dongxiongshi9952
dongxiongshi9952
2015-04-08 07:36

如何在php表单产品表中显示mysql blob图像

已采纳

After two days of search I come here with the hope of finding a solution. I'm trying to display the images uploaded into my database with type BLOB but there is something wrong the image is not displaying just the 64 code for the image here is the code :

<div id="container">

     <img src ="banner.jpg" width="400" height="100"/>


        <div id="menu">

            <h3>
            <li><a href="refresher.php">Home  </a> </li>
            <li><a href="refresher2.html">About  </a> </li>
            <li><a href="refresher3.html">Category  </a>

            <ul>    <li><a href="Boys.php">Boys</a></li> 
                    <li><a href="Girls.php">Girls</a></li> 
                    <li><a href="#">Uni</a></li> 
             </ul>
             </li>

            <li><a href="refresher3.html">Costume Hire  </a> </li>
            <li><a href="refresher3.html">Contact</a> </li>
            </h3>



        <!--   <meta name="ROBOTS" content="NOINDEX, NOFOLLOW" /> -->

<!-- HTML for SEARCH BAR -->
<div>
    <form id="tfnewsearch" method="get" action="http://www.google.com">
            <input type="text" class="tftextinput" name="q" size="21" maxlength="120"><input type="submit" value="search" class="tfbutton">
    </form>
<div class="tfclear"></div>
</div>
</div>

        <div id="content"> 



            <?php
                //connect to the server and create database.
                $host = "";
                $userMS = "";
                $passwordMS = "";
                $connection = mysql_connect($host,$userMS,$passwordMS) or die("Couldn't  connect:".mysql_error());
                $database = "projectDataBase";
                $db = mysql_select_db($database,$connection) or die("Couldn't select database");


                //build the table to show the records.
                echo("<table>");
                echo("<tr>
                        <th>Product ID</th>
                        <th>Product Name</th>
                        <th>Product Image</th>
                        <th>Age</th>
                        <th>Stauts</th>
                        <th>Price</th>
                    </tr>");

                    //database query to show all the records.
                $selectString = "SELECT 
Product.Product_ID,
Product.Product_Name,
Product.Image,
Gender.Gender_Description,
Category.Description,
`Status`.Availability,
 Product.Price,
 Age.Age_Description


    FROM
    Product
JOIN
    Age ON Product.Age_ID = Age.Age_ID
JOIN
    Category ON Product.Category_ID = Category.Category_ID
JOIN
    Gender ON Product.Gender_ID = Gender.Gender_ID
JOIN
    `Status` ON Product.Status_ID = `Status`.Status_ID
                              ";


                $result = mysql_query($selectString);

                    while ($row = mysql_fetch_assoc($result))
                    {
                    echo("<tr>");
                    foreach($row as $index=>$value)
                    //to show the Product image.
                    if($index == "Image")
                    {

                        echo("<td><img src = $value alt = 'Image'></td>");

                    }
                    else{
                        echo("<td>$value</td>");                    

                    }

                    $self = htmlentities($_SERVER['PHP_SELF']);
                        echo("<form action = '$self' method='POST'>");
                        echo "<input type='hidden' name='athID' value='$row[Product_ID]' >";
                        /*echo("<td><input type='submit' name='delete' value = 'Delete'/></td>");*/
                        echo ("</form>");
                        echo("</tr>");  

                    echo("</tr>");  
                }
                echo("</table>");



                /*
                echo("<table>");
        echo("<tr>
                <th>Country Code</th>
                <th>Country Name</th>
                <th>Population</th>
                <th>Image</th>
            </tr>");

        $selectString = "SELECT * from tblCountries";
        $result = mysql_query($selectString);

            while($row = mysql_fetch_assoc($result))
            {
            echo("<tr>");
            foreach($row as $index=>$value)
            //to show the country image.
            if($index == "flag_image")
            {
                echo("<td><img src = $value alt = 'countryImage'></td>");
            }
            else{
                echo("<td>$value</td>");                    

            }
            echo("</tr>");  
        }
        echo("</table>");


        */

                mysql_free_result($result);





            ?>

        </div>



        <footer>Copyright © Hucos Pucos Shop</footer>


</div>

the image is stored in Product Table.

  • 点赞
  • 写回答
  • 关注问题
  • 收藏
  • 复制链接分享
  • 邀请回答

1条回答

  • dp20011 dp20011 6年前

    UPDATE

    just use my example that's right below... so....

    echo '<td><img src="data:image/jpeg;base64,'.base64_encode($value).'" alt="Image" />';
    

    You can drag and drop that directly into your code. I don't know how to make it any easier than that. Please click the checkmark to the left of this answer. And click the Up Arrow as well. Thanks


    PHP display image BLOB from MySQL

    点赞 评论 复制链接分享