php - 如何使用 PHP 从具有其他列文本的表格中获取图像

Question

各位开发者大家好!我正在尝试从我的数据库中检索图像以将其包含在我创建的这个表中。我在 Google 上查找的所有示例都是用于从仅包含图像的 1 个表中检索图像，但在这种情况下我无法让它工作。

<?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);     
                        }   
                ?> 

我试过了，还是不行:

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

Answer 1

你不能做你想做的事。您需要将逻辑分成两个脚本。确实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 标签被提供了一个不是原始数据的 SRC，而是要求服务器提供图像。

在您生成 HTML 的当前脚本中，您只需要让 IMG 标记将 SRC 引用为一个新脚本，该脚本将执行检索图像数据的工作。像这样的东西:

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键。然后编写另一个脚本 display_image.php，它只获取图像数据并使用适当的 header 来显示它:

$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;