PHP - Retrieving Images
i have this script called view.php
Code: [Select] <?php include 'db.inc'; $file = clean($file, 4); if (empty($file)) exit; if (!($connection = @ mysql_pconnect($hostName, $username, $password))) showerror(); if (!mysql_select_db("Php_test", $connection)) showerror(); $query = "SELECT mime, data FROM file WHERE id = $file"; if (!($result = @ mysql_query ($query,$connection))) showerror(); $data = @ mysql_fetch_array($result); if (!empty($data["data"])) { // Output the MIME header header("Content-Type: {$data["mime"]}"); // Output the image echo $data["data"]; } ?>after i have db.inc Code: [Select] <?php // These are the DBMS credentials $hostName = "localhost"; $username = "root"; $password = "oxioxi"; // Show an error and stop the script function showerror() { if (mysql_error()) die("Error " . mysql_errno() . " : " . mysql_error()); else die("Could not connect to the DBMS"); } // Secure the user data by escaping characters // and shortening the input string function clean($input, $maxlength) { $input = substr($input, 0, $maxlength); $input = EscapeShellCmd($input); return ($input); } ?>and i have my page witch part of it is this Code: [Select] <?php include 'db.inc'; $query = "SELECT id, name, mime FROM file"; if (!($connection = @ mysql_pconnect($hostName, $username, $password))) showerror(); if (!mysql_select_db("php_test", $connection)) showerror(); if (!($result = @ mysql_query ($query, $connection))) showerror(); ?> <h1>Image database</h1> <?php if ($row = @ mysql_fetch_array($result)) { ?> <table> <col span="1" align="right"> <tr> <th>Short description</th> <th>File type</th> <th>Image</th> </tr> <?php do { ?> <tr> <td><?php echo "{$row["name"]}";?></td> <td><?php echo "{$row["mime"]}";?></td> <td><?php echo "<img src=\"view.php?file={$row["id"]}\">";?></td> </tr> <?php } while ($row = @ mysql_fetch_array($result)); ?> </table> <?php } // if mysql_fetch_array() else echo "<h3>There are no images to display</h3>\n"; ?>i have do it right but for some reason it doesnt displays the pictures the type and the name are displayed normally but not the picture!! instead of them there are small icons of not existing pictures.. please i want help i really need to do it Similar TutorialsI've searched all over and have found some scripts but none of them work anymore. Is there a way to fetch Google images using PHP? There's no Google Image API (only the regular Google API) Hi Everyone! Just learning php. Hope someone can help with this. Trying to design a product page of photo images with price, description, etc. At this point I am ignoring the design and just trying to get the fields to populate. They all work through my "while" loop except I cannot find the syntax for the image and am just getting the placeholder. My images are just "image".jpg in my root folder. Ex. building.jpg. And my table column is "picture_image". Appreciate any help. Here is the pertinent code which works fine except for the image $ROW. I have bolded the trouble block. Thanks! while($row = mysqli_fetch_assoc($result)) { // display row for each picture echo "<tr>\n"; echo " <td>{$row['Picture_ID']}</td>\n"; echo " <td style='font-weight: bold; font-size: 1.1em'>{$row['Picture_name']}</td>\n"; echo " <td>{$row['Picture_description']}</td>\n"; echo " <td>{$row['Picture_price']}</td>\n"; echo "<td><a href={$row['picture_image']}' border='0'> <img src='{$row['picture_image']}' border='0' width='100' height='80' /> </a></td>\n"; Hello Guys, I need help on this problem. output: display images for the values that exist in the table: input: the values in a table's column where they're in the format of "0,0" and the image corresponding to it is named 0,0.png this is my attempt $q is retrieving "0,0" however in order for me to retrieve its image i need to add ".png" to it in order to find it in the directory that the image is located. It tried without ".png" and it also doesn't work. $q = 'SELECT ID FROM table'; $dirname = "directory/"; $images = glob($dirname.$q.".png"); foreach($images as $image) { echo '<img src="'.$image.'" /><br />';
Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this? thanks There's many options available for downloading a URL - but I'm stuck. I've looked through all the ones I know, but none seem to pay attention to partial content. I'm trying to retrieve a URL that gives the following header: HTTP/1.1 206 Partial Content Content-Range: bytes 0-100000/631723 As you can see it dishes out the file in 100,000 byte chunks. Trouble is, when I use any method in PHP, ie file_get_contents, fopen, or even cURL, none of these continue on after retrieving the 100,000 bytes. End result, I have a 100,000 byte file. What I need is to get the PHP script to grab all the data, in the example above, all 631,723 bytes. How can I do this? Hi, so i have a problem. I'm using a script to enable and disable checkboxes, and i'm id-ing them fetching values from a database: while($data = mysqli_fetch_array($sql)) { if ($count % 5 == 0) { echo '</tr><tr>'; $count = 0; } echo '<td><input name="subject_level[]" class="subject_a" disabled="disabled" type="checkbox" id="subject_level_'.$data['subject_level_id'].'" value="'.$data['subject_level_id'].'"/>'; echo '<label for="subject_level_'.$data['subject_level_id'].'" class="subject_1">'.$data['subject_name'].'</label></td>'; $count++; //Increment the count } The names of the checkboxes come out fine, just that, is there any way to retrieve the values of the checkboxes into an array of some sort so that i could build a search query later? if($_POST(subject_level)) or if($_POST(subject_level[])) doesn't return true. does it make sense? im new to php so please forgive my ignorance. Help appreciated. Thanks! i have had a difficult time trying to work this out.I need to do some pattern matching for certain urls and retrieve information from it. For example, $url=http://www.test.com/showpic.php?do=showpic&u=89165&a=34933 if $url contains the value showpic.php,then i need to retrieve the following { $u=value of u(i.e. 89165 in this case) $a=value of a(i.e. 34933 in this case) } else do nothin.. the format of the url will always be the same as above if it contains showpic.php i am trying to retrieve messages with some code. It seems to be a bit more complicated that i thought it would be. so here it goes: Code: [Select] $query = "SELECT * FROM memberMail WHERE userIDFrom='92' AND unread='1'"; $result = mysql_query($query) or die(mysql_error()); if (mysql_num_rows($result) > 0) { $row = mysql_fetch_array($result) or die(mysql_error()); $num_rows = mysql_num_rows($result); echo "You have (" . $num_rows . ") unread message(s)."; $i = 0; for($i; $i<$num_rows; ++$i) { if($row[$i[unread]] == "1") { echo $row[$i[message]] . "<p>"; } } } what i am trying to do, is print out the messages from the records that are unread, and are from a certain user. i don't think this for statement will pick up the record numbers that got matches though. and i am not entirely sure about the $row statements in the for statement either. ok im back quicker than i thought.... i got my drop box sorted and i got it reloading the page. so it all works correctly. but how do i get the page to display information regarding the film i have selected in the drop box.? i have no code for this at the mo. also i would like the drop box to display the selected item at top of box when it refreshes code for drop box: Code: [Select] <FORM> <?php $result = mysql_query( "SELECT * FROM movie_info ORDER BY title ASC ") ; echo "<select name= Film onChange='submit()' >film name</option>"; while ($nt=mysql_fetch_array($result)){ ?> <?php echo "<option value='$nt[id]'>$nt[title] </option>"; } ?> </select> </FORM> any help would be great Hi, I need to get data from an XML file, like the format below: Code: [Select] <ISBNdb server_time="2011-05-31T19:19:42Z"> <BookList total_results="1" page_size="10" page_number="1" shown_results="1"> <BookData book_id="build_your_own_database_driven_web_site_using_php_mysql_4th_" isbn="0980576814" isbn13="9780980576818"> <Title> Build Your Own Database Driven Web Site Using PHP & MySQL, 4th Edition </Title> <TitleLong/> <AuthorsText>Kevin Yank, </AuthorsText> <PublisherText publisher_id="sitepoint">SitePoint</PublisherText> <Details change_time="2009-04-02T00:30:10Z" price_time="2011-05-31T17:46:10Z" edition_info="Paperback; 2009-06-15" language="" physical_description_text="360 pages" lcc_number="" dewey_decimal_normalized="5" dewey_decimal="005"/> </BookData> </BookList> </ISBNdb> I need to know how you get the details from <title>,<AuthorsText><PublisherText> and Bookdata. Tried using the example on W3C, using simplexml_load_file but it only returns: Code: [Select] ISBNdb BookList: Code I'm currently using is: Code: [Select] $xml = simplexml_load_file("http://isbndb.com/api/books.xml?access_key=----&results=details&index1=isbn&value1=".$isbn); echo $xml->getName() . "<br />"; foreach($xml->children() as $child) { echo $child->getName() . ": " . $child . "<br />"; } In my script, users login with their Username & Password. However, I'd like to be able to echo the email address used on their account.
I've tried adding the email to the session I'm not having much luck... Here's a piece of the login code(untouched); $username = $_POST['name']; $passwd = $_POST['passwd']; $query = "SELECT name,passwd FROM users WHERE CONCAT('0x', hex(passwd)) = '{$salt}'"; $result = mysql_query($query); $login_ok = false; if(mysql_num_rows($result) > 0) { $login_ok = true; } if($login_ok) { $row = mysql_fetch_array($result, MYSQL_NUM); $_SESSION['user'] = $row;I've also tried messing around with this piece below in a few different ways but still nothing. <?php echo htmlentities($_SESSION['user']['email'], ENT_QUOTES, 'UTF-8'); ?>Any help is greatly appreciated.. I'm making a game I have finished but have spent days and still can't figure out this part or how it should be done: Gamer will type in keyword on the form box on my page then click submit the page will connect to yahoo search then save/put the address bar URL search result of the keywords into a variable onto my page but the gamer will never have his page directed to yahoo but continue staying on my page while this all happens. Any Help in the right direction or some code would be appreciated? Hi there I am trying to use cURL to retrieve XML from a gateway. The code I am using works (ie. returns the HTTP response) for any URL that I use except the one that I need! The URL is below, and when using that I get a blank response every time. I've tested and can telnet to that IP address on that port from my server. I'm not sure if it is because of the format/structure of the URL? Could it have something to do with the port of the gateway? Any assistance would be much appreciated. http://196.11.120.190:8080/mtnusa/client.jsp?command=<usareq NODE="tHTTP" USERNAME="HTTP" PASSWORD="1234" TRANSFORM="SMPP"><command><submit_sm><a_number>278200703520709</a_number><b_number>27824411926</b_number><service_type/><message>Test</message><registered_delivery/></submit_sm></command></usareq> The function I am using (from http://davidwalsh.name) is: /* gets the data from a URL */ function get_data($url) { $ch = curl_init(); $timeout = 5; curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout); $data = curl_exec($ch); curl_close($ch); return $data; } $returned_content = get_data('http://196.11.120.190:8080/mtnusa/client.jsp?command=<usareq NODE="HTTP" USERNAME="HTTP" PASSWORD="1234" TRANSFORM="SMPP"><command><submit_sm><a_number>278200703520709</a_number><b_number>27824411926</b_number><service_type/><message>Test</message><registered_delivery/></submit_sm></command></usareq>'); $qID = ''; $question = 'Question not set'; $answerA = 'unchecked'; $answerB = 'unchecked'; $answerC = 'unchecked'; $answerD = 'unchecked'; $answerE = 'unchecked'; ?> <html> <head> <script type="text/javascript"> function show_alert() { alert("Please Click OK to proceed!"); } </script> </head> <form action="Process1.php" method="POST"> <table> <tr> <td> <?php $SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.answer1, answer_type.answer2, answer_type.answer3, answer_type.answer4, answer_type.answer5 FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id "; $result = mysql_query($SQL); while($db_field = mysql_fetch_assoc($result)){ $qID = $db_field['question_id']; $question = $db_field['question']; $A = $db_field['answer1']; $B = $db_field['answer2']; $C = $db_field['answer3']; $D = $db_field['answer4']; $E = $db_field['answer5']; print $question; ?> </td> </tr> <tr> <td> <INPUT TYPE = 'Radio' Name ='q' value= 'A' <?PHP echo $answerA; ?>><?PHP echo $A; ?> <INPUT TYPE = "Hidden" Name = "h" VALUE = <?PHP print $qID; ?>> <?php echo $qID;?> </td> </tr> <tr> <td> <INPUT TYPE = 'Radio' Name ='q' value= 'B' <?PHP echo $answerB; ?>><?PHP echo $B; ?> <?php echo $qID;?> </tr> </td> <tr> <td> <INPUT TYPE = 'Radio' Name ='q' value= 'C' <?PHP echo $answerC; ?>><?PHP echo $C; ?> </tr> </td> <tr> <td> <INPUT TYPE = 'Radio' Name ='q' value= 'D' <?PHP echo $answerD; ?>><?PHP echo $D; ?> </tr> </td> <tr> <td> <INPUT TYPE = 'Radio' Name ='q' value= 'E' <?PHP echo $answerE; ?>><?PHP echo $E; ?> <br> <?php } mysql_close(); ?> </tr> </td> <tr> <td> Please add any comments you may have:</br> <textarea rows="3" cols="60" name="comments" id="comments"> </textarea> </td> </tr> </table> <input type="submit" onclick="show_alert()" value="Submit" /> </form> </html> I ran to some problems here. After filling every question up with the answers. When i process the form, i only managed to get QID of 20 and the answer of 1st question as my database have only 20questions. How do i go about in solving, so that i'm able to retrieve all result and update my database correctly $selected_radio = $_POST['q']; $qID = $_POST['h']; echo $selected_radio; echo $qID; $SQL = "UPDATE answers SET $selected_radio = $selected_radio + 1 WHERE question_id='$qID'"; $result = mysql_query($SQL); mysql_close(); print "Thanks for voting!"; ?> THis is my process page, and i echo out the answer to be B and qid as 20. I have a basic table where I am trying to retrieve records that are filtered by a form request. Here is the code I have... $ps = $pdo->prepare("SELECT * FROM `Products` Where `Vendor` LIKE concat('%',?,'%'); $ps->execute(array($_POST['Vendor']));; echo "post=" . $_POST['vendor']; ////Displays correct data from form request $count = $ps->rowCount(); echo "Count=" . $count; ////Count = 0 although I know for a fact that there is 1 record that should be in there ////Used for display of records foreach ($ps as $row){ echo $row, PHP_EOL . "xxx<br>"; } Where is the incorrect code? I am new to php and pdo. Thank you in advance K Code: [Select] $query = mysql_query("SELECT a.*, b.* FROM friendlist a INNER JOIN friendlist b ON (a.friendemail=b.friendemail) INNER JOIN users c ON (b.friendemail = c.EmailAddress) WHERE a.email = 'asdf@gmail.com' AND c.Username LIKE '%carol%' GROUP BY a.id ORDER BY count(*) DESC"); Code: [Select] while ($showfriends = mysql_fetch_array($query)) { echo $showfriends['Username']; } and I would get nothing. It produces the correct number of <div> so i know it's getting through, but it's having trouble displaying the entries? Hello, I have a few questions. First, this is what I am trying to accomplish - I am trying to take values that are entered into forms. And then store them into a database. My plan was to take them into an array, and create a loop that wrote the values to the database. So, the first step I thought would be to learn how to take info from a form, then display it onto the screen. Theoretically, if I could do that, I could just learn the MySQL commands to write to the database, and paste them instead of printing to the screen. So, I gave it a go. It didn't work. The way the 'Key' is chosen with arrays in PHP was different than I thought. I gave my sample code to a friend and asked for his help. He gave me a working copy back, but signed off before I could ask questions This is what I am working with: Code: [Select] <html> <body> <form name = "barInfo" method = "post"> Establishment name: <input type = "text" name = "EstablishmentName"> <br> Street Address: <input type = "text" name = "StreetAddress"> <br> City: <input type = "text" name = "City"> <br> State: <input type = "text" name = "State"> <br> Zip: <input type = "text" name = "Zip"> <br> <input type="submit" name="Submit" value="Submit"> </form> <?php if (!empty($_POST['EstablishmentName'])) { <br> print "Now, lets see if this shit works..."; <br> print "Establishment Name:" . $_POST['EstablishmentName']; <br> print "Street Address:" . $_POST['StreetAddress']; <br> print "City:" . $_POST['City']; <br> print "State:" . $_POST['State']; <br> print "Zip:" . $_POST['Zip']; <br> } ?> </body> </html> Ok, first off I get a parse error or something like that when I try it. My book has '<br>' thrown all over the place inside '<?php>' things. How can I do an endl; type thing in PHP? Obv <br> does not work., Code: [Select] print "\n";Does nothing, either....? Second, the reason my code didn't display anything, but his did, was because I had a loop that displayed $details[$x], and 1 was added to $x every time it looped. Am I right in saying that during the html part of the code, whatever value is assigned to 'name = ', is the key pointing to that value in my array? If that is true, how would I make that become an integer and use it with a loop. Do I have to put: Code: [Select] name = "1", name = "2", etc...? Thanks in advance for the help! Sorry if I am not referring to terms correctly, I just started looking at HTML/PHP last week... I would like to display weather conditions on my website and store them in mysql database. I wonder if it is possible to load web page into php for parsing so the required info could be found and used? Is it doable with php and what functions could you recommend for this functionality? How can I have program running in the background on the server which would be triggered every so often to perform this task? Another question is if I want to trigger some action once a week like Tuesday at noon, is there function in php which could be used to check what day of the week and time it is ? I am new to website design so any help will be greatly appreciated. Hello all, I'm currently working on a project and it is the first time I have had to store/retrieve an array with PHP and MySQL. Basically its a website that has a list of discounts/coupons on it. People can register and submit their discounts and such. I am storing an array for who is eligible for the discount, as well as which categories the discount falls under. I did some googling and there seems to be tons of thoughts on the best way to do this. Some say to use implode/explode, some say to serialize/unserialize, I was just wondering if there is a common, concrete way to do this? I will need to be able to search the array for its contents. So if children are eligible for the discount, I need to be able to store that in a database, retrieve it later, and search it for only "Children" so I can display the discounts available only to children. If someone could please advise the best way to do this and post some examples, it would be greatly appreciated. I learn best with examples that are explained properly. Hi All, small question here. I was just playing around with a small script I made to obtain an ip-address, remove the dots and than glue the parts together. Not for any use but for practice. But I am having difficulties to accessing the array values. But maybe that because its the first time i am using explode() and print_r. If someone has a spare minute I would be pleased to be enlightened. <?php $ip=$_SERVER['REMOTE_ADDR']; echo $ip.'<br />'; $cleaned_ip = substr($ip,0,6); echo $cleaned_ip.'<br />'; $cleaned_ip2 = print_r(explode('.', $ip, 6)).'<br />'; // this is the point where I dont know how to get the array values. // echo $cleaned_ip2[0].[1].[2]; that doesnt work. ;( ?> |