Search Results

Search found 8646 results on 346 pages for 'echo flow'.

Page 156/346 | < Previous Page | 152 153 154 155 156 157 158 159 160 161 162 163  | Next Page >

  • PHP & MySQL username validation and storage problem.

    - by php
    For some reason when a user enters a brand new username the error message <p>Username unavailable</p> is displayed and the name is not stored. I was wondering if some can help find the flaw in my code so I can fix this error? Thanks Here is the PHP code. if($_POST['username'] && trim($_POST['username'])!=='') { $u = "SELECT * FROM users WHERE username = '$username' AND user_id <> '$user_id'"; $r = mysqli_query ($mysqli, $u) or trigger_error("Query: $u\n<br />MySQL Error: " . mysqli_error($mysqli)); if (mysqli_num_rows($r) == TRUE) { echo '<p>Username unavailable</p>'; $_POST['username'] = NULL; } else if(isset($_POST['username']) && mysqli_num_rows($r) == 0 && strlen($_POST['username']) <= 255) { $username = mysqli_real_escape_string($mysqli, $_POST['username']); } else if($_POST['username'] && strlen($_POST['username']) >= 256) { echo '<p>Username can not exceed 255 characters</p>'; } }

    Read the article

  • A stupid question about wordpress and php

    - by bubdada
    It may seem stupid question, but i've a serious problem... if you could check out orcik.net the thumbnail images does not appear. I figured out the reason but I don't know how to solve.. http://orcik.net/projects/thumb/orcikthumb.php?src=http://orcik.net/wp-content/uploads/2010/05/mac-safari-search-cache.png If you go to the above link you will get page not found error. However, if you go to the link below you'll get the thumbnail version of the image... http://orcik.net/projects/thumb/orcikthumb.php?src=/wp-content/uploads/2010/05/mac-safari-search-cache.png I'm using this piece of code on wordpress and the line appears like <a href="<?php the_permalink() ?>" rel="bookmark"> <img src="<?php bloginfo('template_directory'); ?>/includes/orcikthumb.php?src=<?php get_thumbnail($post->ID, 'full'); ?>&amp;h=<?php echo get_theme_mod($height); ?>&amp;w=<?php echo get_theme_mod($width); ?>&amp;zc=1" alt="<?php the_title(); ?>" /> </a> Thus, I believe I can't change the directory of image. But I could not figure out why I am getting page not found error. Is that might be CHMOD'es??? or something else?? Thanks

    Read the article

  • Why isn't the pathspec magic :(exclude) excluding the files I specify from git log's output?

    - by Jubobs
    This is a follow-up to Ignore files in git log -p and is also related to Making 'git log' ignore changes for certain paths. I'm using Git 1.9.2. I'm trying to use the pathspec magic :(exclude) to specify that some patches should not be shown in the output of git log -p. However, patches that I want to exclude still show up in the output. Here is minimal working example that reproduces the situation: cd ~/Desktop mkdir test_exclude cd test_exclude git init mkdir testdir echo "my first cpp file" >testdir/test1.cpp echo "my first xml file" >testdir/test2.xml git add testdir/ git commit -m "added two test files" Now I want to show all patches in my history expect those corresponding to XML files in the testdir folder. Therefore, following VonC's answer, I run git log --patch -- . ":(exclude)testdir/*.xml" but the patch for my testdir/test2.xml file still shows up in the output: commit 37767da1ad4ad5a5c902dfa0c9b95351e8a3b0d9 Author: xxxxxxxxxxxxxxxxxxxxxxxxx Date: Mon Aug 18 12:23:56 2014 +0100 added two test files diff --git a/testdir/test1.cpp b/testdir/test1.cpp new file mode 100644 index 0000000..3a721aa --- /dev/null +++ b/testdir/test1.cpp @@ -0,0 +1 @@ +my first cpp file diff --git a/testdir/test2.xml b/testdir/test2.xml new file mode 100644 index 0000000..8b7ce86 --- /dev/null +++ b/testdir/test2.xml @@ -0,0 +1 @@ +my first xml file What am I doing wrong? What should I do to tell git log -p not to show the patch associated with all XML files in my testdir folder?

    Read the article

  • post values to external page explicitly PHP

    - by JPro
    Hi, I want to post values to a page through a hyperlink in another page. Is it possible to do that? Let's say I have a page results.php which has the starting line, <?php if(isset($_POST['posted_value'])) { echo "expected value"; // do something with the data } else { echo "no the value expected"; } If from another page say link.php I place a hyperlink like this: <a href="results.php?posted_value=1"> , will this be accepetd by the results page? If instead if I replace the above starting line with if(isset($_REQUEST['posted_value'])), will this work? I believe the above hyperlink evaluates to GET, but since the only visibility difference between GET and POST that is you can see parameters in the address bar with GET But, is there any other way to place a hyperlink which can post values to a page? or can we use jquery in the place of hyperlink to POST the values? Can anyone please suggest me something on this please? Thanks.

    Read the article

  • Session in php are not enough clear to me

    - by Lulzim
    I find sessions in php kind of confusing, can anybody of you explain those to me. I have an example which is not working in my case: I register sessions this way, would you please tell me is this the right way of registering sessions //this is the page from where i register myusername in sessions if($count==1){ session_start(); $_SESSION['myusername'] = $_POST['myusername']; include("enterpincover.php"); } else { echo "Wrong Pin"; } here i check first whether the username is registered in sessions in oder to open his account , otherwise open again login. It works, if user is not loged in, it will show login page which is right, if user is loged it shows welcome message but not the Welcome the name of the user as I want. for ex: Welcome David <?php session_start(); if(isset($_SESSION['myusername'])) { echo 'Welcome '.$_SESSION['myusername']; } else { include("leftmodules.php"); include("rightmodules.php"); include("login.php"); } ?>

    Read the article

  • Why doesn't sed's automatic printing deliver the expected results?

    - by CodeGnome
    What Works This sed script works as intended: $ echo -e "2\n1\n4\n3" | sed -n 'h; n; G; p' 1 2 3 4 It takes pair of input lines at a time, and swaps the lines. So far, so good. What Doesn't Work What I don't understand is why I can't use sed's automatic printing. Since sed automatically prints the pattern space at the end of each execution cycle (except when it's suppressed), why is this not equivalent? $ echo -e "2\n1\n4\n3" | sed 'h; n; G' 2 1 2 4 3 4 What I think the code says is: The input line is copied to the hold space. The next line is read into the pattern space. The hold space is appended to the pattern space. The pattern space (line1 + newline + line2) is printed automatically because we've reached the end of the execution cycle. Obviously, I'm wrong...but I don't understand why. Can anyone explain why this second example breaks, and why print suppression is needed to yield the correct results?

    Read the article

  • Trouble defining method for Javascript class definition.

    - by btoverdrive
    I'm somewhat new to object oriented programming in Javascript and I'm trying to build a handler object and library for a list of items I get back from an API call. Ideally, I'd like the library functions to be members of the handler class. I'm having trouble getting my class method to work however. I defined as part of the class bcObject the method getModifiedDateTime, but when I try to echo the result of the objects call to this method, I get this error: Error on line 44 position 26: Expected ';' this.getModifiedDateTime: function(epochtime) { which leads me to believe that I simply have a syntax issue with my method definition but I can't figure out where. response( { "items": [ {"id":711,"name":"Shuttle","lastModifiedDate":"1268426336727"}, {"id":754,"name":"Formula1","lastModifiedDate":"1270121717721"} ], "extraListItemsAttr1":"blah", "extraListItemsAttr2":"blah2" }); function response(MyObject) { bcObject = new bcObject(MyObject); thing = bcObject.getModifiedDateTime(bcObject.videoItem[0].lastModifiedDate); SOSE.Echo(thing); } function bcObject(listObject) { // define class members this.responseList = {}; this.videoCount = 0; this.videoItem = []; this.responseListError = ""; // instantiate members this.responseList = listObject; this.videoCount = listObject.items.length; // populate videoItem array for (i=0;i<this.videoCount;i++) { this.videoItem[i] = listObject.items[i]; } this.getModifiedDateTime: function(epochtime) { var dateStringOutput = ""; var myDate = new Date(epochtime); dateStringOutput = myDate.toLocaleString(); return dateStringOutput; }; }

    Read the article

  • 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.

    Read the article

  • Show Color Picker Value Code Within Input Text Field in Wordpress

    - by Shwan Namiq
    Hi i have options page for my theme i used color picker jquery plugin in my options page as show in image below. i want when i change the color from color picker automatically show the color value code within the text field.how can do this? this is the code within my options page related to appearing the color picker and text field function to register the options setting function YPE_register_settings_sections_fields() { register_setting ( 'YPE_header_option_group', 'YPE_header_option_name', 'YPE_sanitize_validate_callback' ); add_settings_section ( 'YPE_header_section', 'Header Section', 'YPE_header_section_callback', 'YPE_menu_page_options' ); add_settings_field ( 'YPE_header_bg', 'Header Background', 'YPE_header_bg_callback', 'YPE_menu_page_options', 'YPE_header_section' ); } add_action('admin_init', 'YPE_register_settings_sections_fields'); function to appear the text field and color picker function YPE_header_bg_callback() { $YPE_options = get_option('YPE_header_option_name'); $YPE_header_bg = isset($YPE_options['YPE_header_bg']) ? $YPE_options['YPE_header_bg'] : ''; ?> <div class="input-group color-picker"> <input class="form-control" style="width:80px;" name="YPE_header_option_name[YPE_header_bg]" id="<?php echo 'YPE_header_bg'; ?>" type="text" value="<?php echo $YPE_header_bg; ?>" /> <span class="input-group-btn"> <div id="colorSelector"> <div nam style="background-color: #0000ff"> </div> </div> </span> </div> <script> $("#colorSelector").ColorPicker({ color: '#0000ff', onShow: function (colpkr) { $(colpkr).fadeIn(500); return false; }, onHide: function (colpkr) { $(colpkr).fadeOut(500); return false; }, onChange: function (hsb, hex, rgb) { $('#colorSelector div').css('backgroundColor', '#' + hex); }); </script> <?php }

    Read the article

  • How to get values from SQL query made by php?

    - by Ole Jak
    So I made a query like this global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); in my DB there are filds ID, UID, SID, TIME (all INT type exept time) So I am triing to print query relult into form <form> <select class="multiselect" multiple="multiple" name="SIDs"> <?php global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); $streams_count = mysql_num_rows($streams_set); for ($count=1; $count <= $streams_count; $count++) { echo "<option value=\"{$count}\""; echo ">{$count}</option>"; } ?> </select> <br/> <input type="submit" value="Submit Form"/> </form> How to print out as "option" "values" SID's from my sql query?

    Read the article

  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

    Read the article

  • Parameter error with Mysqli

    - by Morgan Green
    When I run this Query I recieve Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 188 SELECT * FROM characters WHERE id=5 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 194 The Query is running and it is strying to select the correct information, but for on the actual output it's giving me a fetch_array error; if anyone can see where the error lies it'd be much appreciated. Thank you. <?php $adminid= $admin->get_id(); $characterdb= 'characters'; $link = mysqli_connect("$server", "$user", "$pass", "$characterdb"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM characters WHERE id=$adminid"; $result = mysqli_query($link, $query); while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo $query; echo $row['name']; } mysqli_free_result($result); mysqli_close($link); ?>

    Read the article

  • How to write data option in jQuery.ajax() function when it include in a mysql_query?

    - by cj333
    I modify a php comment system. I want add it after every article witch are query from database. this is the php part <?php ... while($result = mysql_fetch_array($resultset)) { $article_title = $result['article_title']; ... ?> <form id="postform" class="postform"> <input type="hidden" name="title" id="title" value="<?=$article_title;?>" /> <input type="text" name="content" id="content" /> <input type="button" value="Submit" class="Submit" /> </form> ... <?php } ?> this is the ajax part. $.ajax({ type: "POST", url: "ajax_post.php", data: {title:$('#title').val(), content:$('#content').val() ajax_post.php echo $title; echo $content; How to modify the ajax data part that each article's comment can send each data to the ajax_post.php? thanks a lot.

    Read the article

  • Javascript variables are not working

    - by linkcool
    Hi, my problem is that my variables are not working in javascript. all variables need names without some character at the beginning, this is the stupid thing...Anyways, im trying to make a funtion that makes "select all checkboxes". It is not working so i looked at the page source/info and found out that the variables were not changing. this is my input: echo "<input onclick='checkAll(1);' type='checkbox' name='master'/><br/>"; My function: function checkAll(i) { for(var i=1; i < <?php echo $num; ?>; i++) { if(document.demo.master[i].checked == true) { document.demo.message[i].checked = true; } else { document.demo.message[i].checked = false; } } } so yes that's it. I can tell you that i also tried without the <i> in: checkAll("i") Thanks for the help.

    Read the article

  • How to get aggregate days from PHP's DateTime::diff?

    - by razic
    $now = new DateTime('now'); $tomorrow = new DateTime('tomorrow'); $next_year = new DateTime('+1 year'); echo "<pre>"; print_r($now->diff($tomorrow)); print_r($now->diff($next_year)); echo "</pre>"; DateInterval Object ( [y] => 0 [m] => 0 [d] => 0 [h] => 10 [i] => 17 [s] => 14 [invert] => 0 [days] => 6015 ) DateInterval Object ( [y] => 1 [m] => 0 [d] => 0 [h] => 0 [i] => 0 [s] => 0 [invert] => 0 [days] => 6015 ) any ideas why 'days' shows 6015? why won't it show the total number of days? 1 year difference means nothing to me, since months have varying number of days.

    Read the article

  • PHP rewrite an included file - is this a valid script?

    - by Poni
    Hi all! I've made this question: http://stackoverflow.com/questions/2921469/php-mutual-exclusion-mutex As said there, I want several sources to send their stats once in a while, and these stats will be showed at the website's main page. My problem is that I want this to be done in an atomic manner, so no update of the stats will overlap another one running in the background. Now, I came up with this solution and I want you PHP experts to judge it. stats.php <?php define("my_counter", 12); ?> index.php <?php include "stats.php"; echo constant("my_counter"); ?> update.php <?php $old_error_reporting = error_reporting(0); include "stats.php"; define("my_stats_template",' <?php define("my_counter", %d); ?> '); $fd = fopen("stats.php", "w+"); if($fd) { if (flock($fd, LOCK_EX)) { $my_counter = 0; try { $my_counter = constant("my_counter"); } catch(Exception $e) { } $my_counter++; $new_stats = sprintf(constant("my_stats_template"), $my_counter); echo "Counter should stand at $my_counter"; fwrite($fd, $new_stats); } flock($fd, LOCK_UN); fclose($fd); } error_reporting($old_error_reporting); ?> Several clients will call the "update.php" file once every 60sec each. The "index.php" is going to use the "stats.php" file all the time as you can see. What's your opinion?

    Read the article

  • Delete file using a link

    - by user329394
    Hi all, i want to have function like delete file from database by using link instead of button. how can i do that? do i need to use href/unlink or what? Can i do like popup confirmation wther yes or no. i know how to do that, but where should i put the code? this is the part how where system will display all filename and do direct upload. Beside each files, there will be a function for 'Remove': $qry = "SELECT * FROM table1 a, table2 b WHERE b.id = '".$rs[id]."' AND a.ptkid = '".$rs[id]."' "; $sql = get_records_sql($qry); foreach($sql as $rs){ ?> <?echo '<a href="download.php?f='.$rs->faillampiran.'">'. basename($rs->faillampiran).'</a>'; ?><td><?echo '<a href=""> [Remove]</a>';?></td><? ?><br> <? } ?> thankz all

    Read the article

  • PHP class extends not working why and is this how to correctly extend a class?

    - by Matthew
    Hi so I'm trying to understand how inherteince works in PHP using object oriented programming. The main class is Computer, the class that is inheriting is Mouse. I'm extedning the Computer class with the mouse class. I use __construct in each class, when I istinate the class I use the pc type first and if it has mouse after. For some reason computer returns null? why is this? class Computer { protected $type = 'null'; public function __construct($type) { $this->type = $type; } public function computertype() { $this->type = strtoupper($this->type); return $this->type; } } class Mouse extends Computer { protected $hasmouse = 'null'; public function __construct($hasmouse){ $this->hasmouse = $hasmouse; } public function computermouse() { if($this->hasmouse == 'Y') { return 'This Computer has a mouse'; } } } $pc = new Computer('PC', 'Y'); echo $pc->computertype; echo $pc->computermouse;

    Read the article

  • How to process a large post array in PHP where item names are all different and not known in advance

    - by Salnajjar
    I have a PHP page that queries a DB to populate a form for the user to modify the data and submit. The query returns a number of rows which contain 3 items: ImageID ImageName ImageDescription The PHP page titles each box in the form with a generic name and appends the ImageID to it. Ie: ImageID_03 ImageName_34 ImageDescription_22 As it's unknown which images are going to have been retrieved from the DB then I can't know in advance what the name of the form entries will be. The form deals with a large number of entries at the same time. My backend PHP form processor that gets the data just sees it as one big array: [imageid_2] => 2 [imagename_2] => _MG_0214 [imageid_10] => 10 [imagename_10] => _MG_0419 [imageid_39] => 39 [imagename_39] => _MG_0420 [imageid_22] => 22 [imagename_22] => Curly Fern [imagedescription_2] => Wibble [imagedescription_10] => Wobble [imagedescription_39] => Fred [imagedescription_22] => Sally I've tried to do an array walk on it to split it into 3 arrays which set places but am stuck: // define empty arrays $imageidarray = array(); $imagenamearray = array(); $imagedescriptionarray = array(); // our function to call when we walk through the posted items array function assignvars($entry, $key) { if (preg_match("/imageid/i", $key)) { array_push($imageidarray, $entry); } elseif (preg_match("/imagename/i", $key)) { // echo " ImageName: $entry"; } elseif (preg_match("/imagedescription/i", $key)) { // echo " ImageDescription: $entry"; } } array_walk($_POST, 'assignvars'); This fails with the error: array_push(): First argument should be an array in... Am I approaching this wrong?

    Read the article

  • Image error with wordpress and php

    - by bubdada
    It may seem stupid question, but i've a serious problem... if you could check out orcik.net the thumbnail images does not appear. I figured out the reason but I don't know how to solve.. http://orcik.net/projects/thumb/orcikthumb.php?src=http://orcik.net/wp-content/uploads/2010/05/mac-safari-search-cache.png If you go to the above link you will get page not found error. However, if you go to the link below you'll get the thumbnail version of the image... http://orcik.net/projects/thumb/orcikthumb.php?src=/wp-content/uploads/2010/05/mac-safari-search-cache.png I'm using this piece of code on wordpress and the line appears like <a href="<?php the_permalink() ?>" rel="bookmark"> <img src="<?php bloginfo('template_directory'); ?>/includes/orcikthumb.php?src=<?php get_thumbnail($post->ID, 'full'); ?>&amp;h=<?php echo get_theme_mod($height); ?>&amp;w=<?php echo get_theme_mod($width); ?>&amp;zc=1" alt="<?php the_title(); ?>" /> </a> Thus, I believe I can't change the directory of image. But I could not figure out why I am getting page not found error. Is that might be CHMOD'es??? or something else?? Thanks

