You haven’t got anything in your template called blogposts. You need to use keyword arguments to pass the data: return render_template(‘index.html’, blogposts=blogposts) Also note you should really do that query inside the function, otherwise it will only ever execute on process start and you’ll always have the same three posts. 4 solved Flask App will … Read more
mysqli_query does not return your number directly. It returns a mysqli_result as you can see here. To get the row you have parsed, you should fetch it first: $row = mysqli_fetch_assoc($result) A lot of information on using mysqli can be found here. 6 solved Object of class mysqli_result could not be converted to int – … Read more
Try this, It should must work. SELECT name, salary FROM employees WHERE salary = (SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees)) Or you can try this one as well. /* looking for 2nd highest salary — notice the ‘=2’ */ SELECT name,salary FROM employees WHERE salary = (SELECT DISTINCT(salary) FROM employees … Read more
You can try to force it to be a start of a string or start of a word: $sql = “SELECT * FROM articles WHERE title LIKE ‘% “.$term.”%’ OR title LIKE ‘”.$term.”%'”; 5 solved MySQL search is too “loose” [closed]
You are messing up your single and double quotes. Try this: <?php include(“db_config.php”); $sql=”SELECT * FROM brojevi”; $result=mysqli_query($connection,$sql); echo ‘<select name=”dropdown” size=”1″>’; echo ‘<option value=”choose”>-choose-</option>’; while($row=mysqli_fetch_array($result)){ $id = $row[‘id’]; $broj = $row[‘brojevi’]; echo ‘<option value=”‘ . $id . ‘”>’ . $broj . ‘</option>’; } echo ‘</select>’; ?> When using html tags inside your echo, try … Read more
Use implode(). $yourArray = array_map(“mysql_real_escape_string”, $yourArray); $query = “SELECT * FROM user_detail WHERE user_id='”; $query .= implode($yourArray, “‘ OR user_id='”); $query .= “‘”; Or indeed, use the SQL IN keyword: $yourArray = array_map(“mysql_real_escape_string”, $yourArray); $query = “SELECT * FROM user_detail WHERE user_id IN (‘”; $query .= implode($yourArray, “‘,'”); $query .= “‘)”; 0 solved Mysql Query … Read more
The error: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given is almost always because you’ve tried to execute a query and it’s failed for some reason, but you’ve continued on blindly anyway. Upon failure, mysqli_query will return false rather than a mysqli_result and, if you then attempt to use that boolean false value … Read more
Your query is badly formed. Try $query = mysql_query(“SELECT * FROM users WHERE email=”” . $username . “” AND password='” . $password . “‘);” Also, note that the mysql_ functions are deprecated. Update your code to mysqli or PDO. 1 solved Log in Script not returning the values from MYSQL [closed]
Check documentation of ResultSet.getString:http://docs.oracle.com/javase/7/docs/api/java/sql/ResultSet.html#getString%28int%29 String getString(int columnIndex) throws SQLException Returns: the column value; if the value is SQL NULL, the value returned is null According to this your code is vulnerable to null pointer exception at following places: if (!results.getString(“level”).isEmpty()) { ……. } else if (!results.getString(“thesisTitle”).isEmpty() || !results.getString(“thesisAdvisor”).isEmpty()) { ………. } else if (!results.getString(“company”).isEmpty()) { … Read more
Assuming each of the articles has its ID. Change the link to go to a dynamic page, passing that ID: “<div class=\”title\”><a href=\”dynamic_page.php?id=$result[id]\”>$resultphp dynamically generate new web page from link [closed]</a></div>” Then create a dynamic_page.php that accepts that ID and generates the article as follows: if (isset($_GET[‘id’])) { $id = mysql_real_escape_string($_GET[‘id’]); $q = “SELECT * … Read more
Do you mean something like that? SELECT * From ProviderDxCptCodes WHERE CodeType=”CPT” AND name LIKE ‘%flu%’; 4 solved How to get rows of a specific type matching a specific value?
you ca us this code $mat=$this->Materiel->Paiement->query(” select *,materiel_id , SUM(paiements.quantite) as t from paiements ,materiels where paiements.materiel_id=materiels.id GROUP BY materiel_id ORDER BY sum(paiements.quantite) desc “); Ajust this code for your model 1 solved execute query with cakephp [closed]
Update your 3rd query to – SELECT Patient.Name, Patient.age from Patient where patient.age < ’50’ and EXISTS ( SELECT Prescription.DoctorID from Prescription where Patient.Primary_DoctorID = Prescription.DoctorID and Prescription.P_ID = ( select Prescription_Medicine.P_ID FROM Prescription_Medicine where Prescription_Medicine.Tradename=”Vitamin” ) ); And update your 4th query to – SELECT dname FROM doctor d JOIN (SELECT speciality, MAX(doctor.experience) FROM … Read more
CheckLogin.php if ( $count == 1 ) { $_SESSION[‘login_id’] = $row[‘id’]; $_SESSION[‘username’] = $row[‘username’]; // added if ( $_SESSION[‘login_id’] != ” || $_SESSION[‘login_id’] > 0 ) { // edited header(“location: index.php”); } else { header(“location: login3.html”); } } Index.php <?php require_once ‘UserSessionAdmin.php’; // edited $username = $_SESSION[‘username’]; // added ?> <!DOCTYPE html PUBLIC “-//W3C//DTD XHTML … Read more
Add value to select box options echo “<option value=”$id”>$id</option>”; 7 solved Failed syntax UPDATE id in MariaDB