如何使用PHP从包含其他列文本的表中获取图像 [英] How to get images from a table with other column texts using PHP

问题描述

你好同志开发者！我正在尝试从我的数据库中检索图像，以便将其包含在我创建的这个表中。我在Google上查看的所有示例都是从仅包含图像的1个表中检索图像，但在这种情况下我无法使其正常工作。

Hello There Fellow Devs! I'm trying to retrieve an image from my database to include it with this table I created. All the Examples I looked up on Google are for retrieving images from 1 table alone that contains only images, but in this case I can't get it working.

<?php
                    $Con = mysql_connect('localhost','root','');
                      if($Con === false)
                      {
                        echo "Not Connected";
                      }

                      else
                      {


                        $select = mysql_select_db("symfony");
                        $TableName = "main";
                        $SQLstring = "SELECT * FROM $TableName ";
                        $QueryResult = mysql_query($SQLstring);
                        $Row = mysql_fetch_row($QueryResult);
                            do {


                                echo "<div class='art-content-layout'>";
                                echo "<div class='art-content-layout-row'>";
                                echo "<div class='art-layout-cell' style='width: 33%'>";
                                echo"   <p><img width='259' height='194' class='art-lightbox' name='image' src='../images/3.jpg'><br></p>";
                                echo "</div>";
                                echo "<div class='art-layout-cell' style='width: 67%'>";
                                echo "<p></p>";
                                echo "<table border>";
                                echo "<tbody>";
                                echo "<tr>";
                                echo "<tr>";
                                    echo "<th colspan='3' align='left'><b> Owner : $Row[0]</b></th>";
                                echo "</tr>";
                                echo "<tr>";
                                    echo "<td colspan='3'><b>$Row[1]:</b>";

                                   echo  "</td>";
                                echo "</tr>";

                                echo "<tr>";
                                    echo "<td><b>Price:$Row[9] US Dollar </b></td>";
                                echo "</tr>";
                                echo "<tr>";
                                    echo "<td><b> City: $Row[4] </br> Hood: $Row[4] </br> Qdr: $Row[5] </br> Street:$Row[6] </br> Property Number :$Row[7] </br> Room Number : $Row[8] </b></td>";
                                    echo" <td><b>Description : $Row[10] </b></td>";

                                echo "</tr>";
                                echo"<tr>";
                                    echo" <td><b>Type : $Row[12] </b></td>";
                                    echo "<td><b>Contact : $Row[1] </b></td>";
                                echo "</tr>";
                                echo "</tr>";
                                echo "</tbody>";
                                echo "</table> <br><p></p>";

                                echo "</div>";
                                echo "</div>";
                                echo "</div>";
                                $Row = mysql_fetch_row($QueryResult);
                                } while ($Row);     
                        }   
                ?> 

我试图这样做，但仍无效：

I tried to do this, it still didn't work :

$img = $Row[15];
//column 15 is the Blob the image
                        $img = mysql_fetch_array($QueryResult);
                            $content = $img['15'];
                            //header('Content-type: image/jpg');

推荐答案

你无法做你想做的事。您需要将逻辑分成两个脚本。实际上没有办法在与其他数据相同的传递中获取图像数据，因为IMG标记被提供的SRC不是原始数据，而是要求服务器提供图像。

You can't do what you are trying to do. You need to separate your logic into two scripts. There really isn't a way to get the image data in the same pass as your other data because the IMG tag is fed a SRC that is not raw data, but instead asks the server to serve the image.

在生成HTML的当前脚本中，您只需要让IMG标记将SRC引用为执行检索图像数据工作的新脚本。类似于：

In your current script where you generate the HTML you just need to have your IMG tag reference the SRC as a new script that does the work of retrieving your image data. Something like:

echo"   <p><img width='259' height='194' class='art-lightbox' name='image' src='display_image.php?id=" . $Row[0] . "'><br></p>";

我假设$ Row [0]持有当前记录的唯一密钥。然后你编写另一个脚本，display_image.php，它只获取图像数据并使用正确的标题来显示它：

I'm assuming there that $Row[0] holds the unique key for the current record. Then you write another script, display_image.php that fetches just the image data and uses the proper headers to display it:

$currentId = $_REQUEST['id'];
//  Your query code would be here using the $currentId to just retrieve the desired record
$SQLstring = "SELECT your_image_column_name FROM $TableName WHERE id = $currentId";
$QueryResult = mysql_query($SQLstring);
$img = mysql_fetch_array($QueryResult);
$content = $img['your_image_column_name'];
header('Content-type: image/jpg');
echo $content;

这篇关于如何使用PHP从包含其他列文本的表中获取图像的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！