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  • change php variable on click event

    - by Claudiu
    I want to change php variable ($start and $end called by var b1 and var b2) if the user clicks on button. Now I know that php is server side, but how can I do it? I read something about using $get but I don't know how to implement it: <?php if ( is_user_logged_in() ) { ?> <input type="submit" value="Start Chat" id="start_chat" style="position: absolute; top: 30px; left: 10px;" /> <?php } ?> <script> jQuery('#start_chat').click(function(){ $(this).data('clicked', true); }); var b1 = '<?php echo $start; ?>'; var b2 = '<?php echo $end; ?>'; if(jQuery('#start_chat').data('clicked')) { // change var b1 and b2 } else { // do not change anything } </script> <div id="eu_la"> <?php $start = strtotime('9:30'); $end = strtotime('12:30'); $timenow = date('U'); if((date('w') == 4) && ($timenow >= $start && $timenow <= $end)) { // day 2 = Tuesday include('facut_mine.php'); } else { do_shortcode('[upzslider usingphp=true]'); } ?>

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  • Reference problem when returning an object from array in PHP

    - by avastreg
    I've a reference problem; the example should be more descriptive than me :P I have a class that has an array of objects and retrieve them through a key (string), like an associative array: class Collection { public $elements; function __construct() { $this->elements = array(); } public function get_element($key) { foreach($this->elements as $element) { if ($element->key == $key) { return $element; break; } } return null; } public function add_element ($element) { $this->elements[] = $element; } } Then i have an object (generic), with a key and some variables: class Element { public $key; public $another_var; public function __construct($key) { $this->key = $key; $this->another_var = "default"; } } Now, i create my collection: $collection = new Collection(); $collection->add_element(new Element("test1")); $collection->add_element(new Element("test2")); And then i try to change variable of an element contained in my "array": $element = $collection->get_element("test1"); $element->another_var = "random_string"; echo $collection->get_element("test1")->another_var; Ok, the output is random_string so i know that my object is passed to $element in reference mode. But if i do, instead: $element = $collection-get_element("test1"); $element = null; //or $element = new GenericObject(); $element-another_var = "bla"; echo $collection-get_element("test1")-another_var; the output is default like if it lost the reference. So, what's wrong? I have got the references to the variables of the element and not to the element itself? Any ideas?

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  • Unable to parse variable length string separated by delimiter

    - by Technext
    Hi, I have a problem with parsing a string, which consists only of directory path. For ex. My input string is Abc\Program Files\sample\ My output should be Abc//Program Files//sample The script should work for input path of any length i.e., it can contain any no. of subdirectories. (For ex., abc\temp\sample\folder\joe) I have looked for help in many links but to no avail. Looks like FOR command extracts only one whole line or a string (when we use ‘token’ keyword in FOR syntax) but my problem is that I am not aware of the input path length and hence, the no. of tokens. My idea was to use \ as a delimiter and then extract each word before and after it (), and put the words to an output file along with // till we reach the end of the string. I tried implementing the following but it did not work: @echo off FOR /F "delims=\" %%x in (orig.txt) do ( IF NOT %%x == "" echo.%%x//output.txt ) The file orig.txt contains only one line i.e, Abc\Program Files\sample\ The output that I get contains only: Abc// The above output contains blank spaces as well after ‘Abc//’ My desired output should be: Abc//program Files//sample// Can anyone please help me with this? Regards, Technext

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  • Require reasonably random results from an SQL SELECT query within a Joomla article (Cache enabled)