    Read the article

  • Array values changing unexpectedly

    - by Lizard
    I am using cakephp 1.2 and I have an array that appears to have a value change even though that variable is not being manipulated. Below is the code to that is causing me trouble. PLEASE NOTE - UPDATE Changing the variable name makes no difference to the outcome, The values get changed somewhere between the two print_r calls, and I can't see why the $this-find would do this . echo "Start of findCountByString()"; print_r($myArr); $test = $this->find('count', array( 'conditions' => $conditions, 'joins' => array('LEFT JOIN `articles_entities` AS ArticleEntity ON `ArticleEntity`.`article_id` = `Article`.`id`'), 'group' => 'Article.id' )); echo "End of findCountByString()"; print_r($myArr); I am getting the following output: Start of findCountByString() Array ( [0] => 4bdb1d96-c680-4c2c-aae7-104c39d70629 [1] => 4bdb1d6a-9e38-479d-9ad4-105c39d70629 [2] => 4bdb1b55-35f0-4d22-ab38-104e39d70629 [3] => 4bdb25f4-34d4-46ea-bcb6-104f39d70629 ) End of findCountByString() Array ( [0] => 4bdb1d96-c680-4c2c-aae7-104c39d70629 [1] => 4bdb1d6a-9e38-479d-9ad4-105c39d70629 [2] => 4bdb1b55-35f0-4d22-ab38-104e39d70629 [3] => '4bdb25f4-34d4-46ea-bcb6-104f39d70629' # This is now in inverted commas ) The the value in my array have changed, and I don't know why? Any suggestions?

    Read the article

  • PHP the SELECT FROM WHERE col IN no working using array

    - by Sam Ram San
    I'm trying to pull some data from MySQL using an Array that was fetch from a first query. Everything's fine all the way to the implode after that, it's been a headache for me. Can someone help me? <?php $var = 'somedata'; include("config/conect.php"); $zip="SELECT * FROM table WHERE firstcol LIKE '%$var%' ORDER BY seconcol"; $results = $connetction->query($zip); while ($row = $results->fetch_array()){ $mycode[]=$row['zip']; } array_pop($mycode); $mycode = implode(', ',$mycode); //print_r ($mycode); echo '<br /><br /><br />'; $usr="SELECT * FROM reg_temp WHERE zip IN('".join("','", $mycode)."')"; $results = $asies->query($usr); while ($row = $results-> fetch_arra()) { $name = $row['name']; echo $name; } ?>

    Read the article

  • Find multiple patterns with a single preg_match_all in PHP

    - by Mark
    Using PHP and preg_match_all I'm trying to get all the HTML content between the following tags (and the tags also): <p>paragraph text</p> don't take this <ul><li>item 1</li><li>item 2</li></ul> don't take this <table><tr><td>table content</td></tr></table> I can get one of them just fine: preg_match_all("(<p>(.*)</p>)siU", $content, $matches, PREG_SET_ORDER); Is there a way to get all the <p></p> <ul></ul> <table></table> content with a single preg_match_all? I need them to come out in the order they were found so I can echo the content and it will make sense. So if I did a preg_match_all on the above content then iterated through the $matches array it would echo: <p>paragraph text</p> <ul><li>item 1</li><li>item 2</li></ul> <table><tr><td>table content</td></tr></table>

    Read the article

  • why my pagination link doesnt appear ?

    - by udaya
    This is my script which i have used to paginate ,,The datas are restricted to 4 but the pagination link doesn't appear <? require_once ('Pager/Pager.php'); $connection = mysql_connect( "localhost" , "root" , "" ); mysql_select_db( "ssit",$connection); $result=mysql_query("SELECT dFrindName FROM tbl_friendslist", $connection); $row = mysql_fetch_array($result); $totalItems = $row['total']; $pager_options = array( 'mode' => 'Sliding', // Sliding or Jumping mode. See below. 'perPage' => 4, // Total rows to show per page 'delta' => 4, // See below 'totalItems' => $totalItems, ); $pager = Pager::factory($pager_options); echo $pager->links; list($from, $to) = $pager->getOffsetByPageId(); $from = $from - 1; $perPage = $pager_options['perPage']; $result = mysql_query("SELECT * FROM tbl_friendslist LIMIT 5 , $perPage",$connection); while($row = mysql_fetch_array($result)) { echo $row['dFrindName'].'</br>'; } ?>

    Read the article

  • 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 :)

    Read the article

< Previous Page | 152 153 154 155 156 157 158 159 160 161 162 163  | Next Page >