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PHP - How To Get Pictures To Start On A New Line
Hello there,
I am working on a pagination script and have hit a bit of a wall. Here is the script: Code: [Select] <?php include("header.php"); ?> <?php $host = "localhost"; // $user = ""; //username to connect to database $pass = ""; //password to connect to database $db = ""; //the name of the database mysql_connect($host,$user,$pass) or die("ERROR:".mysql_error()); mysql_select_db($db) or die("ERROR DB:".mysql_error()); $max = 25; //amount of articles per page. $p = $_GET['p']; if(empty($p)) { $p = 1; } $limits = ($p - 1) * $max; //view the article! if(isset($_GET['act']) && $_GET['act'] == "view") { $id = $_GET['id']; $sql = mysql_query("SELECT * FROM 3dpics WHERE id = '$id'"); while($r = mysql_fetch_array($sql)) { $title = $r['title']; $url = $r['url']; echo "<div align='center'><img src='$url' width='500' height='500'><br />$title<br /><br /><a href='test.php'>Back</a></div>"; } }else{ //view all the articles in rows $sql = mysql_query("SELECT * FROM 3dpics LIMIT ".$limits.",$max") or die(mysql_error()); //the total rows in the table $totalres = mysql_result(mysql_query("SELECT COUNT(id) AS tot FROM 3dpics"),0); $totalpages = ceil($totalres / $max); echo "<table name='myTable' cellpadding='5' cellspacing='5'>"; echo "<tr>"; //the table while($r = mysql_fetch_array($sql)) { $id = $r['id']; $title = $r['title']; $url = $r['url']; echo "<div id='piccontentalign' align='center'>"; echo "<td><img src='$url' width='225' height='225'><br /><div align='center'><a href='test.php?act=view&id=$id'>$title</a></div></td>"; } echo "</tr></table></div>"; //close up the table for($i = 1; $i <= $totalpages; $i++){ //this is the pagination link echo "<a href='test.php?p=$i'>$i</a>|"; } } ?> All this does is pull the url's of the images out of the database. But, it all shows up on 1 line across the screen! I tried echoing the rows instead, but then it just shows up as 1 line going horizontal. So i'm trying to figure out a way to get 5 pictures per row. This is the code that displays them on the main page: Code: [Select] echo "<table name='myTable' cellpadding='5' cellspacing='5'>"; echo "<tr>"; //the table while($r = mysql_fetch_array($sql)) { $id = $r['id']; $title = $r['title']; $url = $r['url']; echo "<div id='piccontentalign' align='center'>"; echo "<td><img src='$url' width='225' height='225'><br /><div align='center'><a href='test.php?act=view&id=$id'>$title</a></div></td>"; } echo "</tr></table></div>"; //close up the table Thanks for all your help! Similar TutorialsI keep having the same problem when i run my script on the server. the error is Code: [Select] Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/alliapop/public_html/ekloges2010/ekloges2010/eisagogi.php:1) in /home/alliapop/public_html/ekloges2010/ekloges2010/includes/session.php on line 1I know i have to put session start at he very first line and i did but the problem keeps showing up. What else should i check?
try {
echo "<br>";
foreach($dbh->query("SELECT * FROM test_shot WHERE sold=1 ORDER BY year ASC") as $row) {
if($row['picture'] != "" && $row['picture'] != null)
{
echo "<div class='image-holder'><img src ='".$row['picture']."' width=300px /><br>";
}
if($row['year'] != "" && $row['year'] != null)
{
echo $row['year'];
}
if($row['description'] != "" && $row['description'] != null)
{
echo $row['description'];
}
if($row['sold'] == 1) {
echo "<img src='images/sold1.png'><br>";//Add your image code here
}
elseif ($row['sold'] == 0) {
echo "</div><br>";
}
}
}
catch (PDOException $e) {
print $e->getMessage();
}
?>
what im trying to do is take a youtube embed code find the URL code for that video and remove all other parts of the code keeping only the URL of the video after i get the URL by it self then replace the http://www.youtube.com/ part of the URL with http://i2.ytimg.com/vi/ to do this i know that i need something like this to get the URL of the video http://www.youtube.com/((?:v|cp)/[A-Za-z0-9\-_=]+) but how to only return just the URL is something idk how to do then use str_replace http://www.youtube.com/(?:v|cp)/" with http://i2.ytimg.com/vi/ so in the end im asking if anyone know how to remove all other codes i dont want and only keep the URL this may have to be done using "regex" im not sure as i dont know to much about regex and what it can do but does sound like it would help lol i have to read a single line from a csv, its a really big file and i only need one column.
i need the response to be a string ,i made a search and found the following code but i dont have any idea how to get a single line from a single string per run .
<?php
$row = 1;
//open the file
if (($handle = fopen("file.csv", "r")) !== FALSE)
{
while (($data = fgetcsv($handle, 0, ",")) !== FALSE)
{
$num = count($data);
echo "<p> $num fields in line $row: <br /></p>\n";
$row++;
for ($c=0; $c < $num; $c++)
{
echo $data[$c] . "<br />\n";
}
}
fclose($handle);
}
?>
Edited by bores_escalovsk, 16 May 2014 - 06:38 PM. Hi. I want a simple textbox, that when submited, will replace very every new line, with the <br> tag. What will happen, when it submits, it will take the contents of a textbox, add the <br> tag where a new line is suposed to be, and save the string to a MySQL Database. I think this is the easiest way of allowing a user to edit what appears on a website when logged in, but if there is a easier way, then please tell me. What I am trying to do, is a login page, and once logged in, you enter new text into the textbox, click submit, and on the website homepage, the main text will change to what was submitted. But if there is a new line, I think the only way in HTML to make one is to put a <br> tag, so I want the PHP to but that tag wherever there is a new line. Sorry if I am confusing, I am not that advanced at PHP, but I would be very happy if you could supply me with the correct code. Time is running out... If you do not understand me, please tell me -- PHPLeader (not) Dear All Good Day, i am new to PHP (a beautiful server side scripting language). i want to send a mail with line by line message i tried with different types like by placing the things in table and using <br /> but the thing is the tags are also visible in the message of the mail. Here is my code: $message1 = "Name :". $_REQUEST['name']."<br />"; $message1 .= "Surname :". $_REQUEST['surname']."<br />"; $message1 .= "Cellphone :". $_REQUEST['mobileno']."<br />"; $message1 .= "Telephone :". $_REQUEST['landno']."<br />"; $message1 .= "Fax :". $_REQUEST['fax']."<br />"; $message1 .= "Company :". $_REQUEST['company']."<br />"; $message1 .= "Email :". $_REQUEST['email']."<br />"; $message1 .= "Country :". $_REQUEST['country']."<br />"; $message1 .= "Enquity :". $_REQUEST['enquiry']."<br />"; $message1 .= "Date&Time :". $date."<br />"; For this code if try to print/echo it it is working fine it is displaying line by line, but using this variable ($message1) in the mail these <br /> are also visible. Can any one guide me to resolve(to remove these tags from the message part) this issue. Thanks in Advance. :confused: Hello, like I promised, I'm back with another problem. This one will be harder xD so squeeze your brain
Now I've got a webpage, with a database shown into a table, under that I've got page numbers. So I got my 20 rows for each page.
https://www.dropbox..../stillnotok.jpg
Great/awesome so far
Now I've got a new problem and it's hard for me to explain this too.
In my database I got allot of tables, in my webpage I'm using 2 of them, called 'artikel' and 'images".
Everything from my artikel table is ok. So we don't need to look at that.
But my problem is with the table 'images'.
In that table I got 2 items.
Like you will see in my code, I need the I_ARTCODE and I_FILE
in I_ARTCODE stands: 14 and for the other image 15
in I_FILE stands: 5.jpg and for the other image 6.jpg
With the code I've got now, I display the word 5.jpg and 6.jpg.
But I need to display the picture behind that, the person who gave me this task sayed, I normally can do this with the pad but it's not a real url, I'll give you guys a picture how my database looks like:
https://www.dropbox....uqplxd/path.jpg
Over here the pad is this: 000000-001000
It doesn't seems normal to me and I can't find a sollution for this on the internet.
Can someone plz help me?
Ooh ya, almost forgot, this is my code:
<?php
include('connect-mysql.php');
if (!empty($_GET["page"])) {
$page = $_GET["page"];
} else {
$page=1;
};
$start_from = ($page-1) * 20;
$sqlget = "SELECT *
FROM artikel, images
LIMIT $start_from, 20
";
$sqldata = mysqli_query($dbcon, $sqlget) or die('error getting');
echo "<table>";
echo "<tr><th>A_ARTCODE</th><th>A_NUMMER</th><th>A_OMSCHRN</th><th>A_REFLEV</th><th>A_WINKEL</th><th>I_ARTCODE</th><th>I_FILE</th></tr>";
while($row = mysqli_fetch_array($sqldata)){
echo "<tr><td align='right'>";
echo $row['A_ARTCODE'];
echo "</td><td align='left'>";
echo $row['A_NUMMER'];
echo "</td><td align='left'>";
echo $row['A_OMSCHRN'];
echo "</td><td align='left'>";
echo $row['A_REFLEV'];
echo "</td><td align='right'>";
echo $row['A_WINKEL'];
echo "</td><td align='right'>";
echo $row['I_ARTCODE'];
echo "</td><td align='right'>";
echo $row['I_FILE'];
//echo "<img src='000000-001000".$row['I_ID']."' />";
echo "</td></tr>";
}
echo "</table>";
$sql = "SELECT COUNT(A_ARTCODE) FROM artikel";
$rs_result = mysqli_query($dbcon, $sql) or die ("mysqli query dies");
$row = mysqli_fetch_row($rs_result) or die ("mysqli fetch row dies");
$total_records = $row[0];
$total_pages = ceil($total_records / 20);
for ($i=1; $i<=$total_pages; $i++) {
echo "<a href='index.php?page=".$i."'>".$i."</a> ";
};
?>
if you have the urls to pictures in a database how would you display the actual pictures on a webpage? Here 's the code. Code: [Select] <?php require_once("functions.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <style type="text/css"> td { border-top-style: solid; border-right-style: solid; border-bottom-style: solid; border-left-style: solid; border-top-color: #30C; border-right-color: #30C; border-bottom-color: #30C; border-left-color: #30C; } </style> </head> <body> <?php navBar(); echo "<form action=\"\" method=\"post\" name=\"catalog\">"; DatabaseConnection(); $query = "SELECT * FROM treats"; $result_set = mysql_query($query) or die(mysql_error()); $i = 0; echo "<table>"; while ($row = mysql_fetch_array($result_set)) { echo"<tr><td width=\"400px\">($row['product_pic']). <br />{$row['product_id']}.<br />{$row['product_title']}.<br /> {$row['product_Description']}.<br />{$row['product_price']}</td></tr>"; } echo "</table>"; ?> </form> </body> </html> Hi,
I have a crwal engine and every time when I try to take a picture from a specific website, it also take some bullshit from superfish, creating iframe and div class, that I can`t deleted using jquery after my page is load.
I was looking in my code and I found out that the only thing that trigger that is the <img src="<?php h(getPicture($itemRow['site'], if I remove 'site', I don`t see any other superfish code but also don`t see the picture. What can I do to take only the .jpg picture , without any of the other crap?
<?php for ($i = 0; $i < count($itemRow['pics']); $i++) { ?> Hi, i would like to know what is the best method to store a picture. Currently what happens, is each user has a different avatar, which when uploaded comes in a folder called Avatar, in my website folder and is also stored as a BLOB in mysql. I would like to know is this ok to store all avatars in one folder or some other step should be used and if so could some one guide me to a good tutorial. As I would also like to make an option, where people could upload their pictures, is it safe to store pictures in the same folder for all users or some how create a different folder for each user. Hi, I cant quite figure out how to assign an unique id to the picture I upload to my page, all help would be appreciated. Here's the upload code: //This is the directory where images will be saved $target = "images/"; $target = $target . basename( $_FILES['photo']['name']); //This gets all the other information from the form $name=$_POST['name']; $email=$_POST['email']; $phone=$_POST['phone']; $pic=($_FILES['photo']['name']); // Connects to your Database mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()) ; mysql_select_db("Database_Name") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO `employees` VALUES ('$name', '$email', '$phone', '$pic')") ; //Writes the photo to the server if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } ?> I am working on this pictures to upload and when I submit it gives the message below for success but I get this error message which is on the line of $place file Warning: move_uploaded_file(members/5/image01.jpg) [function.move-uploaded-file]: failed to open stream: $check_pic = "members/$id/image01.jpg"; $default_pic = "members/0/image01.jpg"; $newname = "image01.jpg"; $place_file = move_uploaded_file( $_FILES['fileField']['tmp_name'], "members/$id/".$newname); $success_msg = '<font color="#009900">Your image has been updated, it may take a few minutes for the changes to show... please be patient.