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我使用以下代码通过我网站的管理区域将多个图像添加到我的数据库中，这工作正常，但我想显示为该特定车辆 ID 上传的多个图像......基本上我正在尝试创建汽车经销商的网站，当管理员添加库存时，他们可以选择多个图像添加到列表中，这是添加库存代码

<?php

if(isset($_POST['addstock'])){

$vehicle_make = $_POST['vehicle_make'];
$veh_model = $_POST['veh_model'];
$veh_mileage = $_POST['veh_mileage'];
$veh_description = $_POST['veh_description'];
$veh_price = $_POST['veh_price'];
$veh_gearbox = $_POST['veh_gearbox'];
$veh_engine_size = $_POST['veh_engine_size'];
$veh_fuel_type = $_POST['veh_fuel_type'];

foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){

$file_name = $key.$_FILES['files']['name'][$key];
$file_size = $_FILES['files']['size'][$key];
$file_tmp = $_FILES['files']['tmp_name'][$key];
$file_type = $_FILES['files']['type'][$key];


$addstock = "insert into stock (veh_make,veh_model,veh_mileage,veh_description,veh_gearbox,veh_engine_size,veh_fuel_type,veh_price,file_name,file_size,file_type) values('$vehicle_makee','$veh_model','$veh_mileage','$veh_description','$veh_gearbox','$veh_engine_size','$veh_fuel_type','$veh_price','$file_name','$file_size','$file_type')";

        move_uploaded_file($file_tmp,"image_uploads/".$file_name);
        }

$addsto = mysqli_query($con, $addstock);

if($addsto){

    echo "<script>alert('Vehicle has been added')</script>";
    echo "<script>window.open('addstock.php','_self')</script>";
}
}

代码成功将多张图片添加到目录文件夹中，

这是管理区域外的代码，在车辆详细信息页面上，我上传的多张图片中只有一张图片显示，我希望显示为车辆ID上传的所有图片，最终将其样式化为图片画廊，

<?php
if(isset($_GET['vehicle_id'])){

$vehicle_id = $_GET['vehicle_id'];

$get_veh = "select * from stock where vehicle_id='$vehicle_id'";

$run_veh = mysqli_query($con, $get_veh);

while($row_veh=mysqli_fetch_array($run_veh)){

    $vehicle_id = $row_veh['vehicle_id'];
    $veh_make = $row_veh['veh_make'];
    $veh_model = $row_veh['veh_model'];
    $veh_mileage = $row_veh['veh_mileage'];
    $veh_price = $row_veh['veh_price'];
    $veh_gearbox = $row_veh['veh_gearbox'];
    $veh_description = $row_veh['veh_description'];
    $file_name = $row_veh['file_name'];

    echo "

        <div id='single_vehicle'>


            <div id='box1'>$veh_make $veh_model</div>
            <div id='box3'>£$veh_price</div>
        </div>


        <div id='single_vehicle2'>

            <div id='box2'><img src='admin/image_uploads/$file_name' width='600' height='450' /></div>
            <div id='box4'>
            <div id='clickbox1'><a href='#'>Book Test Drive</a></div>
            <div id='clickbox1'><a href='#'>Send Enquiry</a></div>
            <div id='clickbox1'><a href='#'>Print this Page</a></div>
            <div id='clickbox1'><a href='#'>Email this Page</a></div>

            </div>



        </div>

        <div id='desription_area'>
        <div id='desription_area1'><span style='text-decoration:underline'>Vehicle Description</span><br>$veh_description</div>
        <div id='desription_area2'><span style='text-decoration:underline'>Specification</span><br>Make: $veh_make<br>$veh_model<br>$veh_mileage</div>
        </div>

        ";
}
}
?>

我是编码新手，仍在学习，我做错了什么？任何帮助深表感谢

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1 回答 1

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嗨，也许你的代码有错误，请试试这个并检查你的代码

用于上传图片

<?php
if(count($_FILES) > 0) {
if(is_uploaded_file($_FILES['userImage']['tmp_name'])) {
mysql_connect("localhost", "root", "");
mysql_select_db ("phppot_examples");
$imgData =addslashes(file_get_contents($_FILES['userImage']['tmp_name']));
$imageProperties = getimageSize($_FILES['userImage']['tmp_name']);
$sql = "INSERT INTO output_images(imageType ,imageData)
VALUES('{$imageProperties['mime']}', '{$imgData}')";
$current_id = mysql_query($sql) or die("<b>Error:</b> Problem on Image Insert<br/>" . mysql_error());
if(isset($current_id)) {
header("Location: listImages.php");
}}}
?>
<HTML>
<HEAD>
<TITLE>Upload Image to MySQL BLOB</TITLE>
<link href="imageStyles.css" rel="stylesheet" type="text/css" />
</HEAD>
<BODY>
<form name="frmImage" enctype="multipart/form-data" action="" method="post" class="frmImageUpload">
<label>Upload Image File:</label><br/>
<input name="userImage" type="file" class="inputFile" />
<input type="submit" value="Submit" class="btnSubmit" />
</form>
</div>
</BODY>
</HTML>

并显示图像保存在数据库中

<?php
$conn = mysql_connect("localhost", "root", "");
mysql_select_db("phppot_examples") or die(mysql_error());
if(isset($_GET['image_id'])) {
$sql = "SELECT imageType,imageData FROM output_images WHERE imageId=" . $_GET['image_id'];
$result = mysql_query("$sql") or die("<b>Error:</b> Problem on Retrieving Image BLOB<br/>" . mysql_error());
$row = mysql_fetch_array($result);
header("Content-type: " . $row["imageType"]);
echo $row["imageData"];
}
mysql_close($conn);
?>

对于列表图像

<?php
$conn = mysql_connect("localhost", "root", "");
mysql_select_db("phppot_examples");
$sql = "SELECT imageId FROM output_images ORDER BY imageId DESC"; 
$result = mysql_query($sql);
?>
<HTML>
<HEAD>
<TITLE>List BLOB Images</TITLE>
<link href="imageStyles.css" rel="stylesheet" type="text/css" />
</HEAD>
<BODY>
<?php
while($row = mysql_fetch_array($result)) {
?>
<img src="imageView.php?image_id=<?php echo $row["imageId"]; ?>" /><br/>
<?php       
}
mysql_close($conn);
?>
</BODY>
</HTML>

为了向浏览器显示 BLOB 图像，我们必须创建一个 PHP 文件来执行以下操作。

从 MySQL BLOB 获取图像数据和类型 使用 PHP header() 将内容类型设置为图像 (image/jpg, image/gif, ...)。打印图像内容。

祝你好运并尝试

于 2016-01-11T20:14:59.223 回答