Question

我正在创建一个时事通讯，我希望将name插入的数据库图像显示到我的数据库中。当我尝试显示图像时，但不显示图像。

$date = $_POST['date'];

$qry_banner = "INSERT INTO `banner_tbl`(`banner_img`, `date`) VALUES ('$img_banner','$date')";
$res_banner = mysql_query($qry_banner);
$banner_tbl_id = mysql_insert_id();


$sec_id = $_POST['sectionID'];
$sec_active_image = $_POST['activeimage'];
$sec_title = $_POST['section_title'];


$total_sec = count($sec_id);

for($i=0;$i<$total_sec;$i++) 
{
$qry_section1 = "INSERT INTO `section_lt_tbl`(`banner_id`, `sectionID`, `activeimage`, `sectiontitle`) VALUES 
('$banner_tbl_id','$sec_id[$i]','$sec_active_image[$i]','$sec_title[$i]')";

 $nav_1_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '1' limit 1");
 $nav_2_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '2' limit 1");
 $nav_3_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '3' limit 1");
 $nav_4_img = mysql_query("SELECT `activeimage` FROM `section_lt_tbl` WHERE sectionID = '4' limit 1");

$res_section1 = mysql_query($qry_section1);
$sec_tbl_id1 = mysql_insert_id();

$section_generated_id[] = $sec_tbl_id1;

}
 $nav_1_img_val = mysql_fetch_assoc ($nav_1_img);
 $nav_2_img_val = mysql_fetch_assoc ($nav_2_img);
 $nav_3_img_val = mysql_fetch_assoc ($nav_3_img);
 $nav_4_img_val = mysql_fetch_assoc ($nav_4_img);

下面是我的html代码，我正在尝试显示图像：

<td bgcolor="#dbdbdb" align="center"><table border="0" cellpadding="0" cellspacing="0" width="550" align="center">
       <tr>

       <td width="25%"><a href="#featuredstartup"><img src="uploads/<?php echo $nav_1_img_val ?>_blue.jpg" width="125px" height="134px" />
</a></td>
       <td width="25%"><a href="#healthcaredeals"><img src="uploads/<?php echo $nav_2_img_val ?>_grey.jpg" width="125px" height="134px" /></a></td>
       <td width="25%"><a href="#healthcarenews"><img src="uploads/<?php echo $nav_3_img_val ?>_grey.jpg" width="125px" height="134px" /></a></td>
       <td width="25%"><a href="#healthcaremarkets"><img src="uploads/<?php echo $nav_4_img_val ?>_grey.jpg" width="125px" height="134px" /></a></td>

       </tr>
       </table>
       </td>

Answer 1

将img src标记修改为以下内容。

<img src="uploads/<?php echo $nav_1_img_val['activeimage'] ?>_blue.jpg" width="125px" height="134px" />

mysql_fetch_assoc（）将返回一个关联数组，您必须获取相应的键值。

Answer 2

您必须使用$nav_1_img_val['activeimage']而不是$nav_1_img_val，因为mysql_fetch_assoc正在返回一个数组。

您还应该考虑转移到mysqli或PDO，因为mysql已被弃用。

Answer 3

mysql_fetch_assoc返回一个关联数组。

您需要使用名称来访问数组的值列是关键。

 $nav_1_img_result = mysql_fetch_assoc ($nav_1_img);
 $nav_1_img_val = $nav_1_img_result['activeimage'];

显示图像，但显示一个数组

3 个答案: