如何以表格（水平）显示数据库的多个图像。 [英] How to show multiple images of database in a form (Horizontally).
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问题描述
嗨朋友们，
我有图像存储在数据库和文件夹中，我想从搜索查询中逐个以横向格式显示这些图像。
我能够显示垂直格式，但如果你们有任何想法，我无法弄清楚如何做到这一点请帮助我。
i只需图片（信息不是强制性的）。
这是我的代码
< center >
< table width = 910 border = 0 cellspacing = 0 cellpadding = 3 样式 = 颜色：＃303030; font-family：Verdana，Arial，Helvetica，sans-serif;字体大小：11像素;行高：18像素; text-align：left; >
< tr bgcolor = ＃e5e5e5 >
< td colspan = 5 class = mainname > < span class = 有效 > < strong <搜索结果< / strong > < / span > < / td < span class =code-keyword>>
< / tr >
< tr >
< td >
<？ php
if（isset（$ _ POST ['Search']））
{
/ * echo < script > alert（'hello'）< / script > ; * /
$ gender = $ _ POST ['gender'];
$ agemin = $ _ POST ['agemin'];
$ agemax = $ _ POST ['agemax'];
$ hmin = $ _ POST ['hmin'];
$ hmax = $ _ POST ['hmax'];
$ religion = $ _ POST ['religion'];
$ marital_status = $ _ POST ['marital_status'];
$ city = $ _ POST ['city'];
$ state = $ _ POST ['state'];
$ country = $ _ POST ['country'];
include（db.php）;
$ query_for_result = mysql_query（select * from matri_user_info where（gender ='$ gender'）and（age agemin $ $ agemax）和（hmin与$ hmax之间的高度）和（宗教信仰） ='$ religion'）和（marital_status ='$ marital_status'）和（city ='$ city'）和（state ='$ state'）和（country ='$ country'）;）;
// $ query_for_result = mysql_query（select * from matri_user_info where（gender ='$ gender'）;）;
$ num = mysql_num_rows（$ query_for_result）;
echo < << end
< table align = right >
< tr >
< ; td bgcolor < span class =code-keyword> = ＃6fa34f style = font-family：Verdana，Arial，Helvet ica，sans-serif;字体大小：12像素;填充：5px 5px 5px 5px;颜色：#FFFFFF; > 总结果搜索< / td > < td > & nbsp;& nbsp; < / td > < td bgcolor = ＃044396 style = font-family：Verdana，Arial，Helvetica，sans-serif;字体大小：12像素;填充：5px 5px 5px 5px;颜色：#FFFFFF; > $ num < / td >
< / tr >
< / table >
end;
mysql_close（）;
？> ;
< div style = height：500px;溢出：scroll; overflow-x：hidden;宽度：910px; >
< table width = 910 cellspacing = 0 cellpadding = 10 样式 = font-family：Verdana，Arial，Helvetica，sans-serif;字体大小：11像素;颜色：＃000000;背景：＃f3f3f3;填充：1px 1px 10px 1px; text-align：left; >
< tr style = font-size：12px; text-align：left; >
< td bgcolor = ＃CCCCCC > 照片< / td >
< td bgcolor = ＃CCCCCC > 名称< / td > ;
< td bgcolor = ＃CCCCCC > 宗教< / td < span class =code-keyword>>
< td bgcolor = ＃CCCCCC > 城市< / td >
< td bgcolor = ＃CCCCCC > 州< / td >
< td bgcolor = ＃CCCCCC > 电子邮件< / td > ;
< / tr >
<！ - < / table> - >
< ; php
$ i = 0;
while （$ i < $ num）
{
$ f3 = mysql_result（$ query_for_result，$ i， candi_pic_1）;
$ f4 = mysql_result（$ query_for_result，$ i， name）;
$ f5 = mysql_result（$ query_for_result ，$ i， religion）;
$ f6 = mysql_result（$ query_for_result，$ i， city）;
$ f7 = mysql_result（$ query_for_result，$ i， state）;
$ f8 = mysql_result（ $ query_for_result，$ i， email_id）;
？ >
< tr valign = top >
< td style = border-bottom：#cdcdcd solid; border-bottom-width：1px; > <？ php echo < img src =' pic /\".$ f3。' height =' 120' width =' 100' > ;？> < < span class =code-leadattribute> / td >
< td style = border-bottom：#cdcdcd solid; border-bottom-width：1px; > <？ php echo $ f4; ？ > < / td >
< td < span class =code-attribute> style = border-bottom：#cdcdcd solid; border-bottom-width：1px; > <？ php echo $ f5; ？ > < / td >
< td < span class =code-attribute> style = border-bottom：#cdcdcd solid; border-bottom-width：1px; > <？ php echo $ f6; ？ > < / td >
< td < span class =code-attribute> style = border-bottom：#cdcdcd solid; border-bottom-width：1px; > <？ php echo $ f7; ？ > < / td >
< td < span class =code-attribute> style = border-bottom：#cdcdcd solid; border-bottom-width：1px; > <？ php echo $ f8; ？ > < / td >
< / tr >
<？ php
$ i ++;
}
}
else { < span class =code-summarycomment>
echo < script > alert（ 未找到数据）< / script > ;
}
？> < / table > < / div >
< / td >
< / tr > < / table >
< / center >
还有一个我的查询不适合搜索，所以如果你们中的任何人有任何想法相关的婚姻网站serch代码比pls分享那个。
谢谢
解决方
_POST ['Search']））
{
/ * echo <脚本 > alert（'hello'）< / script > ; * /
gender =
_POST [ '性别'];
Hi friends,
I have images stored in database and in a folder and i want to show these images in Horizontal format one by one from a search query.
I am able to show vertical format,But i am not able to figure out how to do this if any of you have any idea then pls help me out.
i only need images(information is not compulsory).
This my code
<center>
<table width="910" border="0" cellspacing="0" cellpadding="3" style="color:#303030; font-family:Verdana, Arial, Helvetica, sans-serif; font-size:11px; line-height:18px; text-align:left;">
<tr bgcolor="#e5e5e5">
<td colspan="5" class="mainname"><span class="active"><strong>Search Result</strong></span></td>
</tr>
<tr>
<td>
<?php
if(isset($_POST['Search']))
{
/*echo "<script> alert('hello')</script>";*/
$gender=$_POST['gender'];
$agemin=$_POST['agemin'];
$agemax=$_POST['agemax'];
$hmin=$_POST['hmin'];
$hmax=$_POST['hmax'];
$religion=$_POST['religion'];
$marital_status=$_POST['marital_status'];
$city=$_POST['city'];
$state=$_POST['state'];
$country=$_POST['country'];
include("db.php");
$query_for_result=mysql_query("select * from matri_user_info where (gender='$gender') and (age between $agemin and $agemax) and (height between $hmin and $hmax) and (religion='$religion') and (marital_status='$marital_status') and (city ='$city') and (state='$state') and (country='$country');");
//$query_for_result=mysql_query("select * from matri_user_info where (gender='$gender');");
$num=mysql_num_rows($query_for_result);
echo <<<end
<table align="right" >
<tr>
<td bgcolor="#6fa34f" style="font-family:Verdana, Arial, Helvetica, sans-serif; font-size:12px; padding:5px 5px 5px 5px; color:#FFFFFF;" >Total Result Search</td><td> </td><td bgcolor="#044396" style="font-family:Verdana, Arial, Helvetica, sans-serif; font-size:12px; padding:5px 5px 5px 5px; color:#FFFFFF;">$num</td>
</tr>
</table>
end;
mysql_close();
?>
<div style=" height:500px;overflow: scroll;overflow-x:hidden; width:910px; ">
<table width="910" cellspacing="0" cellpadding="10" style="font-family:Verdana, Arial, Helvetica, sans-serif; font-size:11px; color:#000000; background:#f3f3f3; padding:1px 1px 10px 1px; text-align:left;">
<tr style="font-size:12px; text-align:left;">
<td bgcolor="#CCCCCC">Photo</td>
<td bgcolor="#CCCCCC">Name</td>
<td bgcolor="#CCCCCC">Religion</td>
<td bgcolor="#CCCCCC">Cities</td>
<td bgcolor="#CCCCCC">State</td>
<td bgcolor="#CCCCCC">E-Mail</td>
</tr>
<!--</table>-->
<?php
$i=0;
while ($i < $num)
{
$f3=mysql_result($query_for_result,$i,"candi_pic_1");
$f4=mysql_result($query_for_result,$i,"name");
$f5=mysql_result($query_for_result,$i,"religion");
$f6=mysql_result($query_for_result,$i,"city");
$f7=mysql_result($query_for_result,$i,"state");
$f8=mysql_result($query_for_result,$i,"email_id");
?>
<tr valign="top" >
<td style="border-bottom:#cdcdcd solid; border-bottom-width:1px;"><?php echo "<img src='pic/".$f3."' height='120' width='100'>";?></td>
<td style="border-bottom:#cdcdcd solid; border-bottom-width:1px;"><?php echo $f4; ?></td>
<td style="border-bottom:#cdcdcd solid; border-bottom-width:1px;"><?php echo $f5; ?></td>
<td style="border-bottom:#cdcdcd solid; border-bottom-width:1px;"><?php echo $f6; ?></td>
<td style="border-bottom:#cdcdcd solid; border-bottom-width:1px;"><?php echo $f7; ?></td>
<td style="border-bottom:#cdcdcd solid; border-bottom-width:1px;"><?php echo $f8; ?></td>
</tr>
<?php
$i++;
}
}
else {
echo "<script> alert('No Data Found')</script>";
}
?></table></div>
</td>
</tr></table>
</center>
One more thind my query is not appropriate to search so if any of you have any idea related matrimonial website serch code than pls share that too.
thanks
解决方案
_POST['Search'])) { /*echo "<script> alert('hello')</script>";*/
gender=
_POST['gender'];
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