Question

由于isset()错误，想要存储和检索数据库中的数据，包括图像，能够存储图像但无法显示图像。

以下是我使用的文件。

HTML表单

<html>
    <title>reg</title>
    <style type="text/css">
                body {
                    background-color: rgb(200,200,200);
                    color: white;
                    padding: 20px;
                    font-family: Arial, Verdana, sans-serif;}
                h4 {
                    background-color: DarkCyan;
                    padding: inherit;}
                h3 {
                    background-color: #ee3e80;
                    padding: inherit;}
                p {
                    background-color: white;
                    color: rgb(100,100,90);
                    padding: inherit;}
    </style>

    <form method="POST" action="login_back.php" enctype="multipart/form-data"></br>
    &nbsp<font color="DarkCyan"> Choose a user name:</font> <input type="text" name="username">
    </br></br>
    &nbsp<font color="DarkCyan"> First name:</font> <input type="text" name="firstname"/>
    </br></br>
    &nbsp<font color="DarkCyan">  Last name:</font><input type="text" name="lastname"/>
    </br></br>
    &nbsp<font color="DarkCyan"> File: <input type="file" name="image"></font>
    </br></br>
    <input type="submit" value="Save and Proceed">
    </form>
    </html>

用于存储数据库和从数据库检索数据的PHP文件

 <?php
    #echo "<body style='background-color:rgb(200,200,200)'>";
    session_start();
    if( isset($_POST['username']) && isset($_FILES['image']) )
    {
    $_SESSION['username']=$_POST['username'];
    $_SESSION['firstname']=$_POST['firstname'];
    $lastname=$_POST['lastname'];
    $file=$_FILES['image']['tmp_name'];
    $image_size=getimagesize($_FILES['image']['tmp_name']);
    if(!isset($file))
        echo"please select an image";
    else
    {
    $image_name=$_FILES['image']['name']; //grabing image name
    $image_size=getimagesize($_FILES['image']['tmp_name']); //getting image size
    }
    echo "</br>";


    #connection to db
    mysql_connect("localhost","root","")or die(mysql_error());
    mysql_select_db("wordgraphic")or die(mysql_error());



    #checking the available username
    $query = mysql_query("SELECT * FROM userdata WHERE username = '" . $_SESSION['username'] . "'" ); 
    $ans=mysql_num_rows($query);
    if ($ans > 0) 
    { 
         echo "Username already in use please try another.";

    }

    else if($image_size==FALSE)
    {
    echo"That's not an image.";

    }

    else
    {
    #Insert data into mysql

    #1.Inserting user name & image into db
    $sql="INSERT INTO userdata(username, firstname, lastname, image)VALUES('" . $_SESSION['username'] . "', '" . $_SESSION['firstname'] . "', '$lastname','$image')";
    $result1=mysql_query($sql);

    if($result1)
    {
    echo "</br>";
    echo "Registration successful";
    echo "</br>";

    //displaying image
    $lastid=mysql_insert_id();//get the id of the last record
    echo "uploaded image is :";
    echo "<img src='get.php?id=".$lastid."'>";   >  this command has some mistake

    }#if insertion into db successful
    else 
    {
    echo "Problem in database operation";
    }
    }# else block of unique username n img
    }#end of isset
    ?>

得到。 php，输出数据的附加文件，此文件已在<img src=get.php?id=$lastid>中使用，此文件中使用$_REQUEST。控制流不会通过isset()创建问题。

$image=mysql_query("SELECT * FROM userdata WHERE id=$mid") or die("Invalid query: " . mysql_error());
$image=mysql_fetch_assoc($image);
$image=$image['image'];
header("Content-type: image/jpeg");
echo $image;
}
else
    {
echo"error";
     }
?>

Answer 1

这是我第一次在论坛上发布任何内容，问题应该是，这段代码可能有什么问题：

$lastid=mysql_insert_id();//get the id of the last record
 echo "<img src='get.php?id=".$lastid."'>";

其中get.php是：

<?php
        #connection to db
        mysql_connect("localhost","root","")or die(mysql_error());
        mysql_select_db("wordgraphic")or die(mysql_error());
        if(isset($_REQUEST['id']) ) > this block of code is not runninng
        {
        $mid=(int)($_REQUEST['id']);

        $image=mysql_query("SELECT * FROM userdata WHERE id=$mid") or die("Invalid query: " . mysql_error());
        $image=mysql_fetch_assoc($image);
        $image=$image['image'];
        header("Content-type: image/jpeg");
        echo $image;
        }
        else
            {
        echo"error";
             }
?>

加载存储在数据库中的图像时出错

1 个答案: