If you are looking to search, you need to use the LIKE syntax, not WHERE $sql=”SELECT * FROM members WHERE FirstName LIKE ‘”. $fname .”%'”; The LIKE & % make it a wildcard. Now you said you want to search by first and last, but you only pass in your first variable and you only … Read more
Try this, it is simple SELECT id FROM `table_name` WHERE is_parent= 0 1 solved How to get the numbers of a column in phpmyadmin? [closed]
Go to the config.inc.php file In the file, search for the line $cfg[‘Servers’][$i][‘password’] and change it to $cfg[‘Servers’][$i][‘password’]=’root’ Hope it helps 1 solved Error : (1045) Access denied for user ‘root’@’localhost’ (using password: NO)
Error: you have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use solved Error: you have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use
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]
after you sort out the connection you might want to fix your table echo “<table border=”1″>”; while($row=mysqli_fetch_array($result)){ $col_name = $row[‘Field’]; $click = “<a href=”https://stackoverflow.com/questions/43335024/Column_details.php?mv=”.$col_name.””>” .$col_name. “</a>”; echo “<tr>”; echo “<td>” . $col_name . “</td>”; echo “<td>” . $click . “</td>”; } echo “</table>”; 4 solved sql query not executing on local host
$query = mysql_query(“select `subject` from `markssub1` where `sname`=’$user’ AND `sid`=’$id'”); $subjects = []; while($row = mysql_fetch_array($query)) { $subjects[] = $row[‘subject’]; } foreach($subjects as $subject){ echo $subject . ‘\n’; } 3 solved Can I store query result of one field into different variables? if not then what should I do for that?
There are certain free database hosting provider which provides remote access to their database. Remote access here signifies all the requests made to the database are originated from their own hosting server (eg, using their hosting service for php, jsp etc) or any other hosting service provider. But if remote access is not allowed means … Read more
You have forgotten to close the } for the if clause above. Try closing it after HTML. <?php $i=$i+1; }?> </table> </body> </html> <?php } ?> 2 solved parse error in PHP page [duplicate]
You should better a try var_dump($row) or print_r($row) first to see if the array contains the datas at right keys. $lang[$row[0]] doesnt work because $row[0] is empty. Your are probably not assigning Hello to $row[0]. So try print_r($row); to see whats stored in the whole array. Add : $lang[$row[0]] will give you $lang[hello]. It should … Read more
The design of your table is going to depend a lot on the requirements you need fulfilled. In the situation where a member can only ever be assigned to a single room, but rooms can have many members, then the answer that Double H gave is exactly what you’re looking for: CREATE TABLE members ( … Read more
Depending upon what hashing you are using to store the password, your model confitions would be something like this: $this->db->select(‘*’)->from(‘membership’); $this->db->where(‘username’, $username); $this->db->where(‘password’, MD5($password)); Also, there is a bug in your model for checking if a user is valid. If number of rows returned from DB is 1, then the user is valid, otherwise not. … Read more
In addition to the supplied answer(s), I would like to suggest doin the following: // Before anything else, start your session if (!isset($_SESSION)) { session_start(); } // Next, check if your form is actually submitted if ($_SERVER[‘REQUEST_METHOD’] == ‘POST’) { if (isset($_POST[‘…’])) { } elseif (isset($_POST[‘…’])) { } } ?> <!– Keep PHP and html … Read more
I am trying to create a sql table in phpmyadmin, I’m getting 4 errors. This code works in phpmyadmin on my pc at my work but not on the pc at home [closed] solved I am trying to create a sql table in phpmyadmin, I’m getting 4 errors. This code works in phpmyadmin on my … Read more
open “config.inc.php” on your PMA folder. add this line to prompt the authentication before login $cfg[‘Servers’][$i][‘auth_type’] = ‘cookie’; /* $cfg[‘Servers’][$i][‘user’] = ‘root’; */ /* $cfg[‘Servers’][$i][‘password’] = ”; */ OR manually define your authentication by this line $cfg[‘Servers’][$i][‘auth_type’] = ‘config’; $cfg[‘Servers’][$i][‘user’] = ‘root’; $cfg[‘Servers’][$i][‘password’] = ‘yourpassword’; solved Access denied with localhost [closed]