    - by Shrinivas
    Setup: Joomla website on LAMP stack I have a MySQL table containing some records, these are queried by a simple SELECT on the Joomla article, as pasted below. This specific Joomla website has Caching turned on in Joomla's Global Configuration. I need to randomize the order in which I display the resultset, each time the page is loaded. Regular php/mysql would offer me two approaches for this: 1. use 'order by RAND()' or any of a number of methods to allow a SELECT query to return reasonably random results. 2. once php gets the result from the SELECT into an array, shuffle the array to get a reasonably random order of array items. However, as this Joomla instance has Caching turned ON in its Global Configuration, either of the above approaches fails. The first time I load the page the order is randomized, however any further reloads do not cause the order to change, as the page is delivered from cache. The instant the Cache is disabled, both approaches (shuffle/order by rand) work perfectly. What am I missing? How do I override the Global Cache for this specific article? A very simple requirement, that is met by both php and mysql reasonably well, is blocked by the Joomla Cache that I cannot turn off. The php that returns results from the database. <pre> $db = JFactory::getDBO(); $select = "SELECT id FROM jos_mytable;"; //order by RAND() $db->setQuery($select); echo $db->getQuery(); //Show me the Query! $rows = $db->loadObjectList(); //shuffle($rows); foreach($rows as $row) { echo "$row->id"; }

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  • how to pass value to controller??

    - by rajesh
    When I try to pass url value to controller action, action is not getting the required value. I'm sending the value like this: function value(url,id) { alert(url); document.getElementById('rating').innerHTML=id; var params = 'artist='+id; alert(params); // var newurl='http://localhost/songs_full/public/eslresult/ratesong/userid/1/id/27'; var myAjax = new Ajax.Request(newurl,{method: 'post',parameters:params,onComplete: loadResponse}); //var myAjax = new Ajax.Request(url,{method:'POST',parameters:params,onComplete: load}); //alert(myAjax); } function load(http) { alert('success'); } and in the controller I have: public function ratesongAction() { $user=$_POST['rating']; echo $user; $post= $this->getRequest()->getPost(); //echo $post; $ratesongid= $this->_getParam('id'); } But still not getting the result. I am using zend framework.

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  • Why won't the following PDO transaction won't work in PHP?

    - by jfizz
    I am using PHP version 5.4.4, and a MySQL database using InnoDB. I had been using PDO for awhile without utilizing transactions, and everything was working flawlessly. Then, I decided to try to implement transactions, and I keep getting Internal Server Error 500. The following code worked for me (doesn't contain transactions). try { $DB = new PDO('mysql:host=localhost;dbname=database', 'root', 'root'); $DB->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $dbh = $DB->prepare("SELECT * FROM user WHERE username = :test"); $dbh->bindValue(':test', $test, PDO::PARAM_STR); $dbh->execute(); } catch(Exception $e){ $dbh->rollback(); echo "an error has occured"; } Then I attempted to utilize transactions with the following code (which doesn't work). try { $DB = new PDO('mysql:host=localhost;dbname=database', 'root', 'root'); $DB->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $dbh = $DB->beginTransaction(); $dbh->prepare("SELECT * FROM user WHERE username = :test"); $dbh->bindValue(':test', $test, PDO::PARAM_STR); $dbh->execute(); $dbh->commit(); } catch(Exception $e){ $dbh->rollback(); echo "an error has occured"; } When I run the previous code, I get an Internal Server Error 500. Any help would be greatly appreciated! Thanks!

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  • Password protect user image album

    - by Poland Rocks
    Hi Im making i photo sharint site I want to give the ability for my users to prevent the public to acces their albums with a password. Then they can give the password to the ones they want to be able to see it. To password protect their albums. Im thinking something like this, cant test it on this computer, should work ok. but as im a php/mysql beginner i wanna hear what you experts think if theres a better way/approach Albums id name owner password (if it isnt null the album is considered password protected) The code $id = isset($_GET['albumID']) ? intval($_GET['albumID']) : 0; $result = mysql_query("SELECT * FROM albums WHERE id = $id"); $row = mysql_fetch_object($result); // IS it password protected? if ($row->password != NULL) { echo "This album is password protected."; // User pressed "Enter" if (!empty($_POST['password'])) { $result = mysql_query("SELECT password FROM albums WHERE password = '".mysql_real_escape_string($_POST['password'])."'"); // Was It right password? if (mysql_num_rows($result) == 1) { $authed=1; } echo << <form method="post"> <input type="text" name="password" /> EOT; exit; } else $authed=1; if $authed==1 { // render albumimages etc }

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  • What is wrong with this php loop?

    - by Mark R
    I made a loop but it doesn't work. here's what I did: <?php if(is_tree('4')) { ?> <?php $show_after_p = 2; $content = apply_filters('the_content', $post->post_content); if(substr_count($content, '<p>') > $show_after_p) { $contents = explode("</p>", $content); $p_count = 1; foreach($contents as $content) { echo $content; if($p_count == $show_after_p) { ?> YOUR AD CODE GOES HERE < ? } echo "</p>"; $p_count++; } } ?> <?php else : ?> <?php the_content('<p class="serif">Read the rest of this page &raquo;</p>'); ?> <?php endif; } ?> I need to make it work but don't know how. I'm guessing it's a simple syntax error I'm not seeing?

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  • PHP: parse $_FILES[] data in multidimesional array

    - by superUntitled
    Example form here: http://jsfiddle.net/superuntitled/uaTtx/1/ I have a form that allows for dynamic duplication of the form fields. The form allows for file uploads and text input, so the data is sent in both $_POST and $_FILES arrays. The the initial set of inputs look like this: <input type="text" name="question[1][text]" /> <input type="file" name="question[1][file]" /> <input type="text" class="a" name="answer[1][text][]" /> <input type="file" name="answer[1][file][]" /> When duplicated the fields are incremented, they look like this: <input type="text" name="question[2][text]" /> <input type="file" name="question[2][file]" /> <input type="text" class="a" name="answer[2][text][]" /> <input type="file" name="answer[2][file][]" /> To complicate matters, the "answer" form fields can also be duplicated (thus the [] at the end of the 'answer' name array. How can I parse the posted $_FILES array? I have tried something like this: foreach ($_FILES['question'] as $p_num) { echo $p_num['file']['name']; foreach ($_FILES['answer'] as $a_num) { echo $a_num['file']['name']; } } but I get an "Undefined index: file... " error. How can I parse out the posted values.

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  • PHP Dropdown menu [closed]

    - by rShetty
    <br><h2>Select a Tag</h2></br> <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("portal", $con); $query = "SELECT tag_name FROM tags"; $result = mysql_query($query); ?> <select name="tag_name" id="abc"> <option size=30 selected>Select</option> <?php while($array = mysql_fetch_assoc($result)){ ?> <option value ="<?php echo $array['tag_name'];?>"><?php echo $array['tag_name'];?> </option> <?php } ?> </select> <br><br> This is a snippet of code for getting the dropdown menu in the page. I have a database named portal and table named tags with tag_name as the attribute. Do help me to find the error in the program. I am not getting the tag_names in the dropdown menu

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  • How to evaluate json member using variable ?

    - by Miftah
    Hi i've got a problem evaluating json. My goal is to insert json member value to a function variable, take a look at this function func_load_session(svar){ var id = ''; $.getJSON('data/session.php?load='+svar, function(json){ eval('id = json.'+svar); }); return id; } this code i load session from php file that i've store beforehand. i store that session variable using dynamic var. <?php /* * format ?var=[nama_var]&val=[nilai_nama_var] */ $var = $_GET['var']; $val = $_GET['val']; $load = $_GET['load']; session_start(); if($var){ $_SESSION["$var"] = $val; echo "Store SESSION[\"$var\"] = '".$_SESSION["$var"]."'"; }else if($load){ echo $_SESSION["$load"]; } ?> using firebug, i get expected response but i also received error uncaught exception: Syntax error, unrecognized expression: ) pointing at this eval('id = json.'+svar); i wonder how to solve this

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  • 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>'; } }

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  • Problem with RSA-encrypting a string in PHP and passing it to .NET service

    - by jonasaxelsson
    I need to call a web service that requires a login url containing an RSA encrypted, base_64 encoded and url-encoded login data. I don't have a formal php training, but even for me it seems like an easy task, however when calling the service I get an 'Invalid Format' response. What am I doing wrong and is there another way to come up with the encrypted string? Code example below. Thank you for your help! http://www.edsko.net/misc/ for encryption. $message = '?id=112233&param1=hello&[email protected]&name=Name'; $keyLength = '2048'; $exponent = '65537'; $modulus = '837366556729991345239927764652........'; $encryptedData = rsa_encrypt($message, $exponent, $modulus, $keyLength); $data = urlencode(base64_encode($encryptedData)); $loginurl = 'http://www.somedomain.com/LoginWB.aspx?Id=9876&Data='.$data; echo '<iframe src="'.$loginurl.'" width="570px" height="800px">'; echo '</iframe>'; ?

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  • Variable won't store in session

    - by Mittens
    So I'm trying to store the "rank" of a user when they log in to a control panel which displays different options depending on the given rank. I used the same method as I did for storing and displaying the username, which is displayed on the top of each page and works just fine. I can't for the life of me figure out why it won't work for the rank value, but I do know that it is not saving it in the session. Here is the bit that's not working; $username = ($_POST['username']); $password = hash('sha512', $_POST['password']); $dbhost = 'mysql:host=¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦;dbname=¦¦¦¦¦¦¦¦¦¦¦'; $dbuser = '¦¦¦¦¦¦¦¦¦¦¦'; $dbpassword = '¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦'; try { $db = new PDO($dbhost, $dbuser, $dbpassword); $statement = $db->prepare("select password from users where email = :name"); $statement->execute(array(':name' => $username)); $result = $statement->fetch(); $pass = $result[password]; $rank = $result[rank];} catch(PDOException $e) {echo $e->getMessage();} if ($password == $pass) { session_start(); $_SESSION['username'] = $username; $_SESSION['rank'] = $rank; header('Location: http://¦¦¦¦¦¦¦¦¦.ca/manage.php'); } else{ include'../../includes/head.inc'; echo '<h1>Incorrect username or password.</h1>'; include'../../includes/footer.inc'; } I'm also new to the whole PDO thing, hence why my method of authenticating the password is pretty sketchy.

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  • 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?

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  • remove row on click on specific td of that row by jquery

    - by I Like PHP
    i have an table <table class="oldService"> <thead> <th>name</th> <th>age</th> <th>action</th> </thead> <tbody> <?php foreach($array as $k=>$v){ ?> <tr> <td><?php echo $k[name] ?></td> <td><?php echo $k[age]?></td> <td id="<?php $k[id]" class="delme">X</td> </tr> <?php } ?> </tbody> <table> now i want to delete any row by clicking on X of each row except first and last row, and also need to confirm before deletion. i used below jquery <script type="text/javascript"> jQuery(document).ready(function(){ jQuery('table.oldService>tbody tr').not(':first').not(':last').click(function(){ if(confirm('want to delete!')){ jQuery(jQuery(this).addClass('del').fadeTo(400, 0, function() { jQuery(this).remove()})); jQuery.get('deleteService.php', {id:jQuery(this).attr('id')}); } else return false;}); }); </script> this is working perfect,but it execute by click on that row( means any td), i want that this event only occour when user click on X(third td) . please suggest me how to modify this jquery so that the event occur on click of X.

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  • removing a case clause: bash expansion in sed regexp: X='a\.b' ; Y=';;' sed -n '/${X}/,/${Y}/d'

    - by ChrisSM
    I'm trying to remove a case clause from a bash script. The clause will vary, but will always have backslashes as part of the case-match string. I was trying sed but could use awk or a perl one-liner within the bash script. The target of the edit is straightforward, resembles: $cat t.sh case N in a\.b); #[..etc., varies] ;; esac I am running afoul of the variable expansion escaping backslashes, semicolons or both. If I 'eval' I strip my backslash escapes. If I don't, the semi-colons catch me up. So I tried subshell expansion within the sed. This fouls the interpreter as I've written it. More escaping the semi-colons doesn't seem to help. X='a\.b' ; Y=';;' sed -i '/$(echo ${X} | sed -n 's/\\/\\\\/g')/,/$(echo ${Y} | sed -n s/\;/\\;/g')/d t.sh And this: perl -i.bak -ne 'print unless /${X}/ .. /{$Y}/' t.sh # which empties t.sh and eval perl -i.bak -ne \'print unless /${X}/ .. /{$Y}/' t.sh # which does nothing

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  • Using curl to submit/retrieve a forms results

    - by Jason
    I need help in trying to use curl to post data to a page and retrieve the results after the form has been submitted. I created a simple form: <form name="test" method="post" action="form.php"> <input type="text" name="name" size="40" />e <input type="text" name="comment" size="140" /> <input type="submit" name="submit" value="submit" /> </form> In addition, I have php code to handle this form in the same page. All it does is echo back the form values. The curl that I have been using is this: $h = curl_init(); curl_setopt($h, CURLOPT_URL, "path/to/form.php"); curl_setopt($h, CURLOPT_POST, true); curl_setopt($h, CURLOPT_POSTFIELDS, array( 'name' = 'yes', 'comment' = 'no' )); curl_setopt($h, CURLOPT_HEADER, false); curl_setopt($h, CURLOPT_RETURNTRANSFER, 1); $result = curl_exec($h); When I launch the page with the curl code in it, I get the form.php page contents but it doesn't not show the variables that PHP should have echo'd when the form is submitted. would appreciate any help with this. Thanks.

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  • retrieve only sub-pages (wordpress)

    - by Radek
    I want to list all sub-pages only one level though of one particular page. I was reading Function Reference/get pages and thought that $pages = get_pages( array( 'child_of' => $post->ID, 'parent' => $post->ID)) ; will do the trick but it is not working. It lists all pages on the same level like the page I call that code from. If I omit parent option I will get all pages even with sub-pages that I want. But I want only sub-pages. The whole function is like function about_menu(){ if (is_page('about')){ $pages = get_pages( array( 'child_of' => $post->ID, 'parent' => $post->ID)) ; foreach($pages as $page) { ?> <h2><a href="<?php echo get_page_link($page->ID) ?>"><?php echo $page->post_title ?></a></h2> <?php } } } and mine is second one

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  • 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 }

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  • 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"); } ?>

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  • jquery - DOM traversal question

    - by David
    Hi, Basically what I have here is a button that adds a new list item to an unordered list. Within this list item is a dropdown box (.formSelector) that when changed triggers an ajax call to get the values for another dropdown box (.fieldSelector) in the same list item. The problem is on this line: $(this).closest(".fieldSelector").append( If I change it to $(".fieldSelector").append( the second box updates correctly. The problem with this is that when I have multiple list items it updates every dropdown box on the page with the class fieldSelector, which is undesirable. How can I traverse the DOM to just pick the fieldSelector in the current list item? Any help is greatly appreciated. Complete code: $("button", "#newReportElement").click(function() { $("#reportElements").append("<li class=\"ui-state-default\"> <span class=\"test ui-icon ui-icon-arrowthick-2-n-s \"></span><span class=\"ui-icon ui-icon-trash deleteRange \"></span> <select name=\"form[]\" class=\"formSelector\"><? foreach ($forms->result() as $row) { echo "<option value='$row->formId'>$row->name</option>"; }?></select><select class=\"fieldSelector\" name=\"field[]\"></select></li>"); $(".deleteRange").button(); $('.formSelector').change(function() { var id = $(this).val(); var url = "<? echo site_url("report/formfields");?>" + "/" + id; var atext = "ids:"; $.getJSON(url, function(data){ $.each(data.items, function(i,item){ $(this).closest(".fieldSelector").append( $('<option></option>').val(item.formId).html(item.formLabel) ); }); }); }); return false; });

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  • 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; } ?>

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  • 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?

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  • 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>

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