[ad_1] Based on the link you’ve shared, if you want a popup over any desktop application through the browser, use the W3C Notification API. For example window.onload = function () { if (Notification.permission !== “granted”) Notification.requestPermission(); }; function notifyMe() { if (!Notification) { alert(‘Desktop notifications not available in your browser. Try Chromium.’); return; } if … Read more
[ad_1] Do it via .htaccess, as @Droid said. All you have to do is to replace file.php with the unique file you want to allow and add more of them if you want so. Order deny,allow Deny from all <Files “file.php”> Allow from all </Files> 1 [ad_2] solved Restrict whole WP site access except few … Read more
[ad_1] Use two variables to store the positioncolumn values of ids having 1 and 2. Then use a CASE expression to update the table accordingly. Query set @a := (select `position` from `your_table_name` where id = 1); set @b := (select `position` from `your_table_name` where id = 2); update `your_table_name` set `position` = ( case … Read more
[ad_1] If you create the variable in config.php then it is loaded when you include the file in index.php like this: config.php <?php $config = [ ‘key1’ => ‘value1’, ‘key2’ => [ ‘subkey1’ => ‘value3’ ], ‘key3’ => ‘value4’ ]; ?> index.php include “config.php”; $value4 = $config[‘key3’]; // ‘value4’ made some assumptions because your question … Read more
[ad_1] It seems like you want to encode your PHP array to JSON. Use json_encode() for that. json_encode($array); Returns a string containing the JSON representation of value. [ad_2] solved How to change data php to javascript json [closed]
[ad_1] is this what you are looking for? foreach([100, 500, 543, 1000, 5000, 51000, 500000] as $my_friends) echo ‘5 in ‘. getScoreOf($my_friends) . “<br>”; function getScoreOf($my_friends){ $of = 5; $total = 5e5; //that’s 500,000 😉 $step = 100; //minimum step, so output is not “4604” but “4600” $out_of = $total / $my_friends * $of; return … Read more
[ad_1] As you can already find in the other answer, the bar is called the debug bar. You will not find this in the twig templates. Instead you can remove it by disabling it in your configuration. This setting is in app/config/config_dev.yml: web_profiler: toolbar: true intercept_redirects: false [ad_2] solved Symfony2 default installation keeps on adding … Read more
[ad_1] You can use the download attribute to force download and JS to automatically click that link. document.getElementById(‘download’).click() <a href=”https://scontent.cdninstagram.com/t51.2885-15/s640x640/sh0.08/e35/23279358_143343893084349_1681002671546302464_n.jpg” download id=”download”>Download begins</a> 3 [ad_2] solved How can I download pictures from a URL to a local computer without asking permission?
[ad_1] require_once() is a statement that imports another php script file in another one. Your warning and Error syas that : the “DB.PHP” file dose not exist in the path specified . that import occured in config.php file . try find db.php file and put in the paths specified by error description. 1 [ad_2] solved … Read more
[ad_1] Try this… $conn = mysql_connect(db_server, db_user, db_pass); if ($conn !== false) { mysql_select_db(db_name); $result = mysql_query(“SELECT *, COUNT(track.usr_id) AS usr_views FROM user, track WHERE track.usr_id = $usr_id AND track.timesin = $les_tag GROUP BY $usr_id ORDER BY usr_views LIMIT 1”); if ($result !== false && mysql_num_rows($result) > 0) { $usercheck = mysql_fetch_row($result); $Userstats = $usercheck; … Read more
[ad_1] $db is usually for database connection. It provides connection link to database, identifies the connection and provides database object. When you use mysql_* functions, you can define last parameter as connection variable. You use it to indentify connection, so multiple mysql connections don’t mix up. [ad_2] solved What can I do with the $db … Read more
[ad_1] Heres my suggestion for you. Hopefully this makes sense and works for you. Please try to understand what is happening rather than just copy and hope it works. Your form, on submit, gets sent to the JS file. This grabs the form values, uses JSON to send them to your php file which processes … Read more
[ad_1] These links may help. Submit a form with Ajax and jQuery: http://net.tutsplus.com/tutorials/javascript-ajax/submit-a-form-without-page-refresh-using-jquery/ Validate a form: http://jqueryvalidation.org [ad_2] solved Submit a Form with Validations using AJAX and PHP [closed]
[ad_1] Generally you should do it the following way: Attach an event to the button just as @Marc B suggested In the function for the event do a jQuery.ajax request, sending the users name (type to POST would be a good idea and the users name inside data) On the server create a PHP file … Read more
[ad_1] $fields = array_keys($_POST); if (!empty($fields)) { $names = implode(‘`, `’, $fields); $values = implode(‘, :’, $fields); $sql = “INSERT INTO customers ( `”.$names.”` ) VALUES ( :”.$values.” )”; print_r($sql); $addRandom = $pdo->prepare($sql); foreach ($fields as $field) { $addRandom->bindValue(“:{$field}”, $_POST[$field]); } $boolean = $addRandom->execute(); if ($boolean){ echo ‘INSERTED’; } else { echo ‘FAILED’; } } … Read more