Question

所以我创建了一个表单，它带有两个文件按钮（用于图像），一个用于缩略图，另一个用于主图像，它已成功获取文件并将其放入数据库，它还获得了正确的目录。 ..但我有一个问题...当它回应他们没有出现的图像时...有些请帮助我

这里是表单和代码（都在同一个文件中，名为addimage.php）：

<?php
include'includes/connect.php';

if($_POST['submit'])
 {
$img="main/";
$img= $img . basename($_FILES['photo1']['name']);

$tmb="thumb/";
$tmb= $tmb . basename($_FILES['photo2']['name']);


$imagename = mysql_real_escape_string($_POST['imagename']);
$content = mysql_real_escape_string($_POST['content']);
$links = mysql_real_escape_string($_POST['links']);
$pic1 = "main/" .(mysql_real_escape_string($_FILES['photo1']['name']));
$pic2 = "thumb/" .(mysql_real_escape_string($_FILES['photo2']['name']));

if(move_uploaded_file($_FILES['photo1']['tmp_name'], $img));{
    if(move_uploaded_file($_FILES['photo2']['tmp_name'], $tmb));{
{

    mysql_query("INSERT INTO `portweb` ( `imagename`, `content`, `links`, `photo1`, `photo2`)  VALUES('$imagename', '$content', '$links', '$pic1', '$pic2')");
    echo "files upload success";
}
}
}
/*else
{
    echo "upload failed";   
}*/

}

?>

<html>
<head>
<title>upload pictures portfolio web</title>
</head>
<body>
<a href="portweb.php">Back</a>
 <br />
 <form action="addportweb.php" method="post" enctype="multipart/form-data">
 <label>Main Image:</label>
 <input type="file" name="photo1" /><br />
<label>Thumb:</label>
<input type="file" name="photo2" /><br />
<label>Name:</label> <input type="text" name="imagename" class="text_input" maxlength="100" /><br />
<label>Description:</label>
<textarea name="content" style="width: 300px; height:80px; padding: 5px; resize:none;" ></textarea>
<br />

<label>Link:</label> <textarea name="links" style="width: 100px; height:50px; padding: 5px; resize:none;" ></textarea><br />
<input type="submit" name="submit" value="upload" /> 
</form>
</body>
</html>

这是帮助输出的功能......还有我的问题：

function getportweb() {
     $query = mysql_query("SELECT * FROM portweb") or die(mysql_error());
      while($post = mysql_fetch_assoc($query)){

        echo "<img src =".$post['photo1']."/>";
            echo "<br>";

        echo "<img src =" . $post['photo2'] . "/>";
            echo"<br>";

        echo "<p>" . $post['imagename'] . "</p>";
            echo "<br>";
        echo "<p>" . $post['content'] . "</p>";
        echo "<br>";
        echo "<p>" . $post['links'] . "</p>";
            echo "<br>";

        echo "<a href=\"deleteportweb.php?id=" . $post['id'] . "\">Delete</a>";
        echo "<a href=\"editportweb.php?id=" . $post['id'] . "\">Edit</a>";
         echo "<br>";
         } 

}

我想相信这是我的问题

echo "<img src =".$post['photo1']."/>";
                echo "<br>";

            echo "<img src =" . $post['photo2'] . "/>";
                echo"<br>";

但我一无所获......请有人帮忙

Answer 1

尝试这样的事情：

echo'<img src="'.$post['photo1'].'" />';

我的数据库中的图像未在我的网站上发布（PHP MySQL）

1 个答案: