What I want to do is that, It should display 3 images on every line and along with their captions.And their could be many rows. How can do that ? here is my code..
while($info=mysql_fetch_array($query))
{
$image=$info['image'];
$cap=$info['caption'];
echo '<img src="'.$image.'" />';
echo $cap;
}
I got the answer.
I don't know hows its working . But it is ....
<table>
<?php
$i=1;
$query=mysql_query("SELECT * FROM publish");
while($info=mysql_fetch_array($query))
{
$inq=$info['ref'];
$imagesrc=$info['image'];
$ti=$info['title'];
if($i%3==1 || $info['id']==1 )
{
echo "<tr></tr>";
}
?>
<td>
<a href="movieview.php?ref=<?php echo $info['ref']; ?>">
<?php
echo '<img src="admin/'.$imagesrc.'" style="width:138px;height:200px;" />';
echo '</a><br />';
echo $ti;
$i++;
}
?>
</td>
</table>
Related
Below is my while loop table, currently it's showing just one item per row. I can't seem to figure out how to get 4 items/columns to show before it shows a new row below. Any help with this would be much appreciated, thank you!
Jerome
<table style="background-color: white">
<?php
while ($row = mysql_fetch_array($rs)) {
?>
Collapse | Copy Code
<tr <?php //if ($i % 2) echo ' style="background-color: #ECECFB;"';?>>
<td class="rows"><font color="4D4D4D" size="3"><? echo '<a type="video/x-matroska" href=file://///'.$row["Link"].' target=blank>'.$row["Name"].'</a>' ?><br />
<? echo '<a type="video/x-matroska" href=file://///'.$row["Link"].' target=blank><img src='.$row["Picture"].' height=210px width=141px></a>' ?><br />
<? echo '('.$row["Type"].')' ?> <? echo $row["Year"] ?> <? echo $row["Rating"] ?> <? echo date('H:i', mktime(0,$row["Length"])); ?><br />
<? echo $row["Genre"] ?><br />
<? if ($row["Queue"]==='x') {
echo
"<center><a title=Remove From Watch List href='deletequeue.php?id=".$row['ID']."'><img align='center' width='20px' src='http://www.ourlittlelucas.net/ourflix/images/Minus.png'></a></td></font></center>";
}
else {
echo "<center><a title=Add To Watch List href='addqueue.php?id=".$row['ID']."'><img align='center' width='20px' src='http://www.ourlittlelucas.net/ourflix/images/Plus.png'></a></td></font></center>";
}
?>
</td></tr></font>
<?php $i++?>
<?php
}
?>
</table>
Try it
<table style="background-color: white">
<?php
while ($row = mysql_fetch_array($rs))
{
?>
Collapse | Copy Code
<tr <?php //if ($i % 2) echo ' style="background-color: #ECECFB;"';?>>
<td class="rows">
<font color="4D4D4D" size="3"><?php echo '<a type="video/x-matroska" href=file://///'.$row["Link"].' target=blank>'.$row["Name"].'</a>'; ?><br />
<?php echo '<a type="video/x-matroska" href=file://///'.$row["Link"].' target=blank><img src='.$row["Picture"].' height=210px width=141px></a>'; ?><br />
<?php echo '('.$row["Type"].')' ?>
<?php echo $row["Year"] ;?>
<?php echo $row["Rating"]; ?>
<?php echo date('H:i', mktime(0,$row["Length"])); ?><br />
<?php echo $row["Genre"] ;?><br />
<?php
if ($row["Queue"]=="x") {
echo
"<center><a title=Remove From Watch List href='deletequeue.php?id=".$row['ID']."'><img align='center' width='20px' src='http://www.ourlittlelucas.net/ourflix/images/Minus.png'></a></td></font></center>";
}
else
{
echo "<center><a title=Add To Watch List href='addqueue.php?id=".$row['ID']."'><img align='center' width='20px' src='http://www.ourlittlelucas.net/ourflix/images/Plus.png'></a></td></font></center>";
}
?>
</td></tr></font>
<?php $i++?>
<?php
}
?>
</table>
i have a pagination code in php ,i tried to show images post_images from my sql to this table in my code but nothing happen ,i tried echo '<img src= images/$post_image />';
,can you tell me please what i can do ?
my code :
<table height="200px" style="border:3px black solid;border-radius:5px" width="550px">
<tr>
<th colspan="2" height="40px" style="border-bottom:3px black solid">posts in this site</th>
</tr>
<?php
$link=mysql_connect("localhost","username","pass");
mysql_select_db("itstuff_paging",$link);
$q="select count(*) \"total\" from posts";
$ros=mysql_query($q,$link) or die(mysql_error());
$row=mysql_fetch_array($ros);
$total=$row['total'];
$dis=4;
$total_page=ceil($total/$dis);
$page_cur=(isset($_GET['page']))?$_GET['page']:1;
$k=($page_cur-1)*$dis;
$q="select * from posts limit $k,$dis";
$ros=mysql_query($q,$link);
while($row=mysql_fetch_array($ros))
{
echo '<tr>';
echo '<td width="10px" style="border-bottom:1px #a1a1a1 solid">'.$row
['post_id'].'.';
echo '<br/>';
echo '<td style="border-bottom:1px #a1a1a1 solid">'.$row['post_title'];
echo '<br/>';
echo ($row['post_content']);
echo '<br/>';
echo '<img src= images/$post_image />';
echo '</tr>';
}
echo '</table>';
echo '<br/>';
if($page_cur>1)
{
echo '<a href="index.php?page='.($page_cur-1).'"
style="cursor:pointer;color:green;" ><input style="cursor:pointer;background-
color:green;border:1px black solid;border-radius:5px;width:120px;height:30px;color:white;font-
size:15px;font-weight:bold;" type="button" value=" Previous "></a>';
}
else
{
echo '<input style="background-color:green;border:1px black solid;border-
radius:5px;width:120px;height:30px;color:black;font-size:15px;font-weight:bold;" type="button"
value=" Previous ">';
}
for($i=1;$i<$total_page;$i++)
{
if($page_cur==$i)
{
echo ' <input style="background-color:green;border:2px black
solid;border-radius:5px;width:30px;height:30px;color:black;font-size:15px;font-weight:bold;"
type="button" value="'.$i.'"> ';
}
else
{
echo '<a href="index.php?page='.$i.'"> <input style="cursor:pointer;background-
color:green;border:1px black solid;border-radius:5px;width:30px;height:30px;color:white;font-
size:15px;font-weight:bold;" type="button" value="'.$i.'"> </a>';
}
}
if($page_cur<$total_page)
{
echo '<a href="index.php?page='.($page_cur+1).'"><input
style="cursor:pointer;background-color:green;border:1px black solid;border-
radius:5px;width:90px;height:30px;color:white;font-size:15px;font-weight:bold;" type="button"
value=" Next "></a>';
}
else
{
echo '<input style="background-color:green;border:1px black solid;border-
radius:5px;width:90px;height:30px;color:black;font-size:15px;font-weight:bold;" type="button"
value=" Next ">';
}
?>
</table>
You never define $post_image before you use it, so you're generating <img src="" />. Perhaps you meant:
echo "<img src=images/{$row['post_image']} />";
instead?
Your code is malformed. Please try:
echo '<img src="images/'.$post_image.'" />';
I have a web page with images and when user clicks on any of the image, it has to derive data of that particular image from MYSQL database. What I am doing is using a simple JavaScript popup and putting the data from database. However I am just getting the first item from database on all images.
This is the code:
$files = glob("admin/images/paintings/*.*");
echo '<div id="painting"><table border="0" style="width:590px;">';
$colCnt=0;
$i = 0;
while($row = mysql_fetch_array($result))
{
if ($colCnt%4==0)
echo '<tr>';
echo '<td width="25%" style="font-size:8.5px; font-family:arial">';
echo($i);
$num = $files[$i];
echo '<img id="indPainting" src="'.$num.'" align="absmiddle" /> <br> <div id="paintingName">';
print $row['name'];
echo '<div id="openModal" class="modalWindow">
<div>
<p>This is a sample modal window that can be created using CSS3 and HTML5.'.$row['name'].'</p>
Ok
</div>
</div>';
echo '</td>';
$colCnt++;
if ($colCnt==4)
{
echo '</tr>';
$colCnt=0;
}
$i++;
}
mysql_close($con);
include 'footer.php';
?>
$row['name'] is just giving out the first name as it is in a while loop. I am not being able to get other names for other images. How can this be done. Any help would be appreciated.
Does one iteration in your while fetch single image data? And what I can understand according to your code is that you are displaying 4 image in a row.
Can you please format your code a bit..its looking too ugly.
I need to know which statement is calling your modal window.
<?php
$files = glob("admin/images/paintings/*.*");
echo '<div id="painting"><table border="0" style="width:590px;">';
$colCnt=0;
$i = 0;
echo '<tr>';
while($row = mysql_fetch_array($result))
{
$num = $files[$i];
echo '<td width="25%" style="font-size:8.5px; font-family:arial">';
echo '<img id="indPainting" src="'.$num.'" align="absmiddle" /> <br>
<div id="paintingName">';
print $row['name'];
echo '<div id="openModal" class="modalWindow"><div><p>This is a sample modal window that can be created using CSS3 and HTML5.'.$row['name'].'</p>Ok</div>
</div></td>';
$colCnt++;
if ($colCnt % 4 == 0)
{
echo '</tr>';
$colCnt=0;
}
$i++;
}
mysql_close($con);
include 'footer.php';
?>
Try this.
Also see how beautiful the code looks if its properly formatted..
try this
<?php
$files = glob("admin/images/paintings/*.*");
echo '<div id="painting"><table border="0" style="width:590px;">';
$colCnt=4;
while($row = mysql_fetch_array($result))
{
for ($i = 0; $i < $colCnt; $i++) {
echo '<tr>';
echo '<td width="25%" style="font-size:8.5px; font-family:arial">';
echo($i);
$num = $files[$i];
echo '<img id="indPainting" src="'.$num.'" align="absmiddle" /> <br> <div id="paintingName">';
print $row['name'];
echo '<div id="openModal" class="modalWindow">
<div>
<p>This is a sample modal window that can be created using CSS3 and HTML5.'.$row['name'].'</p>
Ok
</div>
</div>';
echo '</td>';
}
if ($colCnt==4)
{
echo '</tr>';
$colCnt=0;
}
}
mysql_close($con);
include 'footer.php';
?>
hi i'm disaplyed images from mysql db table but it displays on by one means one row has one image. but i need 3 or 4 image per row. my coding is below. please give some idea.
<?php
include_once("config.php");
$result=mysql_query("SELECT * FROM merchant");
while($res=mysql_fetch_array($result))
{
?>
<?php echo $res['description'];?></p>
<img src="<?php echo $res['image'];?>" width="80" height="80"/>
<?php } ?>
Do it in table like this, You might need to fix it a little bit, but it way how it will work
<table>
<?php
include_once("config.php");
$result=mysql_query("SELECT * FROM merchant");
$count = 0;
while($res=mysql_fetch_array($result))
{
if($count==3) //three images per row
{
print "</tr>";
$count = 0;
}
if($count==0)
print "<tr>";
print "<td>";
?>
<?php echo $res['description'];?></p>
<img src="<?php echo $res['image'];?>" width="80" height="80"/>
<?php
$count++;
print "</td>";
}
if($count>0)
print "</tr>";
?>
</table>
use a <table> to display.
<?php
include_once("config.php");
$result=mysql_query("SELECT * FROM merchant");
$count = 0;
echo '<table>';
while($res = mysql_fetch_array($result))
{
if($count % 2 == 0) echo '<tr>';
?>
<td>
<p><?php echo $res['description'];?></p>
<img src="<?php echo $res['image']; ?>" width="80" height="80"/>
</td>
<?php
if($count % 2 == 0) echo '</tr>';
} ?>
I'm working on a php gallery. I'm displaying image from mysql db using php, but my images are displaying in one by one. which means the first image in first row and second image in second row. but I want to display my image as 3 or 4 per row. what coding changes can I make. my php code as shown below.
<?php
include_once("config.php");
$result=mysql_query("SELECT * FROM images");
while($res=mysql_fetch_array($result)){ ?>
<table width='200'>
<tr>
<td><?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?><?php echo $res['imagename']?><?php echo"</a>"?></td>
</tr>
<tr>
<td>
<div id="news-image">
<?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?>
<?php echo'<img src='.$res['image'].' width="250" height="100">'?><?php echo"</a>"?>
</div>
</td>
</tr>
</table>
<?php } ?>
array_chunk() is a function to split an array into a collection of X items for you to loop through without having to keep counters (you can then use array_pad() on the last item in the list if you need padding)
if($array = array_chunk(mysql_fetch_assoc($result),4))
{
foreach($array as $row)
{
echo '<div class="row">';
foreach($row as $col)
{
echo '<div class="item">' . $col['image'] . '</div>';
}
echo '</div>';
}
}
You're outputting a table per image. At minimum, your code should be more like this:
<table>
<tr>
<?php while($res etc...) { ?>
<td>
<img src="<?php echo ......?>" />
</td>
<?php } ?>
</tr>
</table>
Now you'll get all the images in a single row of a single table. Making it have multiple rows is left as an exercise to the OP.
follow this example
<table>
<tr>
<?php
$i = 1;
do{
echo "<td>" . $i . "</td>";
//Num of Columns
if( $i%3 == 0 ){
echo "</tr><tr>";
}
$i++;
}while($i<=10);
?>
</tr>
<table>
will return the result something like you want..
<?php
include_once("config.php");
$result=mysql_query("SELECT * FROM images");
?>
<table>
<tr>
<?php
$cnt = 0;
while($res=mysql_fetch_array($result))
{
if($cnt == 3){
echo "</tr><tr>";
}
?>
<td>
<table width='200'>
<tr>
<td><?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?><?php echo $res['imagename']?><?php echo"</a>"?></td>
</tr>
<tr>
<td>
<div id="news-image">
<?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?>
<?php echo'<img src='.$res['image'].' width="250" height="100">'?><?php echo"</a>"?>
</div>
</td>
</tr>
</table>
</td>
<?php
$cnt++;
}
?>
</tr>
</table>
Use the following code.
<?php
include_once("config.php");
$result=mysql_query("SELECT * FROM images");
?>
<table>
<tr>
<?
$varcount=0;
while($res=mysql_fetch_array($result))
{
$varcount++;
if($varcount == 4) // Count of images per row. 3 or 4
{
$varcount=0;
?>
</tr><tr>
<?
}
?>
<td>
<table width='200'>
<tr>
<td><?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?><?php echo $res['imagename']?><?php echo"</a>"?></td>
</tr>
<tr>
<td>
<div id="news-image">
<?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?>
<?php echo'<img src='.$res['image'].' width="250" height="100">'?><?php echo"</a>"?>
</div>
</td>
</tr>
</table>
</td>
<?php
}
?>
</tr>
</table>
You could instead output the images as a list of divs, or just divs, and then use CSS to show the images in two columns. Your layout should not be that hardwired.
<style>
div.gallery {
width: 650px;
}
div.gallery ul li {
list-style: none;
float: left;
}
div.image {
height: 500px;
width: 300px;
}
</style>
<div class="gallery">
<ul>
<li>
<div class="image">
<span class="image_title">Some title</span><br/>
<img src="foo.png"/>
</div>
</li>
<li>
<div class="image">
<span class="image_title">Another title</span><br/>
<img src="bar.png"/>
</div>
</li>
<li>
<div class="image">
<span class="image_title">Another title</span><br/>
<img src="foo.png"/>
</div>
</li>
<li>
<div class="image">
<span class="image_title">Another title</span><br/>
<img src="bar.png"/>
</div>
</li>
<li>
<div class="image">
<span class="image_title">Another title</span><br/>
<img src="foo.png"/>
</div>
</li>
</ul>
</div>
Result:
Your code should look something like:
<div class="gallery">
<ul>
<?php
include_once("config.php");
$result = mysql_query("SELECT * FROM images");
while($res = mysql_fetch_array($result)) {
?>
<li>
<div class="image">
<a class="image_title" href="indimage.php?imageid=<?php echo $res['imageid']?>"><?php echo $res['imagename']?></a><br/>
<img src="<?php echo $res['image']?>" />
</div>
</li>
<?php
}
?>
</ul>
</div>
Here is the Solution i think you want that:
there are three pages.
1. index.php (which has the form for uploading the image)
2. upload.php (which save the image in directory and its path in database)
3. showimage.php (Finally, which will show the image)
here is the code
(index.php)
<form method="post" action="upload.php" enctype="multipart/form-data">
<label>Choose File to Upload:</label><br />
<input type="hidden" name="id" />
<input type="file" name="uploadimage" /><br />
<input type="submit" value="upload" />
</form>
(upload.php)
<?php
$target_Folder = "upload/"; // directory where images will be saved
$target_Path = $target_Folder.basename( $_FILES['uploadimage']['name'] );
$savepath = $target_Path.basename( $_FILES['uploadimage']['name'] );
$file_name = $_FILES['uploadimage']['name'];
if(file_exists('upload/'.$file_name))
{
echo "That File Already Exisit";
}
else
{
// Database
$con=mysqli_connect("localhost","user_name","pasword","database"); // Change it if required
//Check Connection
if(mysqli_connect_errno())
{
echo "Failed to connect to database" . mysqli_connect_errno();
}
$sql = "INSERT INTO image (id,image, image_name)
VALUES ('$uid','$target_Folder$file_name','$file_name') ";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added successfully in the database";
echo '<br />';
mysqli_close($con);
// Move the file into UPLOAD folder
move_uploaded_file( $_FILES['uploadimage']['tmp_name'], $target_Path );
echo "File Uploaded <br />";
echo 'File Successfully Uploaded to: ' . $target_Path;
echo '<br />';
echo 'File Name: ' . $_FILES['uploadimage']['name'];
echo'<br />';
echo 'File Type: ' . $_FILES['uploadimage']['type'];
echo'<br />';
echo 'File Size: ' . $_FILES['uploadimage']['size'];
}
?>
Show Image
(showimage.php)
<?php
$con=mysqli_connect("localhost","user_name","password","database_name"); // Change it if required
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM image " );
while($row = mysqli_fetch_array($result))
{
echo '<img src="' . $row['image'] . '" width="200" />';
echo'<br /><br />';
}
mysqli_close($con);
?>
Features
It will check the names of file if that name file already exisit it will not uplad the file and alert the user.
Database Structure
id int(4) Auto Increment - image varchar(100) - image_name varchar(50)