>';
while ($query_row = mysql_fetch_array($query_run)) {
$picture = $query_row['picture'];
echo "</br>";
echo $query_row ['brand/model'];
echo "</br>";
echo "</br>";
//header("content-type: image/jpeg");
echo "<img src=image.php?id=".$row['id']." width=300 height=200/>";
echo "</br>";
}
} else {
echo 'No Results Found.';
}
} else {
echo 'Text field must be more than 3 characters.';
}
} else {
echo 'Text Field Cannot be Empty!';
}
}
?>
我这里有一个 image.php
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
$conn = mysql_connect("localhost","root","");
if(!$conn){
echo mysql_error();
}
$db = mysql_select_db("phsdatabase");
if(!$db){
echo mysql_error();
}
$id = $_GET['id'];
$query = "SELECT `picture` FROM shoes where id='$id'";
$query_run = mysql_query("$query",$conn);
if($query_run){
$row = mysql_fetch_array($query_run);
$type = "Content-type: image/jpeg";
header($type);
echo $row['picture'];
} else {
echo mysql_error();
}
?>
storeinfo.php存储新信息，
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
$conn = mysql_connect("localhost","root","");
if(!$conn)
{
echo mysql_error();
}
$db = mysql_select_db("phsdatabase",$conn);
if(!$db)
{
echo mysql_error();
}
@$brandmodel = $_POST['brand/model'];
@$picture = addslashes (file_get_contents($_FILES['picture']['tmp_name']));
@$image = getimagesize($_FILES['picture']['tmp_name']);//to know about image type etc
//$imgtype = $image['mime'];
if (isset($_POST['brand/model'])){
$brandmodelentry = $_POST['brand/model'];
if (!empty($brandmodelentry)){
if (strlen($brandmodelentry)>=3) {
$query ="INSERT INTO shoes VALUES('','$brandmodel','$picture')";
$query_run = mysql_query($query,$conn);
echo '<br>';
echo "Information Stored Successfully!";
} else {
echo mysql_error();
}
echo '<br>';
echo '<br>';
echo "Thank you for Registering new information to our database!";
} else{
echo 'Text Field cannot be empty!';
}
}
?>
newentry.php 注册新信息
<form enctype="multipart/form-data" action="storeinfo.php" method="POST">
<center>Shoes Information</center>
Brand and Model Name<input type=text name="brand/model">
Picture of Shoes(Acceptable formats:<br>JPEG,JPG,PNG)<input type="file" name="picture" id ="picture">
<input type=submit name="submit" value="Store Information">