[ad_1] For PHP you can do something like this: $string = “one two three four five six seven eight nine One Two Three Four Five Six Seven Eight Nine oNE tWO tHREE fOuR fIVE sIX sEVEN eIGHT nINE ONE TWO THREE FOuR FIVE SIX SEVEN EIGHT NINE”; $replace = array(0,1,2,3,4,5,6,7,8,9); $search = array(‘zero’, ‘one’, ‘two’, … Read more
[ad_1] If you’re adamant you that you don’t want to use PHP, you can always do something like this: (It’s not really considered best practice) <form action=”mailto:[email protected]?Subject=Signup%20Info” enctype=”text/plain” method=”post”> <p><label for=”firstName”>First Name:</label><p> <p><input type=”text” name=”firstName” id=”firstName” required><p><br> <p><label for=”lastName”>Last Name:</label><p> <p><input type=”text” name=”lastName” id=”lastName” required><p><br> <input type=”submit” value=”Submit”> </form> Obviously, the mailto will redirect your … Read more
[ad_1] First of all, paste your code correctly… You are missing </table> and </form> in the end… Second of all, you were missing a simple apostrophe and a semi-colon in the line $result = mysquli_query($con, ‘select * from account where username=”‘.$username.'” and password=”‘.$password.'”) Also, you should use mysqli_query instead of mysqlui_query… Typo! Your code should … Read more
[ad_1] Try by this if($_POST[‘grade’] == ‘G1’ && $_POST[‘district’] == ‘D1’){ //Both Founded }else{ //Not Founded } 3 [ad_2] solved 2 declaration of $_POST not working [closed]
[ad_1] Your SQL query/logic is completely wrong. What you should do is check if that user and password combination exits in database using WHERE clause. But actually you are doing is fetching each row and checking for equality. In such situation you can also get n numbers of wrong credentials. $query = mysql_query(“SELECT * FROM … Read more
[ad_1] Replace your code with this . <?php try{ $db = mysqli_connect (‘localhost’, ‘root’, ”, ‘car_rental’) or die (“SQL is Off”); } catch (Exception $e){ echo “SQL is Off”; exit; } echo “success”; $email = $_POST[“email”]; $pass = $_POST[“pass”]; mysqli_select_db($db,”car_rental”); $result = mysqli_query($db,”SELECT email, users FROM users WHERE email =$email”); //$row = mysql_fetch_array($result); $row = … Read more
[ad_1] You are missing the connection parameter to sqlsrv_query. While you’re at it, you might as well use bind variables instead of string substitution to help guard against SQL-Injection attacks. $serverName = “serverName\instancename”; $connectionInfo = array( “Database”=>”dbName”, “UID”=>”username”, “PWD”=>”password” ); $conn = sqlsrv_connect( $serverName, $connectionInfo); $params = array($username, $password); $stmt = sqlsrv_query ($conn, ‘select * … Read more
[ad_1] $url=”http://example.com”; $html = file_get_contents($url); $doc = new DOMDocument(); @$doc->loadHTML($html); $tags = $doc->getElementsByTagName(‘img’); foreach ($tags as $tag) { echo $tag->getAttribute(‘src’); } Check this code its working for me. [ad_2] solved How to get image from any website with url
[ad_1] If I’ve got your question then you might be looking for the following thing $array = array(‘one’ => Array ( ‘count’ => 2 ), ‘two’ => Array ( ‘count’ => 2 ), ‘three’ => Array ( ‘count’ => 2 )); foreach($array as $key => $value){ echo “<span class=”red”> $key <em>({$value[‘count’]})</em></span><br/>”; } 2 [ad_2] solved … Read more
[ad_1] Put the script after the form tag. It searches dateForm id and until that no form is in the output so it does nothing. When you place that after it, it’ll search the page for that id and it finds that and submits. <?Php if($something):?> <form id=”dateForm” action=”https://www.paypal.com/cgi-bin/webscr” method=”POST”> <input type=”hidden” name=”test” value=”test”> <input … Read more
[ad_1] Try This $i = 0; $j = 1; echo $i; $flag = 1; $inc = 0; for($k = 0; $k <= 10; $k++) { $inc++; if($flag == 0){ echo $i; if($inc == 2){ $flag = 1; $inc = 0; } }elseif($flag == 1){ echo $j; //$flag = 0; if($inc == 2){ $flag = 0; … Read more
[ad_1] Side note, this is bad code: in_array(isset($weekendArr) && $weekendArr,$arr) do it like isset($weekendArr) && in_array($weekendArr,$arr) and in_array is not strict so this in_array(true,array(‘w’,’s’)) will be allways TRUE do it with: in_array(true,array(‘w’,’s’),true) and you see. And you can’t check an array with an array the $needle be an STRING here. The only solution is to … Read more
[ad_1] Assuming your PHP script is producing correct JSON. Try changing this part JSONArray jArray = new JSONArray(result); to this. JSONObject jObject = new JSONObject(result); JSONArray jArray = jObject.getJSONArray(“NomTableau”); 2 [ad_2] solved org.json.JSONException: Value Exception of type java.lang.String cannot be converted to JSONArray
[ad_1] file() by default will keep the line break/feed characters. So instead of I, i.e. chr(73), you have three characters probably chr(73)chr(13)chr(10) or chr(73)chr(10)chr(13). But there is a flag which you can pass as second parameter to have these characters removed (and also ignore otherwise empty lines as well). $test = file(‘test.txt’, FILE_IGNORE_NEW_LINES|FILE_SKIP_EMPTY_LINES ); 2 … Read more
[ad_1] PHP is processed on the server, before sending the result to the client. For example, if you do this: <p> <?php $name = “Dan”; print “hello, $name, “; ?> How are you? </p> The client will get: <p>hello, Dan, How are you?</p> And then, the client will process any javascript included in the code. … Read more