php

[Solved] PHP – How do I convert string number to number [closed]

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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’, ‘three’, … Read more

[Solved] Simple HTML/CSS website requires a form content placement in an email message

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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 user … Read more

[Solved] PHP website debug [duplicate]

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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 be … Read more

[Solved] Sign in form always showing “Wrong Credentials”

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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 table … Read more

[Solved] mysql_select_db() expects parameter 2 to be resource, object given PHP validation

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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 = $result->fetch_array(MYSQLI_ASSOC); … Read more

[Solved] How to convert mysql function into sqlsrv? [closed]

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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 * from … Read more

[Solved] How to get image from any website with url

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$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. solved How to get image from any website with url

[Solved] PHP insert Tag CSS in arrays [closed]

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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 solved PHP insert … Read more

[Solved] PHP form does not submit [closed]

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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 type=”hidden” … Read more

[Solved] Wrong output in php program

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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 do … Read more

[Solved] org.json.JSONException: Value Exception of type java.lang.String cannot be converted to JSONArray

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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 solved org.json.JSONException: Value Exception of type java.lang.String cannot be converted to JSONArray

[Solved] if statement always false even if the value comparing must produce true result

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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 solved … Read more