我想显示来自数据库的图像，但目前它只显示图像的名称。需要帮忙。这是我的代码
提交.php
<?php
$radio = $_POST['RadioGroup1'];
$img_banner = $_FILES['bannerImage']['name'];
$target_file = "uploads/";
if(!empty($img_banner))
{
$errors = array();
$file_name = $_FILES['bannerImage']['name'];
$file_size = $_FILES['bannerImage']['size'];
$file_tmp = $_FILES['bannerImage']['tmp_name'];
$file_type = $_FILES['bannerImage']['type'];
$file_ext = strtolower(end(explode('.',$_FILES['bannerImage']['name'])));
$k = move_uploaded_file($file_tmp,"uploads/".$file_name);
}
$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]')";
$res_section1 = mysql_query($qry_section1);
$sec_tbl_id1 = mysql_insert_id();
$section_generated_id[] = $sec_tbl_id1;
}
$array_section_ids = array();
for($i=1;$i<=$radio;$i++)
{
$j = $i-1 ;
$array_section_ids[$i] = $section_generated_id[$j];
}
for($h=1;$h<=count($array_section_ids);$h++)
{
$news_ids_sec[$h] = $_POST['sec_'.$h.'_new'];
$news_title_sec[$h] = $_POST['sec_'.$h.'_title_news'];
$sec_1_desc[$h] = $_POST['sec_'.$h.'_desc'];
$sec_1_newslink[$h] = $_POST['sec_'.$h.'_newslink'];
$sec_1_news_url[$h] = $_POST['sec_'.$h.'_news_url'];
$sec_1_news_img[$h] = $_FILES['sec_'.$h.'_news_img']['name'];
$temp_file[$h] = $_FILES['sec_'.$h.'_news_img']['tmp_name'];
$section = $array_section_ids[$h];
for($c=0;$c<count($news_ids_sec[$h]);$c++)
{
$id = $news_ids_sec[$h][$c];
// $sec_id = $array_section_ids[$h][$c];
$title= $news_title_sec[$h][$c];
$desc = $sec_1_desc[$h][$c];
$link = $sec_1_newslink[$h][$c];
$url = $sec_1_news_url[$h][$c];
$img = $sec_1_news_img[$h][$c][0];
$tmp_name = $temp_file[$h][$c][0];
$qry_news_insert = "INSERT INTO `news_lt_tbl`(`newsid`, `news_sec_id`, `newstitle`, `Description`, `titleoflink`, `urlofnews`, `news_img`) VALUES ('$id','$section','$title','$desc','$link','$url','$img')";
$res_news = mysql_query($qry_news_insert);
$file_name = $img;
$file_tmp = $tmp_name;
$file_ext = strtolower(end(explode('.',$file_name)));
$k = move_uploaded_file($file_tmp,"uploads/".$file_name);
}
}
?>
<?php
$qry_test = "SELECT * FROM `section_lt_tbl` WHERE sec_id = 162";
$res_tses = mysql_query($qry_test);
$arr = mysql_fetch_assoc($res_tses);
?>
<?php
$query = "SELECT `newsid`, `newstitle`, `Description`, `titleoflink`, `urlofnews`, `news_img` FROM `news_lt_tbl` ";
$result_news = mysql_query($query);
$file_name = $img;
$file_tmp = $tmp_name;
$file_ext = strtolower(end(explode('.',$file_name)));
$k = move_uploaded_file($file_tmp,"uploads/".$file_name);
?>
和
<table align="center" width="600">
<tr>
<tr>
<td bgcolor="#ffffff" valign="bottom" align="left"><a href="http://nrsadvisors.com/"><img border="0" src="http://www.nrsadvisors.com/email-images/mainlogo.jpg" alt="NRS Advisors"/></a></td>
</tr>
<tr>
<td style="border-top-width:2px; border-top-style:solid; border-top-color: #3E7DBD;" bgcolor="#dbdbdb"><img border="0" src="http://www.nrsadvisors.com/email-images/banner-hero-image.png" alt="bannerimg" width="550" /><a name="featuredstartup" style="height:0;margin:0;padding:0"></a></td>
</tr>
<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 $arr['activeimage'] ?>_blue.jpg" width="125px" height="134px" />
</a></td>
<td width="25%"><a href="#healthcaredeals"><img src="uploads/<?php echo $arr['activeimage'] ?>_grey.jpg" width="125px" height="134px" /></a></td>
<td width="25%"><a href="#healthcarenews"><img src="uploads/<?php echo $arr['activeimage'] ?>_grey.jpg" width="125px" height="134px" /></a></td>
<td width="25%"><a href="#healthcaremarkets"><img src="uploads/<?php echo $arr['activeimage'] ?>_grey.jpg" width="125px" height="134px" /></a></td>
</tr>
</table>
</td>
</tr>
<?php
while ($row = mysql_fetch_array($result_news)) {
?>
<table align="center" width="600">
<tr>
<span style="font-size:14px;line-height:17px;font-weight:normal;color:#333333">
<td align="left">
<span style="font-weight:bold">
<?php echo $row['newsid'] ?>
</span>
</td>
</span>
</tr>
<tr>
<td>
<span align='left' style='color:#3E7DBD;font-weight:bold'>
<span style='font-size:14px;line-height:17px;font-weight:normal;color:#333333'><span style='font-weight:bold'>
<?php echo $row['newstitle'] ?>
-</span><br><br>
</td>
</tr>
<tr>
<td><?php echo $row['Description'] ?></td><br><br>
</tr>
<tr>
<td>
<a href="<?php echo $row['urlofnews'] ?>" style='font-size:14px;line-height:17px;color:#3FC47B; text-decoration:none;'><?php echo $row['titleoflink'] ?></a></td><br><br>
</tr>
<tr>
<td><?php echo $row['news_img'] ?></td>
</tr>
</table>
<?php
}
?>
最佳答案
您需要使用img 标签来显示图片。像这样:
<img src="your_upload_path/<?php echo $row['news_img'] ?>" />
关于php - 在 PHP 中显示和检索数据库中的图像，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32687852/