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  • php array problem ..need expert

    - by user295189
    I have tried this in another post but trying my luck again. My current array that I am making produce a different result than I am wanting. I want to have this kind of out put Row: 0: Column: 1: ID 1 Row: 1: Column: 1: ID 1 Row: 0: Column: 2: ID 2 Row: 1: Column: 2: ID 2 Row: 2: Column: 2: ID 2 Row: 3: Column: 2: ID 2 Row: 0: Column: 3: ID 3 Row: 1: Column: 3: ID 3 As you can see the Rows and columns change based on the ID. So if the ID is same it just go to next row in the same column. However if id is changed it goes to next column and rows start. Currently I my code looks like this for($i=0;$i<count($pv->rawData); $i++) { $relative=0; $relativeTypeID = -1; if ($pv->rawData[$i]->relativeTypeID != $relativeTypeID) { $relativeTypeID = $pv->rawData[$i]->relativeTypeID; $iTypeCount++; } if(!empty($pv->rawData[$i]->description)) { $pv->results[$i][$iTypeCount][0] = $pv->rawData[$i]->description; echo "Row: ".$i.": Column: ".$iTypeCount.": ID".$relativeTypeID." <br>"; } } It gives me the following output Row: 0: Column: 1: ID1 Row: 1: Column: 2: ID1 Row: 2: Column: 3: ID2 Row: 3: Column: 4: ID2 Row: 4: Column: 5: ID2 Row: 5: Column: 6: ID2 Row: 6: Column: 7: ID2 Row: 7: Column: 8: ID2 Row: 8: Column: 9: ID2 Row: 9: Column: 10: ID2 Row: 10: Column: 11: ID2 Row: 11: Column: 12: ID2 Row: 12: Column: 13: ID2 …. …. As you can see the Row and Columns are changing but not with ID number. I appreciate your help Thanks

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  • PHP transfer files from server to server in LAN

    - by cheapez
    So, I have 5-6 pages of requirements. I'm trying to build this application in PHP based on the requirements. I want to transfer files from one server to the other server in LAN, and then send a shell command to the other server to find out if the file has been transferred successfully. In php, I can transfer files using FTP, and send shell commands using SSH. Using the methods above, I will need to open connection to the server first, but I don't know the ftp server name, domain name, ip address, or anything like that. I only know the the server ID (I'm not sure what this ID is, but I guess it is like the computer's name). An example of the server ID is: "c23bap234" How do I open a connection with just that server ID? These servers are in the same building, have LAN connection, don't have connection to the outside world. These machines have PHP, Apache, ... installed. If my post doesn't make sense to you, it's because I'm a learner. I hope someone can help me on this. Thanks in advance.

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  • Assign a static function to a variable in PHP

    - by Felipe Almeida
    I would like to assign a static function to a variable so that I can send it around as a parameter. For example: class Foo{ private static function privateStaticFunction($arg1,$arg2){ //compute stuff on the args } public static function publicStaticFunction($foo,$bar){ //works $var = function(){ //do stuff }; //also works $var = function($someArg,$someArg2){ //do stuff }; //Fatal error: Undefined class constant 'privateStaticFunction' $var = self::privateStaticMethod; //same error $var = Foo::privateStaticFunction; //compiles, but errors when I try to run $var() somewhere else, as expected //Fatal error: Call to private method Foo::privateStaticMethod() from context '' $var = function(){ return Foo::privateStaticMethod(); }; } } I've tried a few more variations but none of them worked. I don't even expect this sort of functional hacking to work with PHP but hey, who knows? Is it possible to do that in PHP or will I need to come up with some hack using eval? P.S.: LawnGnome on ##php mentioned something about it being possible to do what I want using array('Foo','privateStaticMethod') but I didn't understand what he meant and I didn't press him further as he looked busy.

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  • PHP: Need a double check on an error in this small code

    - by Josh K
    I have this simple Select box that is suppose to store the selected value in a hidden input, which can then be used for POST (I am doing it this way to use data from disabled drop down menus) <body> <?php $Z = $_POST[hdn]; ?> <form id="form1" name="form1" method="post" action="test.php"> <select name="whatever" id="whatever" onchange="document.getElementById('hdn').value = this.value"> <option value="1">1Value</option> <option value="2">2Value</option> <option value="3">3Value</option> <option value="4">4Value</option> </select> <input type="hidden" name ='hdn' id="hdn" /> <input type="submit" id='submit' /> <?php echo "<p>".$Z."</p>"; ?> </form> </body> The echo call works for the last 3 options (2,3,4) but if I select the first one it doesnt output anything, and even if i change first one it still doesnt output anything. Can someone explain to me whats going on, I think it might be a syntax issue.

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  • Finding occurrences of element before and after the element in Array

    - by user3718040
    I am writing a algorithm, if an array contain 3 does not contain between two 1s. like this int array={5, 2, 10, 3, 15, 1, 2, 2} the above array contain 3, before 3 there is no 1 and after 3 is one 1 it should return True. int array={3,2,18,1,0,3,-11,1,3} in this array after first element of 3 there is two 1 it should return False. I have try following code public class Third { public static void main(String[] args){ int[] array = {1,2,4,3, 1}; for(int i=0;i<array.length;i++) { if(array[i]==3) { for(int j=0;j<array[i];j++) { if(array[j]==1) { System.out.println("I foud One before "+array[j]); }else { break; } System.out.println("yes i found the array:"+array[i]); } for(int z=0;z>array[i];z++) { if(array[z]==1) { System.out.println("I found after 3 is :"+array[z]); } break; } } } } } I am not getting exact result from my above code which i want.

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  • PHP Saving information to MySQL but not using hard coded code

    - by Homer_J
    Hi guys, Thanks everyone for assisting me but know I'm well and truely stuck...hopefully someone can advise? My current code that saves to a MySQL database: $q1 = $_POST["q1"]; $q2 = $_POST["q2"]; $q3 = $_POST["q3"]; $q4 = $_POST["q4"]; $q5 = $_POST["q5"]; $q6 = $_POST["q6"]; $q7 = $_POST["q7"]; $q8 = $_POST["q8"]; $proc = mysqli_prepare($link, "INSERT INTO tresults_bh_main (respondent_id, ip, browser, q1, q2, q3, q4, q5, q6, q7, q8) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?);"); mysqli_stmt_bind_param($proc, "issiiiiiiii", $respondent_id, $ip, $browser, $q1, $q2, $q3, $q4, $q5, $q6, $q7, $q8); At the moment this is all hard coded and I need it to use variables: I have an array which stores the following information: $qs['questions'] - stores, for example q1, q2, q3, q4 etc - obviously these change both in terms of of numbers, ie. q10, q11, q12 and also the amount in the array - so there could be 4 q's or 10 q's stored in the array. What I struggling to get my head around is how I would set-up this hard coded page to work with my variables. If anyone can help, much appreciated. Homer.

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  • PHP download page: file_exists() returns true, but browser returns 'page not found'

    - by Chris
    I'm using PHP for a file download page. Here's a snippet of the problem code: if (file_exists($attachment_location)) { header($_SERVER["SERVER_PROTOCOL"] . " 200 OK"); header("Cache-Control: public"); // needed for i.e. header("Content-Type: application/zip"); header("Content-Transfer-Encoding: Binary"); header("Content-Length:".filesize($attachment_location)); header("Content-Disposition: attachment; filename=file.zip"); readfile($attachment_location); die("Hooray"); } else { die("Error: File not found."); } This code works absolutely fine when testing locally, but after deploying to the live server the browser returns a 'Page not found' error. I thought it might be an .htaccess issue, but all .htaccess files on the live server are identical to their local counterparts. My next guess would be the live server's PHP configuration, but I have no idea what PHP setting might cause this behaviour. The file_exists() function always returns true - I've checked this on the live server and it's always correctly picking up the file, its size etc., so it does have a handle on the file. It just won't perform the download! The main site is a Wordpress site, but this code isn't part of a Wordpress page - it's in a standalone directory within the site root. UPDATE: is_file() and is_readable() are both returning true for the file, so that's not the problem. The specific line that is causing the problem is: readfile($attachment_location) Everything up until that point is super happy.

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  • jQuery and array of objects

    - by sepoto
    $(document).ready(function () { output = ""; $.ajax({ url: 'getevents.php', data: { ufirstname: 'ufirstname' }, type: 'post', success: function (output) { alert(output); var date = new Date(); var d = date.getDate(); var m = date.getMonth(); var y = date.getFullYear(); $('#calendar').fullCalendar({ header: { left: 'prev,next today', center: 'title', right: 'month,basicWeek,basicDay' }, editable: true, events: output }); } }); }); I have code like this and if I copy the text verbatim out of my alert box and replace events: output with events: [{ id: 1, title: 'Birthday', start: new Date(1355011200*1000), end: new Date(1355011200*1000), allDay: true, url: 'http://www.yahoo.com/'},{ id: 2, title: 'Birthday Hangover', start: new Date(1355097600*1000), end: new Date(1355097600*1000), allDay: false, url: 'http://www.yahoo.com'},{ id: 3, title: 'Sepotomus Maximus Christmas', start: new Date(1356393600*1000), end: new Date(1356393600*1000), allDay: false, url: 'http://www.yahoo.com/'},] Everything works just fine. What can I do to fix this problem? I though that using events: output would place the text in that location but it does not seem to be working. Thank you all kindly in advance for any comments or answers!

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  • Array length is zero in jQuery.

    - by James123
    I wrote like this. After submit click loop is not excuting. But I saw value are there, But array lenght is showing '0'. (Please see picture). Why it is not going into loop? and $('#myVisibleRows').val(idsString); becoming 'empty'. $(document).ready(function() { $('tr[@class^=RegText]').hide().children('td'); var list_Visible_Ids = []; var idsString, idsArray; alert($('#myVisibleRows').val()); idsString = $('#myVisibleRows').val(); idsArray = idsString.split(','); $.each(idsArray, function() { if (this != "") { $('#' + this).siblings(('.RegText').toggle(true)); window['list_Visible_Ids'][this] = 1; } }); $('tr.subCategory1') .css("cursor", "pointer") .attr("title", "Click to expand/collapse") .click(function() { //this = $(this); $(this).siblings('.RegText').toggle(); list_Visible_Ids[$(this).attr('id')] = $(this).css('display') != 'none' ? 1 : null; alert(list_Visible_Ids[$(this).attr('id')]) }); $('#form1').submit(function() { idsString = ''; $.each(list_Visible_Ids, function(key, val) { alert(val); if (val) { idsString += (idsString != '' ? ',' : '') + key; } }); $('#myVisibleRows').val(idsString); form.submit(); }); });

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  • Changing PHP variable from an <a href=""> link

    - by aener
    I'm making a site using divs. I have one across the top with a row of links which I want to change what's in the main window. I would prefer to keep strictly to PHP and HTML and use a full page refresh as I am comfortable with them at current, rather than using an Ajax refresh as I don't have any knowledge of Ajax. I don't want to use iframes. I thought the best way to do this would be as follows: <div id="top"> <a href="news.html">News</a> </div> <div id="main"> <?php include($page); ?> </div> What I need is a way of changing the $page variable upon clicking the link. If this means changing the variable and then reloading the page then so be it. A command following the logic of: <a href="news.html" action="<?php $page="news.html"; ?>">News</a> would be ideal! I thought about using a form, but it seems there should be an easier way.

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  • Notification 7 Days Before Payment PHP No Error Message

    - by user1858672
    So I have PHP script running on cron daily. It is suppose to check the date and send an email if it is 7 days before any payments as a reminder. I haven't used PHP in a long time so sorry if the way I did it was ridiculous or something. Thanks a lot. I would say what the problem is but I don't get any error message or anything.. <?php mysqli_connect("xxxxxx", "xxxxxxx", "xxxxxxxxx") or die(mysqli_error()); mysqli_select_db("xxxxxxxxx") or die(mysqli_error()); //Retrieve original payment dates and enter them into an array $result = mysqli_query("SELECT DateCreated FROM UserTable"); $payment_dates = mysqli_fetch_array($result); //Puts original payment date into month-day format $md_payment_dates = substr($payment_dates, 5); //Gets todays date in m-d format and adds 7 days to it $due_date = mktime(0, 0, 0, date("m"), date("d")+7); //Checks if any payment days match today's date. If there are none it script will stop. if (count(preg_grep($due_date, $md_payment_dates)) > 0) { //Retrieves usernames of users that have an invoice due. $users_to_pay = mysqli_fetch_array("SELECT Username IN UserTable WHERE DateCreated = $date"); //Notifies you via email $to = "[email protected]"; $subject = "7 Day Payment Reminder"; $message = "Hi, <br /> The following owe a payment in 7 days : " + $users_to_pay ".<br/> Their payments are due on " + $md_payment_dates " of this year."; mail($to, $subject, $message); exit(); }else{ exit(); } ?>

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  • How can I add one line into all php files' beginning?

    - by Tom
    So, ok. I have many php files and one index.php file. All files can't work without index.php file, because I include them in index.php. For example. if somebody click Contact us the URL will become smth like index.php?id=contact and I use $_GET['id'] to include contacts.php file. But, if somebody find the file's path, for example /system/files/contacts.php I don't want that that file would be executed. So, I figured out that I can add before including any files in index.php line like this $check_hacker = 1 and use if in every files beginning like this if($check_hacker <> 1) die();. So, how can I do it without opening all files and adding this line to each of them? Is it possible? Because I actually have many .php files. And maybe there is other way to do disable watching separate file? Any ideas? Thank you.

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  • Using AJAX to POST data to PHP database, then refresh

    - by cb74656
    Currently I have a button: <ul> <li><button onclick="display('1')">1</button></li> <li><button onclick="display('2')">2</button></li> <li><button onclick="display('3')">3</button></li> </ul> That when pressed, calls a javascript function, and displays PHP based on which button is pressed using AJAX. I figured this out all on my own. The AJAX gets a PHP file with a postgres query that outputs a table of data to a div. Now I want to be able to add, via form, new data and have it refresh (without reloading the page, yannknow?). I've tried a couple of things, and have hit roadblocks every time. My initial idea was to have the form submit the data using a javascript function and AJAX, then call my "display()" function after the query to reload the content. I just can't figure it out using GoogleFu. Based on my current idea, I'd like help with the following: How do I pass the form data to a javascript function. How do I use POST to pass that data to PHP using AJAX? I'm super new to javascript and AJAX. I've looked into jquery as it seems like that's the way to go, but I can't figure it out. If there's a better way to do this, I'm open to suggestions. Please forgive any misuse of nomenclature. EDIT: Once I solve this problem..., I'll have all the tools needed to finish the project preliminarily.

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  • PHP Array to CSV

    - by JohnnyFaldo
    I'm trying to convert an array of products into a CSV file, but it doesn't seem to be going to plan. The CSV file is one long line, here is my code: for($i=0;$i<count($prods);$i++) { $sql = "SELECT * FROM products WHERE id = '".$prods[$i]."'"; $result = $mysqli->query($sql); $info = $result->fetch_array(); } $header = ''; for($i=0;$i<count($info);$i++) { $row = $info[$i]; $line = ''; for($b=0;$b<count($row);$b++) { $value = $row[$b]; if ( ( !isset( $value ) ) || ( $value == "" ) ) { $value = "\t"; } else { $value = str_replace( '"' , '""' , $value ); $value = '"' . $value . '"' . "\t"; } $line .= $value; } $data .= trim( $line ) . "\n"; } $data = str_replace( "\r" , "" , $data ); if ( $data == "" ) { $data = "\n(0) Records Found!\n"; } header("Content-type: application/octet-stream"); header("Content-Disposition: attachment; filename=your_desired_name.xls"); header("Pragma: no-cache"); header("Expires: 0"); print "$data"; Also, the header doesn't force a download. I've been copy and pasting the output and saving as .csv

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  • Building html structure in php or javascript?

    - by Adam
    I've been doing a lot of ajax calls and using the returned data to build html with javascript. However, I've noticed some people are returning the constructed html in the ajax calls since they're doing it all in php. What is the preferred method? I have a bunch of stuff already using javascript, so I guess I would prefer not changing everything to use just php. But, I'm assuming php would be more "secure."? The following is what I've been doing: $main_frag = $("<div class='order-container'/>"); $contact_frag = $("<div class='group'><div class='line-data'>Name: "+data.name+"</div><div class='line-data'>Email: "+data.email+"</div><div class='line-data'>Phone: "+data.phone+"</div></div>"); $address_frag = $("<div class='group'><div class='line-data'>Address 1: "+data.address_one+"</div><div class='line-data'>Address 2: "+address2+"</div><div class='line-data'>City: "+data.city+"</div><div class='line-data'>Province: "+data.province+"</div><div class='line-data'>Postal Code: "+data.postal+"</div></div>"); etc.. I just want to hear the opinions of the community.

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  • Not getting an array as result (calling a webservice by AJAX-JSON)

    - by Pasargad
    I'm trying to get the result of my web service as an array and then loop over the result to fetch all of the data; what I have done so far: In my web service when I return the result I use return json_encode($newFiles); and the result is as following: "[{\"path\":\"c:\\\\my_images\\\\123.jpg\",\"ID\":\"123\",\"FName\":\"John\",\"LName\":\"Brown\",\"dept\":\"Hr\"}]" tehn in my Web application I'm calling the rest web service by the following code in the RestService class: public function getNewImages($time) { $url = $this->rest_url['MyService'] . "?action=getAllNewPhotos&accessKey=" . $this->rest_key['MyService'] . "&lastcheck=" . $time; $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $data = curl_exec($ch); if ($data) { return json_decode($data); } else { return null; } } and then in my controller I have the following code: public function getNewImgs($time="2011-11-03 14:35:08") { $newImgs = $this->restservice->getNewImages($time); echo json_encode$newImgs; } and I'm calling this `enter code here`controller method by AJAX: $("#searchNewImgManually").click(function(e) { e.preventDefault(); $.ajax({ type: "POST", async: true, datatype: "json", url: "<?PHP echo base_url("myProjectController/getNewImgs"); ?>", success: function(imgsResults) { alert(imgsResults[0]); } }); }); but instead of giving me the first object it is just giving me quotation mark (the first charachter of the result) " Why is that? I'm passing in JSON format and in AJAX I mentioned datatype as "JSON" ! Please let me know if you need more clarification! Thanks :)

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  • detect click on submit button in PHP

    - by Remus Rigo
    hi all I have a php file that contains a form (witch contains 2 input boxes and a submit button) for updating a contact. I managed to fill the fields with the contact's data, but I can't detect if the submit button is clicked form looks like this echo "<form action=Contact.php><table>". "<tr><td>First Name</td><td><input type=text size=75% name=FirstName value='".$row['FirstName']."'></td></tr>". "<tr><td>Last Name</td><td><input type=text size=75% name=LastName value='".$row['LastName']."'></td></tr>". "<tr><td colspan=2><input type=submit name=UpdateContact value=Update></td></tr>". "</table></form>"; this code should output a "clicked" message if the button is clicked if (isset($_POST['UpdateContact'])) { echo "<p>clicked"; } else { echo "<p>not clicked"; } can anyone help me out or tell me what i've done wrong (I want from the same php file to fill the contact's data in a from and to update the database)

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  • PHP session token can be used multipletimes?

    - by kornesh
    I got page A which is a normal HTML page and page which is an AJAX response page. And I want to prevent CSRF attacks by tokens. Lets say I use this method for an autocomplete form, is it possible to use same token multiple times (of course the session is only set one time) because i tired this method but the validation keep failing after the first suggestion (obviously the token has changed, somehow) page A <?php session_start(); $token = md5(uniqid(rand(), TRUE)); $_SESSION['token'] = $token; ?> <input id="token" value="<?php echo $token; ?>" type="hidden"></input> <input id="autocomplete" placeholder="Type something"></input> .... The form is autosubmitted every time theres a change using Jquery. page B <?php session_start(); if($_REQUEST['token'] == $_SESSION['token']){ echo 'Im working fine'; } ?>

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  • Load the <?php the_permalink(); ?> with an ajax loader

    - by fxg
    I´m working on a wordpress template. I´m trying to load the single.php of a post using ajax. I´m doing all the load thru a loader.js file that has this: // load single project page $("#project_slider").live("click", function(){ $("#content").hide(); $("#content").load("<?php the_permalink(); ?>", function(){ $(this).fadeIn("slow"); }); }); The problem is that I can´t just put on the .load because it doesn´t works. this is the markup: <div id="project_page" class="item"> <a href="#"> <img src="<?php the_field('artworks_thumbnail'); ?>" alt="" width="240" height="173"> </a> <div class="art_title"> <p>SWEET LIFE</p> </div> <div class="mask"></div> </div> How can I add the permalink via the loader.js?

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  • How to separate date in php

    - by user225269
    I want to be able to separate the birthday from the mysql data into day, year and month. Using the 3 textbox in html. How do I separate it? I'm trying to think of what can I do with the code below to show the result that I want: Here's the html form with the php code: $idnum = mysql_real_escape_string($_POST['idnum']); mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO='$idnum'"); $month = mysql_real_escape_string($_POST['mm']); ?> <?php while ( $row = mysql_fetch_array($result) ) { ?> <tr> <td width="30" height="35"><font size="2">Month:</td> <td width="30"><input name="mo" type="text" id="mo" onkeypress="return handleEnter(this, event)" value="<?php echo $month = explode("-",$row['BIRTHDAY']);?>"> As you can see the column is the mysql database is called BIRTHDAY. With this format: YYYY-MM-DD How do I do it. So that the data from the single column will be divided into three parts? Please help thanks,

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  • PHP - Undefined index error

    - by user1815290
    This is my form with a file input field named "photo": <form action = "spremi-film.php" method = "POST" enctype = "multipart/form-data"> <div class="grid_2">Naslov:</div> <div class="grid_10"><input type="text" name="naslov" value="" /></div> <div class="grid_2">Žanr: </div> <div class="grid_10"><?php izborZanra(); ?></div> <div class="grid_2">Godina: </div> <div class="grid_10"><?php izborGodine(); ?></div> <div class="grid_2">Trajanje:</div> <div class="grid_10"><input type="text" name="trajanje" value="" /></div> <div class="grid_2">Izbor slike:</div> <!--<input type="hidden" name="MAX_FILE_SIZE" value="30000" />--> <div class="grid_10"><input type="file" name="photo" /></div> <div class="grid_12"><input type="submit" value="SPREMI" /></div> </form> After that i put this code: $uploaddir = '/slike'; $uploadfile = $uploaddir . basename($_FILES['photo']['name']); //echo '<pre>'; if(move_uploaded_file($_FILES['photo']['tmp_name'], $uploadfile)) { echo 'Slika je uspješno spremljena!'; } else { echo 'Slika nije spremljena!'; } //print_r($_FILES); //echo '</pre>'; When i run this i have an undefined index: photo notice in my browser. Help, please.

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  • Jquery - $.(post) data response not consistent with PHP

    - by Sasha
    Jquery code: var code = $('#code'), id = $('input[name=id]').val(), url = '<?php echo base_url() ?>mali_oglasi/mgl_check_paid'; code.on('focusout', function(){ var code_value = $(this).val(); if(code_value.length != 16 ) { if ($('p[role=code_msg]').length != 0 ) $('p[role=code_msg]').remove() ; code.after('<p role=code_msg>Pogrešan kod je unešen.</p>'); } else { if ($('p[role=code_msg]').length != 0 ) $('p[role=code_msg]').remove() ; $.post(url, {id : id, code : code_value}, function(data){ if(data != 'TRUE'){ code.after('<p role=code_msg>Uneti kod je neispravan.</p>'); } else { code.after('<p role=code_msg>Status malog oglasa je promenjen.</p>'); code.after(create_image()); code.remove(); } }); } }); PHP (Codeigniter) code: function mgl_check_paid() { $code = $this->input->post('code'); $id = $this->input->post('id'); echo ($this->mgl->mgl_check_paid($code, $id)) ? 'TRUE' : 'FALSE'; } Problem is following: When code is sent and if it is correct, PHP part will echo TRUE, and JS will execute ELSE part (after post), but for some reason it is not doing that (it is executing the first part of the statment)? What is wrong with this code?

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  • Adding different objects to array, but only one object repeating

    - by Carpetfizz
    I have a small piece of PHP code that goes through valid values and fetches information about them. It then pushes it to an array. For some reason, I'm only getting the last item of $row, repeated several times. When I try to print_r at #1 in the code, the expected values are outputted. However, at the end of the loop, or outside of it, when I try to print_r($ipArray), I'm only getting the last value repeated multiple times. Any help would be much appreciated! while($row = mysqli_fetch_array($getIpQuery, MYSQLI_NUM)){ for($x=0;$x<count($row);$x++) { $getIpInfo = mysqli_query($dbcon, "SELECT * FROM ipInfo WHERE address='$row[$x]'"); $retrievedInfo = mysqli_fetch_array($getIpInfo, MYSQLI_NUM); $ipInfo->ipAddress = $retrievedInfo[0]; $ipInfo->portNum = $retrievedInfo[1]; print_r($ipInfo); //#1: Works perfectly fine. array_push($ipArray,$ipInfo); } } print_r($ipArray); //this is where I'm getting an output of only the last element of `$row`. Thanks! ~Carpetfizz

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  • Update php 5.2.0 to 5.2.4 with aptitude

