You can combine these into a single query by joining with a subquery that sets the variable. SELECT CASE WHEN id % 2 == 0 THEN id – 1 WHEN id = @maxid THEN id ELSE id + 1 END AS id, student FROM Seat CROSS JOIN (SELECT @maxid := MAX(id) FROM Seat) ORDER BY … Read more
This type mismatch error which means that you provide “String value ” but needs of integer type value for more information you refere this link Type ‘String’ is not a subtype of type ‘int’ of ‘index’ solved what should I do flutter. _TypeError (type ‘String’ is not a subtype of type ‘int’ of ‘index’) [closed]
As you are trying to add all items to the query you will need to remove the [] from the implode(“,”, $array[]) so it reads implode(“,”, $array). You could try adding echo “INSERT INTO Product_img_col (image, color, pro_ID) VALUES ‘.implode(“,”, $array[])”); to your page so you have the query that you can test in phpMyAdmin … Read more
You’ll have to use css/javascript to do this. There are 3 ways: The more easy way is to print all data, but just not displaying it yet until someone clicks the link. For the hyperlink part. In this example I used a ‘span’ element which is styled like a hyperlink. //place inside your head tag … Read more
Your SQL should look like this: SELECT Breed.Breed, Gender.Sex, Dog.Name FROM Dog JOIN Gender ON Dog.GenderID = Gender.ID JOIN Breed ON Gender.BreedID = Breed.ID WHERE Dog.ID = ‘YourDogIDHere’ And it would return the following: Breed | Sex | Name ——–|——–|——- Collie | Male | Fluffy Collie | Female | Holey Edit: Edited to include breed and … Read more
NOTE: Your question is not clear. You have not included any code or query snippet. I will try my best to answer your question. As per my knowledge I think your users are nothing but the members. Just write a query to pull all the users from the users table and display. NOTE : I … Read more
Sure, just use javascript variables. var ajaxMethod = “POST”; var ajaxUrl = “save.php”; $.ajax({ type: ajaxMethod, url: ajaxUrl, data: dataString, success: function(chat_list) { } }); solved Two URLs with same AJAX structure
select author_id (Select count(book_id) from mytable group by author_id) as booksPerAuthor from mytable where booksPerAuthor > 1; 2 solved How to count a number of duplicate entries in a group using a subquery?
Make sure to add checks when making your connection as well as after the query $server=”127.0.0.1″; $username=”root”; $password = ”; $database=”test”; mysql_connect($server, $username, $password) or die(mysql_error()); mysql_select_db($database) or die(mysql_error()); $TableName = “opportunities”; $Opportunities = array(); $SQLString = “SELECT opportunity_ID, company, city, ” . “start_date, end_date, position, description” . ” FROM $TableName;”; $QueryResult = mysql_query($SQLString); if(mysql_error()) … Read more
The standard way to do this is to send it from the form to your API in cleartext (but using HTTPS, this is essential), then hash it on your backend (seperate issue how to do this), and finally save that hash in the MYSQL database. You do not want to encrypt passwords, you want to … Read more
Could you use a regular expression to match the URL? You didn’t include the query… /^(https?:\/\/)?([\da-z\.-]+)\.([a-z\.]{2,6})([\/\w \.-]*)*\/?$/ The above would be a regular expression to match a URL including the https and double forward slash Numbers, letters, dots, and hyphens are included here. And in MySQL you could use this to match it: http://dev.mysql.com/doc/refman/5.1/en/regexp.html#operator_regexp solved … Read more
You have many errors in your code : $result = mysql_query(“SELECT * FROM utilizatori WHERE username=”” . $loginUsername . “” and parola=””. $loginUsername.”””); You check username and parola on same var. You probably want : $result = mysql_query(“SELECT * FROM utilizatori WHERE username=”” . $loginUsername . “” and parola=””. $loginPassword.”””); You are also affecting vars … Read more
response_id,participant_id,question_id,answer_option_answer_text vs (response_id,participant_id,question_id,answer_option) and your line ” $participant_id=$_POST[‘participant_id’] ” is missing its semicolon (;) also please use if( !mysql_query( $query, $con ) ) { echo(“Failure: ” . mysql_error() ); } to insert data. 2 solved Data inserting in mysql data base not working [closed]
you have to remove the () from values , or it will be considered as one entry . try that: INSERT INTO evraklar(evrak_tipi_grubu, evrak_tipi, evrak_konu, evrak_subeye_gelis_tarihi, evrak_gonderen, evrak_alici, evrak_tarihi, evrak_sayisi, evrak_aciklama, evrak_kurum_icindenmi, gelen_evrak_tarihi, gelen_evrak_sayi, gelen_evrak_etakip, evrak_kaydeden_ybs_kullanici_id,kaydeden_kullanici_birim_id) VALUES (6,43,’Test amaçlı girilen konu’,STR_TO_DATE(’12/12/2012′, ‘%d/%m/%Y’),0,566,STR_TO_DATE(’12/12/2012′, ‘%d/%m/%Y’),’5555555555′,’YBS:110 nolu evrağa cevaben yazılan yazıdır. Alıcısı:Antalya Valiliği – İl Sağlık Müdürlüğü’,0,STR_TO_DATE(’12/12/2012′, ‘%d/%m/%Y’),’5555555555′,777777777,1,685), … Read more
should be if ($insert_stmt = $mysqli->prepare(“INSERT INTO members (username, email, password, salt, privilegio) VALUES (?,?,?,?,?)”) { $insert_stmt->bind_param(‘sssss’, $username, $email, $password, $random_salt, $privilegio); // Execute the prepared query. $insert_stmt->execute(); } solved Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn’t match number of parameters in prepare [closed]