</font>'; I need some help working through this. I have an uploader for files, but I'd like it to create a link of the file uploaded and automatically post it to a page. Each user has a profile page and I'd like these pictures to be posted to their profile page. I guess with a session, it would take the logged in username and post the pictures uploaded on the page " member_profile2.php?username=xxxx " Here is my upload file for reference. uploads.php Code: [Select] <?php $Dir = "temporary"; if (isset($_POST['upload'])) { //if file uploaded if (isset($_FILES['new_file'])) { //if successful if (move_uploaded_file($_FILES['new_file']['tmp_name'], $Dir . "/" . $_FILES['new_file']['name']) == TRUE){ chmod($Dir . "/" . $_FILES['new_file']['name'], 0644); echo "<b>File \"" . htmlentities($_FILES['new_file']['name']) ." was successfully uploaded </b><br />\n"; //displays file info echo "Name: ". htmlentities($_FILES['new_file']['name']) . "<br />". "Size: " . $_FILES['new_file']['size'] . " bytes" . "<br />". "Type: " . $_FILES['new_file']['type']; } //if not successful else { echo "<b>Uploading \"" . htmlentities($_FILES['new_file']['name']) . " was unsuccessful</b><br />\n"; } } } ?> <form action="uploads.php" method="POST" enctype="multipart/form-data"> <input type="hidden" name="MAX_FILE_SIZE" value="250000" /><br /> Select file to upload:<br /> <input type="file" name="new_file" /> 250kB limit!<br /> <input type="submit" name="upload" value="Upload File" /><br /> </form> ?> and my member_profile2.php Code: [Select] <?php /*?><?php error_reporting(E_ALL ^ E_NOTICE); ?><?php */?> <?php # Starting the session session_start(); # Requiring SQL connection require_once 'mysql-connect2.php'; # Setting auser as SESSION['user'] $auser = $_SESSION['user']; # SQL protecting variables $username = (isset($_GET['username']))?mysql_real_escape_string($_GET['username']):$username; # Checking through each query if(isset($auser)) { $sql = mysql_query("SELECT * FROM `user` WHERE `username` = '$username'"); if(mysql_num_rows($sql)) { while($row = mysql_fetch_array($sql)) { $page = "<h1>User Info</h1>". "<b>Username: {$row['username']}<br /><br />". "<b>First Name: {$row['firstname']}<br /><br />". "<b>Last Name: {$row['lastname']}<br /><br />". "<b>Email: {$row['email']}<br /><br />". "<form name=\"backlistfrm\" method=\"post\" action=\"members2.php\">". " <input type=\"submit\" value=\"Back to The List\">". "</form><br />"; } } else { $page = "ERROR: No member found for username: <strong>{$_GET['username']}</strong>."; } } else { $page = "ERROR: Not logged in."; } # Printing the final output print $page; ?> I'd like the pictures to be displayed after the user details. Any help with this would be great. I was also thinking, could I just have the uploader save it as a filename with the user's username? For example, "user1-image1.jpg". And then have their profile page pull and display any files with the name user1-image1-10.jpg (10 being max possible uploads for that user). How would I do this? This sounds maybe easier. Hello. I'm working on a website with an editor that allows image uploading. Ideally I want to be able to develop a framework for this and use it in later projects. What I have in mind is this: Database has an images table Entries contain these fields: id, filename, and a short description Images are referred to by their ids in other parts of the application. That part seems simple to me, but now there are two details I need to determine: the file name, and storing the images. Big sites like Facebook - as far as I know - parse uploads and store them all in the same format. I can see the huge security benefit there. What are the best ways of doing that? As for the file name, I know PHP has a function to generate a file with a unique name. Is there any benefit to doing that over using the id? (i.e. 1.jpg, 2.png, 3.jpg, etc) Hi guys, A beginner at php/mysql here... For a project I'm doing, I would like to display search results with an image, small description, then a 'read more' link. After googling this issue, I can't find much information on it, which surprised me. Someone did mention blobs for images... (whatever they are!) I'm wondering if anyone knows of any tutorials, or has any code to do this? Your help is much appreciated. Thanks. I want to display pictures in a table using a PHP loop or something, for example 3 pictures in one row, and then when that row has filled, move to the next row, So I can have a grid of pictures, (similar to that on facebook, when your viewing photo albums)... But I've no idea how to do it! Hi I am trying to create a dynamic gallery in php with specific order of pictures on the page, but I can't find the function or piece of php code to do so.
Conditions: My code:
$files = glob("layout/gallery/*.jpg");
rsort($files, SORT_NATURAL);
for ($i=0; $i < count($files); $i++) {
for( ; $i<5; $i++){
$one = $files[$i];
echo '<img src="'.$one.'">' . '<br><br>';
}
echo "<br>";
for( ; $i<9; $i++){
$two = $files[$i];
echo '<img src="'.$two.'">' . '<br><br>';
}
}
The code works well, but it just displays 9 pictures obviously. I was unable to make it dynamic displaying 5 pictures first, 4 pictures after and stay this way in a loop till displays all pictures from that folder. Hi all, I have a code which works just fine for adding one picture to my database but when I change the form to add mutliples i get an error because my code is set for multiple pictures as an array. Error says this is a string code. Does anyone know the code for array? Code: [Select] $target = "upload/"; $target = $target . basename( $_FILES['photo']['name']); Error message says Warning: basename() expects parameter 1 to be string, array given in /home/content/19/6550319/html/listingsss.php on line 7 Thanks, Philip How would I write PHP code to echo the number of pictures I have in a folder on my hosting account?
Thanks in advance.
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