    - by Kiva
    Hi guy, I would like to update my php 5 in my server. At this moment, I use php 5.2.0 so I want to update it to php 5.2.4 (not php 5.3). I tried to do this: aptitude update aptitude upgrade 63 packets were updated but not php which is always in 5.0 How can I update my php please ? Here is the output of commands asked by David in another post: aptitude search php5 p libapache-mod-php5 - server-side, HTML-embedded scripting langu i A libapache2-mod-php5 - server-side, HTML-embedded scripting langu i php5 - server-side, HTML-embedded scripting langu p php5-apache2-mod-bt - PHP bindings for mod_bt p php5-auth-pam - A PHP5 extension for PAM authentication i php5-cgi - server-side, HTML-embedded scripting langu p php5-clamavlib - PHP ClamAV Lib - ClamAV Interface for PHP5 p php5-cli - command-line interpreter for the php5 scri i A php5-common - Common files for packages built from the p i php5-curl - CURL module for php5 p php5-dev - Files for PHP5 module development i A php5-gd - GD module for php5 p php5-idn - PHP api for the IDNA library p php5-imagick - ImageMagick module for php5 p php5-imap - IMAP module for php5 p php5-interbase - interbase/firebird module for php5 p php5-json - JSON serialiser for PHP5 p php5-ldap - LDAP module for php5 p php5-mapscript - module for php5-cgi to use mapserver p php5-maxdb - PHP extension to access MaxDB databases fo i A php5-mcrypt - MCrypt module for php5 p php5-memcache - memcache extension module for PHP5 p php5-mhash - MHASH module for php5 p php5-ming - Ming module for php5 i A php5-mysql - MySQL module for php5 p php5-odbc - ODBC module for php5 p php5-pgsql - PostgreSQL module for php5 p php5-ps - ps module for PHP 5 p php5-pspell - pspell module for php5 p php5-radius - PECL radius module for PHP 5 p php5-recode - recode module for php5 p php5-snmp - SNMP module for php5 p php5-sqlite - SQLite module for php5 p php5-sqlite3 - SQLite3 module for php5 p php5-sqlrelay - SQL Relay PHP API p php5-suhosin - advanced protection module for php5 p php5-sybase - Sybase / MS SQL Server module for php5 p php5-tidy - tidy module for php5 p php5-uuid - OSSP uuid module for php5 p php5-xapian - Xapian search engine interface for PHP5 p php5-xcache - Fast, stable PHP opcode cacher p php5-xmlrpc - XML-RPC module for php5 p php5-xsl - XSL module for php5 aptitude show php5 | grep Version Version : 5.2.0-8+etch13 aptitude show php5-cgi | grep Version Version : 5.2.0-8+etch13 php5 --version -bash: php5: command not found php-cgi --version PHP 5.2.0-8+etch13 (cgi-fcgi) (built: Oct 2 2008 08:21:17) Copyright (c) 1997-2006 The PHP Group Zend Engine v2.2.0, Copyright (c) 1998-2006 Zend Technologies

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  • LinkedIn API returns 'Unauthorized' response (PHP OAuth)

