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PHP - Image Display Table
I have the code below displaying images and image names. I want these to display in a table 2 rows high by the needed number of columns to show all the images in the directory. I have no idea what to do. What I am getting now is a single column with each image in its own row.
<?php $path = "./uploaded/"; $dir_handle = @opendir($path) or die("Unable to open folder"); while (false != ($file = readdir($dir_handle))) { if($file == "index.php") continue; if($file == ".") continue; if($file == "..") continue; //show in a table 2 rows by required/needed number of columns echo'<div>'; echo '<table border="1">'; echo "<img src='$path/$file' alt='$file'>"."<img src='$file' alt='$file'>"; echo'</table>'; echo '<div>'; } ?> Similar Tutorialscan anione help me to display imageid along wid each image in table format...here is my code which takes image from user....n d id can b anithing i mean its ur choice u can start wid first image by giving it imageid 1 n can continue til d image ends <?php mysql_connect("localhost","root",""); @mysql_select_db(proj) or die( "Unable to select database"); $query="SELECT * FROM form"; $result=mysql_query($query); //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","100"); //This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['Submit'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Unknown extension!</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>You have exceeded the size limit!</h1>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $newname="upload/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; }}}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { $sql=mysql_query("INSERT INTO image (img) VALUES('$newname')"); if($sql) { echo "<h1>File Uploaded Successfully! Try again!</h1>"; } } ?> <!--next comes the form, you must set the enctype to "multipart/frm-data" and use an input type "file" --> <form name="newad" method="post" enctype="multipart/form-data" action=""> <table> <tr><td><input type="file" name="image"></td></tr> <tr><td><input name="Submit" type="submit" value="Upload image"></td></tr> </table> <p><a href="resizeimg.php">resize the images </a></p> </form> <p> </p> here is my code for displaying d images Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance. BTW, is using php only doable? Is there a simpler or more elegant way to do this? Thanks all! Hello, Five images will be displayed inside a division. There will be a previous and next button/link. If someone click the next button the next image will be added in that div and the first image will be gone from that div. The previous button/link will do the same thing. Is it possible with php? I am confused if it's a javascript or ajax question. Thanks. Hi guys,
I've got the following command im trying to push out how it would look if i ran it in ssh
<?php
if (!function_exists("ssh2_connect")) die("function ssh2_connect doesn't exist");
if(!($con = ssh2_connect("hostname", 22))){
echo "fail: unable to establish connection\n";
} else {
if(!ssh2_auth_password($con, "username", "password")) {
echo "fail: unable to authenticate\n";
} else {
echo "okay: logged in...\n <br>";
if (!($stream = ssh2_exec($con, "showspace" ))) {
echo "fail: unable to execute command\n";
} else {
stream_set_blocking($stream, true);
$data = "";
while ($buf = fread($stream,4096)) {
echo $data .= $buf;
}
fclose($stream);
}
}
}
?>
This displays as:
---Estimated(MB)---- ---Estimated(MB)---- RawFree UsableFree ---Estimated(MB)---- RawFree UsableFree 135770112 67885056
in putty it displays as:
TestRepo cli% showspaceHey everyone, I have a problem here.Now as u see my data is being displayed one after another in vertical manner.But what do i want to do is the entire table being displayed in same page continuously one after another in horizontal manner.how would i do that? (below a screen shot is given how my table looks like.) <code> <?php for($i=0;$i<=25;$i++) { if($i%5==0) { ?> <table> <tr> <th scope="row">U_Id :</th> <td><?php echo 'uid'; ?></td> </tr> <tr> <th scope="row">Name :</th> <td><?php echo "name"; ?></td> </tr> <tr> <th scope="row">Teamname :</th> <td><?php echo "teamname"; ?></td> </tr> <tr> <th scope="row">Coins :</th> <td><?php echo 'coins';?></td> </tr> <tr> <th scope="row">Cash:</th> <td><?php echo 'cash'; ?></td> </tr> </table> <br/> <?php //echo "$i<br/>"; } } ?> </code> This may be in here already and I'm sorry for not being able to find it but.. I just want to display some information that I get from the database in a HTML table only using 2 columns. Do I just set i=1 and run a if statement to see what column I am on? like Code: [Select] $i =0 While ($row = msyql_fetch_array($result)) { if ($i==0) { // start a new row echo "<tr>"; $i = $i++; }else{ //columns echo "<td>info</td>"; $i=$i++; } if ($i < 2) { //end the row reset $i echo "</tr>"; $i =0; } } Or something like that, am I heading in right direction? Thanks Stephen i have a stupid simple problem here, but ive never done this exactly this way before and im having a tough time working it out. was looking for any suggestions. my script is working fine: $table_name = "plan"; $sql = "SELECT id, plan_name FROM $table_name ORDER BY plan_name"; $query = mysql_query($sql); if(mysql_num_rows($query) > 0) { echo "<table>"; while($row = mysql_fetch_array($query)) { echo "<tr><td>" . $row['plan_name'] ."</tr></td>"; } echo "</table>"; ... but i have another table company that are related to plan_name i have the company_id field in my plan table for the relationship. and that id is related to the company id field of course. all i am trying to do is display the company_name next to $row['plan_name'], so i know which plans are related to which company. i think i need to create two seperate queries, but is there a way to include everyone in one query? is there a better way? here is my sql: plan table: `id` int(11) NOT NULL AUTO_INCREMENT, `company_id` int(11) NOT NULL, `plan_name` varchar(255) NOT NULL, ... company table `id` int(11) NOT NULL AUTO_INCREMENT, `plan_id` int(11) NOT NULL, `company_name` varchar(255) NOT NULL,... it also seemed to be overkill to have two while loops running.. i am just thinking out loud on the best approach to this. Hi I have the following code: Code: [Select] $result = mysql_query("SELECT * FROM xbox_games order by gameid"); $row = mysql_fetch_assoc($result); $id = $row["gameid"]; $title = $row["gametitle"]; $cover = $row["cover"]; <?php echo $title;?> Which displays only the first result from my database. How can i change this to display all the results either as a list, or in a table? Thanks Hi, i cant seem to get something working, should be simple but its not working for me. I just need to only display a table if a variable in my table = a certain value. The column in the table is called 'option1_available' and if its value is set to 'Y' i want it to display a table. Appreciate any help, Thanks How to display items like bottom table?
And here is the code
<?
print"
<table style=\"width:100%\" class=\"tableList\">
<tr>
<th style=\"width:35%\">Prize Name</th>
<th style=\"width:12%\">Amount</th>
<th style=\"width:12%\">Points</th>
<th style=\"width:12%\">Available</th>
<th style=\"width:12%\">Redeemed</th>
<th style=\"width:17%\">Action</th>
</tr>";
$giftCardQuery = 'SELECT currency, amount, pointsPrice, instant_gift_cards.id, instant_gift_cards.giftCardName, instant_gift_cards.giftCardImage FROM instant_gift_card_codes INNER JOIN instant_gift_cards ON (instant_gift_card_codes.giftCardId = instant_gift_cards.id) WHERE instant_gift_cards.status = :cardStatus ORDER BY instant_gift_cards.dateCreated DESC';
$giftCard = $db->prepare($giftCardQuery);
$giftCard->bindValue(':cardStatus', 'Enabled', PDO::PARAM_STR);
$giftCard->execute();
if($giftCard->rowCount() > '0'){
while($giftCardRow = $giftCard->fetch(PDO::FETCH_ASSOC)){
$giftCardsAvailableQuery = 'SELECT count(*) FROM instant_gift_card_codes WHERE currency = :currency AND amount = :amount AND pointsPrice = :pointsPrice AND giftCardId = :id AND status = :status';
$giftCardsAvailable = $db->prepare($giftCardsAvailableQuery);
$giftCardsAvailable->bindParam(':currency', $giftCardRow['currency'], PDO::PARAM_STR);
$giftCardsAvailable->bindParam(':amount', $giftCardRow['amount'], PDO::PARAM_STR);
$giftCardsAvailable->bindParam(':pointsPrice', $giftCardRow['pointsPrice'], PDO::PARAM_STR);
$giftCardsAvailable->bindParam(':id', $giftCardRow['id'], PDO::PARAM_INT);
$giftCardsAvailable->bindValue(':status', 'Available', PDO::PARAM_STR);
$giftCardsAvailable->execute();
$gCardsAvailable = $giftCardsAvailable->fetch(PDO::FETCH_COLUMN);
$giftCardsRedeemedQuery = 'SELECT count(*) FROM instant_gift_card_codes WHERE currency = :currency AND amount = :amount AND pointsPrice = :pointsPrice AND giftCardId = :id AND status = :status';
$giftCardsRedeemed = $db->prepare($giftCardsRedeemedQuery);
$giftCardsRedeemed->bindParam(':currency', $giftCardRow['currency'], PDO::PARAM_STR);
$giftCardsRedeemed->bindParam(':amount', $giftCardRow['amount'], PDO::PARAM_STR);
$giftCardsRedeemed->bindParam(':pointsPrice', $giftCardRow['pointsPrice'], PDO::PARAM_STR);
$giftCardsRedeemed->bindParam(':id', $giftCardRow['id'], PDO::PARAM_INT);
$giftCardsRedeemed->bindValue(':status', 'Redeemed', PDO::PARAM_STR);
$giftCardsRedeemed->execute();
$gCardsRedeemed = $giftCardsRedeemed->fetch(PDO::FETCH_COLUMN);
if($giftCardRow['giftCardImage']){
$nameOrImage = '<img src="./images/giftcardrewards/'.$giftCardRow['giftCardImage'].'" alt="'.$giftCardRow['giftCardName'].'" title="'.$giftCardRow['giftCardName'].'">';
}else{
$nameOrImage = $giftCardRow['giftCardName'];
}
if($gCardsAvailable == '0'){
$redeemAction = 'Out of Stock';
}
elseif($userInfo['currentPoints'] < $giftCardRow['pointsPrice']){
$needed = $giftCardRow['pointsPrice'] - $userInfo['currentPoints'];
$redeemAction = 'You need '.$needed.' point(s)';
}
elseif($userInfo['currentPoints'] >= $giftCardRow['pointsPrice']){
$redeemAction = '<input type="button" value="Redeem" onclick="if(confirm(\'Are you sure to redeem this prize?\')){location.href=\'index.php?do=instantGiftCards&action=redeem&cardId='.$giftCardRow['id'].'&amount='.$giftCardRow['amount'].'\';}">';
}
print"
<tr>
<td>".$nameOrImage."</td>
<td style=\"text-align:center\">".$giftCardRow['currency'].$giftCardRow['amount']."</td>
<td style=\"text-align:center\">".$giftCardRow['pointsPrice']."</td>
<td style=\"text-align:center\">".$gCardsAvailable."</td>
<td style=\"text-align:center\">".$gCardsRedeemed."</td>
<td style=\"text-align:center\">".$redeemAction."</td>
</tr>";
}
}else{
print"
<tr>
<td colspan=\"4\" style=\"text-align:center;color:#2B1B17;padding:15px 0\">No prizes added.</td>
</tr>";
}
print"
</table>";
?>
Hello, I want to make a list with 2 columns on one of my pages with link to categories on my page. But they appear in only one column (on below the other) and i want them to be like: Pictures Videos Pictures2 Videos2 These are my codes <li> <a href="user_album_add.php">Pictures</a> <li> <a href="user_album2_add.php">Pictures2</a> <li> <a href="user_video_add.php">Video</a> <li> <a href="user_video2_add.php">Video2</a> What should i do? Thank you! I would like to echo field names in my table on webpage one by one. I know mySQL has "describe" function which will list the complete table, I am looking for a way to display each field name one by one with other stuff in between them like input field or description. Sorry for the beginner question, here I'm trying to retrieve data from the database and display it in the table format. But only table headers are printed, and not the actual values. The count variable is echoing 2 saying that data is present and correctly retrieved. Can anyone help?
<?php
include 'connect.php';
error_reporting(E_ALL ^ E_DEPRECATED);
error_reporting(E_ERROR | E_PARSE);
$sql="SELECT * FROM `resources` as r INNER JOIN `project_resources` as pr
ON r.res_id =pr.res_id WHERE project_id='$_POST[project_id]'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
if($result === FALSE) {
die(mysql_error());
}
echo "$count";
echo '<table>
<tr>
<th>Resource ID</th>
<th>Resource Name</th>
<th>Email</th>
<th>Phone Number</th>
<th>Reporting Manager</th>
<th>Role</th>
<th>Designation</th>
</tr>';
while ($row = mysql_fetch_array($result)) {
echo '
<tr>
<td>'.$row['res_id'].'</td>
<td>'.$row['res_name'].'</td>
<td>'.$row['email'].'</td>
<td>'.$row['phone_number'].'</td>
<td>'.$row['reporting_manager'].'</td>
<td>'.$row['role'].'</td>
<td>'.$row['designation'].'</td>
</tr>';
}
echo '
</table>';
?>
Edited by mac_gyver, 22 September 2014 - 07:25 AM. code tags please Ok the pagination part is all working fine. but i thought id be able to create a heap variables inside the loop then display the images in a table. the only trouble is all variables are grabbing the same img. i need them to grab the 10 different records. thanks Code: [Select] $sql = "SELECT * FROM mongrels_db.gallery ORDER BY id DESC LIMIT $offset, $rowsperpage "; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); // while there are rows to be fetched... while ($list = mysql_fetch_array($result)) { $img1=$list['img']; $img2=$list['img']; $img3=$list['img']; $img4=$list['img']; $img5=$list['img']; $img6=$list['img']; $img7=$list['img']; $img8=$list['img']; $img9=$list['img']; $img10=$list['img']; // echo data } // end while echo "<table><tr>"; echo "<td>"."<img src='../gallery/".$img1 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img2 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img3 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img4 ."' width='100' height='100''> "."</td></tr>"; echo "<tr><td>"."<img src='../gallery/".$img5 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img6 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img7 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img8 ."' width='100' height='100''> "."</td>"; echo "<tr><td>"."<img src='../gallery/".$img9 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img10 ."' width='100' height='100''> "."</td></tr>"; When displaying data from a mysql table, what php code can I use to display the number of columns in said table? Like, say I'm displaying comments by users. I want to be able to put "displaying 'thismany' comments". Any suggestions? I'm trying to display relevant user details when click after user details button.it is css pop up window.i want to know how to catch relevant user when click the button.my primary key is email. Hi
I require a PHP code to get output table.
My Table
Item Qty Date aa-1 2 2014-10-01 aa-2 5 2014-10-01 aa-3 1 2014-10-01 ab-1 2 2014-10-01 ab-2 1 2014-10-01 bb-1 4 2014-10-01 bb-2 3 2014-10-01 bb-3 2 2014-10-01 aa-1 1 2014-10-02 aa-2 2 2014-10-02 aa-3 5 2014-10-02 ab-1 6 2014-10-02 ab-2 1 2014-10-02 bb-1 9 2014-10-02 bb-2 0 2014-10-02 bb-3 4 2014-10-02 aa-1 1 2014-10-03 aa-2 2 2014-10-03 aa-3 5 2014-10-03 ab-1 4 2014-10-03 ab-2 3 2014-10-03 bb-1 1 2014-10-03 bb-2 8 2014-10-03 bb-3 2 2014-10-03 I wrote code as mentioned below.
<?php
$accounts=mysql_connect("localhost", "root", "") or die("could not connect");
mysql_select_db("shops",$accounts) or die("could not find db!");
if(isset($_POST['search']) && ($_POST['from']) && ($_POST['to'])){
$searchq=$_POST['search'];
$searchq=preg_replace("#[^0-9a-z]#i", "" , $searchq);
$from=$_POST['from'];
$to=$_POST['to'];
$dateInput = explode('-',$from);
$fdate = $dateInput[2].'-'.$dateInput[1].'-'.$dateInput[0];
$dateInput = explode('-',$to);
$tdate = $dateInput[2].'-'.$dateInput[1].'-'.$dateInput[0];
for ($date=$fdate; $date<=$tdate; $date++) {
$sql = "SELECT item, SUM(CASE WHEN `date` = '$date' THEN Qty ELSE 0 END)
FROM shop WHERE item LIKE '%$searchq%' GROUP BY item";
$query = mysql_query($sql) or die("could not search!");
echo "<table border='1'>";
echo "<tr>
<td>Item </td>
<td>$date</td>
</tr>" ;
while ($row=mysql_fetch_array($query)) {
echo "<tr>
<td>". $row[0] ." </td>
<td>" . $row[1] . "</td>
</tr>";
}
echo "</table>";
}
}
?>
I am getting result like this as I asked for three days. Item 2014-10-01 aa-1 2 aa-2 5 aa-3 1 Item 2014-10-02 aa-1 1 aa-2 2 aa-3 5 Item 2014-10-03 aa-1 1 aa-2 2 aa-3 5 But I need result like below Item 2014-10-01 2014-10-02 2014-10-03 aa-1 2 1 1 aa-2 5 2 2 aa-3 1 5 5 Can anybody help me to write PHP code to display result as needed. Please help. Thanks in advance. Hello, I know this should be pretty simple to figure out, but everything I try is giving me absolutely no results. I have a mysql query selecting columns from my database and returning results. I have the results printing out right now, so I can see that this part is working. All I want to do is take the results and put them into a table to display on my page. Basically, take what's in the database table and copy it to a table I can put on the web. FYI I am using sourcerer so the "connect" code is taken care of for me in the "JFactory" bit of code. Here is the first part of my code, selecting the information from the database. {source} <?php $db = JFactory::getDbo(); $query = $db -> getQuery(true); $query -> SELECT($db -> quoteName(array('first_dept_name', 'last_name', 'dept', 'position', 'phone_num'))); $query -> FROM ($db -> quoteName('#__custom_contacts')); $query -> ORDER ('first_dept_name DESC'); $query -> WHERE ($db -> quoteName('contact_category')."=".$db -> quote('YTown Employees')); $db -> setQuery($query); $results = $db -> loadObjectList(); print_r($results); Here is where I am trying to print the results into a table. I got this code directly from a PHP book, but I am getting nothing at all returned back to me. I get table headers, but no data. <?php echo "<table><tr><th>Name</th><th>Department</th></tr>"; while ($row = mysqli_fetch_array ($result)){ echo "<tr>"; echo "<td>".$row['last_name']."</td>"; echo "<td>".$row['dept']."</td>"; echo "</tr>"; } echo "</table>"; ?> {/source} I have a table that needs to display data and its formatting should allow a scroll bar after a bit of length. In this case, the below code seems to allow it to continue well past the footer of the page... Did I miss something obvious?
<div class="col-md-12">
<div class="card card-plain">
<div class="header">
<h4 class="title">Current Vendors</h4>
<p class="category">Vendors listed as active within VendorBase.</p>
</div>
<?php
$con=mysqli_connect("localhost","root","test","vendors");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM vendor_data");
echo " <div class='content table-responsive table-full-width'>
<table class='table table-hover'>
<thead>
<th>Name</th>
<th>Type</th>
<th>Company</th>
<th>Email</th>
<th>SOC2 Report</th>
<th>Status</th>
</thead>
<tbody>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['type'] . "</td>";
echo "<td>" . $row['company'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['soc'] . "</td>";
echo "<td>" . $row['status'] . "</td>";
echo "</tr>";
}
echo "</table>";
echo "</div>";
mysqli_close($con);
?>
</table>
</div>
</div>
</div>
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