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PHP - Twitter Api Script Failing To Display
Hello all, I have followed the stop on this page to what I thought was a tee: http://www.inkplant.com/code/pull-tw...-your-site.php but I guess I was wrong because I am getting errors. I have setup the database and table properly and created all files, but when I do step 4 and place the code on my site I am getting this error Code: [Select]
Fatal error: Call to undefined function dbQuery() The api pull went fine and it stored the tweets in the database, but it can't display them. Is there a missing step about connecting to the database somewhere? Thanks for any help Similar TutorialsIf you look at the botton of http://getminecraft.net/ you see my latest tweets. The problem is that it shows http://getminecraft.net/%22http://twitter.com/PC_Gamer%22 instead of the original twitter url, that results to a wrong url and wrong site. How can I fix it? so it shows the original twitter url. Here is the script: <?php $username = "getminecraft"; $prefix = ""; $suffix = ""; $feed = "http://search.twitter.com/search.atom?q=from:" . $username . "&rpp=1"; function parse_feed($feed) { $stepOne = explode("<content type=\"html\">", $feed); $stepTwo = explode("</content>", $stepOne[1]); $tweet = $stepTwo[0]; $tweet = str_replace("<", "<", $tweet); $tweet = str_replace(">", ">", $tweet); return $tweet; } $twitterFeed = file_get_contents($feed); echo stripslashes($prefix) . parse_feed($twitterFeed) . stripslashes($suffix); ?> Hi all, Since twitter changed the way they allow people to post, my simple php script to grab the $_POST data from an input and post it to my twitter profile has broken. I've been reading up on Oauth, but none of the documentation is clear at all. It's quite complex, in fact - they assume you are creating a full app, which I am not - I just want to flick some text to twitter. I just want to pass a simple string variable to my account - can anyone help me with how to do this? All help appreciated. WoolyG Hey folks, looking for a twitter follow/unfollow php script that WORKS. any hints/links? thanks Im trying to make a php program that will grab all the user's tweets they have ever done.. I'm using curl to request the twitter page and scrape the html. It looks as though the maximum amount of tweets I can get is 20, which is the amount on the initial first page. There is some script that loads another 20 tweets once I scroll to the bottom of the page... Does anyone know how to make the page automatically load a certain amount so I can scrape more than the first 20 at once. Well the subject line is pretty explicit. I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database. Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work. Code: [Select] //authenticate user //Start session session_start(); //Connect to database require ('config.php'); $sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($sql); $imagebytes = $row['image']; header("Content-type: image/jpeg"); print $imagebytes; Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which. Can anyone help?? Hello all, I am new to php and was wondering if i could get some guidance here. I am using phpAdmin 2.6.0 and running Mysql 4.1.21. here is my situation.... I have a script that allows us to upload a new product name, product code, category and a PDF file to the data base. There is also a folder on the server that has the PDF files in it. I think I deleted the code on the page (library.php) that displays the files for the client to download. My goal here is after I upload everything I want it to then be displayed on the page with a link to the PDF file. Here is the page that has the links on it. I hope that I explained this correctly. I am not a programmer but do have some idea and have been reading up on php to try and figure this out. I am looking to create the script that would display the links on the library.php page. Any help would be great. The other link is to the script that allows us to upload. www.pennstateind.com/library.php www.pennstateind.com/lib-admin.php Hello I have an array that I would like to display as a family tree... and am looking for an example script that will allow me to do this. I want to be able to create X number of generations and fill the boxes with data from my array.... So, if I want 5 generations I want to generate a tree with 63 elements or boxes Any help would be appreciated Thanks I came up with the following script for displaying an uploaded picture on a webpage but for some reason I can't pinpoint, the picture won't be displayed. I checked my database from php myadmin and sure enough, the picture was successfully uploaded but somehow, the picture won't be displayed. So here is the display script. I named it display_pic.php Code: [Select] <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require('config.php'); $image = stripslashes($_REQUEST[imname]); $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($rs); $imagebytes = $row[image]; header("Content-type: image/jpeg"); print $imagebytes; ?> The tag on the html page that's supposed to display the picture reads something like this <img src="display_pic.php" width="140" height="140"> And just in case this might help, I will include the image upload script below, which I think worked just fine because my values were successfully inserted into the database. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Make sure the user actually // selected and uploaded a file if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { // Temporary file name stored on the server $tmpName = $_FILES['image']['tmp_name']; // Read the file $fp = fopen($tmpName, 'r'); $data = fread($fp, filesize($tmpName)); $data = addslashes($data); fclose($fp); // Create the query and insert // into our database. $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')"; $results = mysql_query($query); // Print results print "Thank you, your file has been uploaded."; } else { print "No image selected/uploaded"; } // Close our MySQL Link mysql_close(); } //End of if statmemnt. ?> So any insights as to why the script fails to display the image? Any help is appreciated. I had posted a similar post a few days back, about an display script which is supposed to retrieve a stored image from my database and displays it on an html page, but fails to do so. Someone suggested that I upload my image onto a folder on the server and then save the image path name in my database. I found this script(below) which does just that. It uploads the image into a folder on the server called images and stores the image path into a table in my database called images and I know this script works because i saw the file saved in the images folder and the path name inserted into the images table. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> Now the display image script accompanying this insert image script is shown below. Code: [Select] <?php // Get our database connector require("config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Dream in code tutorial - List of Images</title> </head> <body> <div> <?php // Grab the data from our people table $sql = "SELECT * from images WHERE image_cartegory = 'main'"; $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<div class=\"picture\">"; echo "<p>"; // Note that we are building our src string using the filename from the database echo "<img src=\"images/ " . $row['filename'] . " \" alt=\"\" />"; echo "</p>"; echo "</div>"; } ?> </div> </body> </html> Well, needless mentioning, the picture isn't displayed. Just a tiny jpeg icon is displayed at the top left hand corner of the page. Figuring out that the problem might be the lack of a header that declares the image type, I add this line header("Content-type: image/jpeg"); but all it does is display a blank white page, without the tiny icon this time. What am I missing? Any clues?? This is my script I want to display but it does not show in the page just blank. <td><object width="220" height="148"><param name="movie" value="http://www.youtube.com/user/THEWORLDOFTRAVEL#p/search/2/uX9nd3xM_MY=1&h1=enUS&color1=0x234900&color3=34900&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/user/THEWORLDOFTRAVEL#p/search/2/uX9nd3xM_MY=1&h1=enUS&color1=0x234900&color3=34900&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="220" height="148"></embed></object></td> Hi, I am planning to display remote images with the following code but it seems to return errors. <?php $image_dir = 'http://www.domain.com/images/' ; $dir_handle = opendir( $image_dir ); $count = 0 ; $display = '' ; while( $filename = readdir($dir_handle)){ if( preg_match( '/$[a-z0-9]{4}_th\.jpg$/' , $filename ) ){ $display .= " <img src='$image_dir$filename' /> " ; $count++ ; if( $count % 10 == 0 ){ $count=0; $display .= "<br />"; } } } closedir( $dir_handle ); echo $display ; ?> I am thinking might be the $image_dir cannot use address format.... Errors is as follow Quote Warning: opendir(http://www.foo.com/images/) [function.opendir]: failed to open dir: not implemented in /home/jch02140/public_html/test.php on line 3 Warning: readdir(): supplied argument is not a valid Directory resource in /home/jch02140/public_html/test.php on line 6 Warning: closedir(): supplied argument is not a valid Directory resource in /home/jch02140/public_html/test.php on line 16 Hello All,
I am working on a project where the client has provided me with the public key file and the private-key is being passed via url, along with 2 params that will be used on my end.
These are the basics of the process that I am to use for verifying. Generate your own plaintext message matching the format of the string provided Create a SHA1withRSA hash of this message using the provided public key (UTF-16LE encode and pass this value) Base64 decode the signature Using a SHA1withRSA validator, verify your hashed message matches the Base64 decoded value in step 3 At this point I have performed steps 1-3 but am having an issue with step 4. The code is failing here. When I say failing I mean it is not being verified. $base64Sig = base64_decode($signature, true); $publickey = getPemKey(); Why is this line of code not sending an e-mail to my gmail account? Code: [Select] mail($email, ' Log-In Issue', $body, 'From: admin@MySite.com <admin@MySite.com>'); I stepped through my code, and $email = 'debbies_email_account@gmail.com' I'm confused?! Debbie I've got an autodownloader for a file that I want my users to download when they click a link and the file is right but when a user downloads the file, the file is corrupted. Code: [Select] <?php header('Content-Type: application/x-rar'); header('Content-Length: ' . strlen($data)); header('Content-Disposition: attachment; filename=Addypk.rar'); header('Content-type: application/octet-stream'); ?> It's not downloading the actual content of the file, rather, it's downloading the shell. Any ideas? Please help with this, it not going to the .html controller and returing the values need, it captures the email form values, bu then fails with a 406 error.
Please help
$('#newsletter').submit(function() {
//E-mail is passed
var values = $(this).serialize();
var formRef = $('form').attr('id');
var dataString = JSON.stringify(values);
//var values = $("#newsletter_email").val();
if($('#newsletter_tc_check').is(':checked')){
$.ajax({
type: 'POST',
url: "./saveNewsletterSignupEmailPost.html",
dataType: 'json',
data: dataString,
contentType: 'application/json',
mimeType: 'application/json',
success: function(data) {
var result = $.parseJSON(data);
if(result.form == formRef){
$('.success.message').contents().find('h4').text(success.text());
VanillaReload.notify.showNotification("success");
$("#newsletter_email").val("");
}else{
$('.error.message').contents().find('h4').text(error.text());
VanillaReload.notify.showNotification(".error");
}
}
});
return false;
}else{
$('.error.message').contents().find('h4').text(error.text());
vanillaeGift.notify.showNotification(".error");
}
return false;
});
I'm now having a problem with PHP uploads on the test site I'm working on. There is a special page called "cart_import.php" that directs CSV files selected for uploading to the "files" directory in the same location as the the page mentioned, and from there works on updating the website's front end with the data in the CSV file. But, when I try and upload the file, the page reports that it's failing to receive the upload entirely. I've confirmed with the host that the php.ini file does allow uploads, the "files" directory is set to write access and upload size allowed is a little over 2MB. Is there perhaps another thing I'm missing to allow uploads, something that I haven't found yet with Google? I am building a whitelist based bbcode sanitiser that allows only permitted bbcode element (the
tag. All other bbcode elements will be rejected regardless of what they are. Also, this system should reject empty bbcode tag This is the function below:
$val = 'This is a post with code, [code]cmmcxm[/code]';
/* The negative lookahead regex contruct in the function below ensures all opening
square brackets must
be followed by this regex inside the lookahead(code\](.)+\[\/code\])
*/
function isSuspect($val)
{
// create a pattern to whitelist allowed bbcode phrase
$pattern = '/[(?!(code\](.)+\[\/code\]))/i';
// if one of the suspect phrases is found, reject \[(?!(code\]\[/code\]))
if (preg_match($pattern, $val))
{
return '<div id="login-alert" class="alert alert-danger col-sm-12">You can not post that, it appears you may have included banned words or code tags in your post.<br/>If you think this is not the case, kindly contact the portal administrator <a href="#"><img class="warning" src="assets/img/icons/mail-black.png" alt="" /> here</a></div>';
}
}
echo isSuspect($val);
This function is not validating $val = 'This is a post with code,
cmmcxm'; and I was thinking it should. I would appreciate inputs to finish this. Thanks. i have a simple class which sees if there are any members in the database with the supplied details, then returns a simple number to show how many results there are, but it isnt working and throwing the error: Code: [Select] Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\classTest\classes\user_process.php on line 52 here is the class: require_once("db_mysql.php"); require_once("./init.php"); class emptyArgs extends Exception { function __toString() { return "<strong>Empty Arguments</strong>"; } } class user_process { public $user_name; public $password; public function login($user_name, $password) { try { if (empty($user_name) || empty($password)) { throw new emptyArgs(); } if ($user_name == "") { throw new emptyArgs(); } } catch (emptyArgs $e) { echo $e; echo "<p>Please Supply All Details Marked *</p>"; } echo $user_name; echo $password; $query = <<<QUERY SELECT COUNT(*) FROM ".TBL_PREFIX."members WHERE user_username = '$user_name' AND user_password = '$password' QUERY; $process = mysql_query($query); $result = mysql_num_rows($process); // Line 52 where the error originates return $result; } } and here is the call: $user = new user_process(); $user->login("username", "password"); All of the database details are correct and the values do match in the database. Where am i going wrong? Note that this class is just for testing and not actually used. Good evening! Another bizzare thing that stopped working all of a sudden... I have a registration form and this code no longer works... // Check for First Name. // Allow letters, space, period, apostrophe, and hyphen. if (preg_match('/^[A-Z\'.-]{2-20}$/i', $_POST['first_name'])){ $fn = mysqli_real_escape_string($dbc, $_POST['first_name']); } else { // Add error-message to array. $reg_errors['first_name'] = 'Please enter your first name!'; } All day when I typed in a two letter answer - to speed up typing and testing - things worked fine, but now I get a built-in error message off to the side of the input box. I typed 'ee' and that always worked before but no luck now. The Reg Ex hasn't changed. Could it be that there is a database connectivity issue that is screwing up my RegEx?? If I comment out all of the RegEx stuff then it works and my INSERT works - thanks to AlexWD TomTees |
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