    - by Jim Greenleaf
    I've been struggling with this one for a few days now. I've got a test app set up to connect to LinkedIn via OAuth. I want to be able to update a user's status, but at the moment I'm unable to interact with LinkedIn's API at all. I am able to successfully get a requestToken, then an accessToken, but when I issue a request to the API, I see an 'unauthorized' error that looks something like this: object(OAuthException)#2 (8) { ["message:protected"]=> string(73) "Invalid auth/bad request (got a 401, expected HTTP/1.1 20X or a redirect)" ["string:private"]=> string(0) "" ["code:protected"]=> int(401) ["file:protected"]=> string(47) "/home/pmfeorg/public_html/dev/test/linkedin.php" ["line:protected"]=> int(48) ["trace:private"]=> array(1) { [0]=> array(6) { ["file"]=> string(47) "/home/pmfeorg/public_html/dev/test/linkedin.php" ["line"]=> int(48) ["function"]=> string(5) "fetch" ["class"]=> string(5) "OAuth" ["type"]=> string(2) "->" ["args"]=> array(2) { [0]=> string(35) "http://api.linkedin.com/v1/people/~" [1]=> string(3) "GET" } } } ["lastResponse"]=> string(358) " 401 1276375790558 0000 [unauthorized]. OAU:Bhgk3fB4cs9t4oatSdv538tD2X68-1OTCBg-KKL3pFBnGgOEhJZhFOf1n9KtHMMy|48032b2d-bc8c-4744-bb84-4eab53578c11|*01|*01:1276375790:xmc3lWhXJvLSUZh4dxMtrf55VVQ= " ["debugInfo"]=> array(5) { ["sbs"]=> string(329) "GET&http%3A%2F%2Fapi.linkedin.com%2Fv1%2Fpeople%2F~&oauth_consumer_key%3DBhgk3fB4cs9t4oatSdv538tD2X68-1OTCBg-KKL3pFBnGgOEhJZhFOf1n9KtHMMy%26oauth_nonce%3D7068001084c13f2ee6a2117.22312548%26oauth_signature_method%3DHMAC-SHA1%26oauth_timestamp%3D1276375790%26oauth_token%3D48032b2d-bc8c-4744-bb84-4eab53578c11%26oauth_version%3D1.0" ["headers_sent"]=> string(401) "GET /v1/people/~?GET&oauth_consumer_key=Bhgk3fB4cs9t4oatSdv538tD2X68-1OTCBg-KKL3pFBnGgOEhJZhFOf1n9KtHMMy&oauth_signature_method=HMAC-SHA1&oauth_nonce=7068001084c13f2ee6a2117.22312548&oauth_timestamp=1276375790&oauth_version=1.0&oauth_token=48032b2d-bc8c-4744-bb84-4eab53578c11&oauth_signature=xmc3lWhXJvLSUZh4dxMtrf55VVQ%3D HTTP/1.1 User-Agent: PECL-OAuth/1.0-dev Host: api.linkedin.com Accept: */*" ["headers_recv"]=> string(148) "HTTP/1.1 401 Unauthorized Server: Apache-Coyote/1.1 Date: Sat, 12 Jun 2010 20:49:50 GMT Content-Type: text/xml;charset=UTF-8 Content-Length: 358" ["body_recv"]=> string(358) " 401 1276375790558 0000 [unauthorized]. OAU:Bhgk3fB4cs9t4oatSdv538tD2X68-1OTCBg-KKL3pFBnGgOEhJZhFOf1n9KtHMMy|48032b2d-bc8c-4744-bb84-4eab53578c11|*01|*01:1276375790:xmc3lWhXJvLSUZh4dxMtrf55VVQ= " ["info"]=> string(216) "About to connect() to api.linkedin.com port 80 (#0) Trying 64.74.98.83... connected Connected to api.linkedin.com (64.74.98.83) port 80 (#0) Connection #0 to host api.linkedin.com left intact Closing connection #0 " } } My code looks like this (based on the FireEagle example from php.net): $req_url = 'https://api.linkedin.com/uas/oauth/requestToken'; $authurl = 'https://www.linkedin.com/uas/oauth/authenticate'; $acc_url = 'https://api.linkedin.com/uas/oauth/accessToken'; $api_url = 'http://api.linkedin.com/v1/people/~'; $callback = 'http://www.pmfe.org/dev/test/linkedin.php'; $conskey = 'Bhgk3fB4cs9t4oatSdv538tD2X68-1OTCBg-KKL3pFBnGgOEhJZhFOf1n9KtHMMy'; $conssec = '####################SECRET KEY#####################'; session_start(); try { $oauth = new OAuth($conskey,$conssec,OAUTH_SIG_METHOD_HMACSHA1,OAUTH_AUTH_TYPE_URI); $oauth->enableDebug(); if(!isset($_GET['oauth_token'])) { $request_token_info = $oauth->getRequestToken($req_url); $_SESSION['secret'] = $request_token_info['oauth_token_secret']; header('Location: '.$authurl.'?oauth_token='.$request_token_info['oauth_token']); exit; } else { $oauth->setToken($_GET['oauth_token'],$_SESSION['secret']); $access_token_info = $oauth->getAccessToken($acc_url); $_SESSION['token'] = $access_token_info['oauth_token']; $_SESSION['secret'] = $access_token_info['oauth_token_secret']; } $oauth->setToken($_SESSION['token'],$_SESSION['secret']); $oauth->fetch($api_url, OAUTH_HTTP_METHOD_GET); $response = $oauth->getLastResponse(); } catch(OAuthException $E) { var_dump($E); } I've successfully set up a connection to Twitter and one to Facebook using OAuth, but LinkedIn keeps eluding me. If anyone could offer some advice or point me in the right direction, I will be extremely appreciative!

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