is it possible to have 2 different sql statement in php? i mean like for example
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "tsukishiro";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT DISTINCT student_no FROM grades_tbl2 WHERE student_no ='C2012-02918'";
$sql = "SELECT * FROM grades_tbl2 WHERE student_no ='C2012-02918' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<tr><td>".$row["student_no"]."</td><td>".$row["last_name"]." ".$row["first_name"]." ".$row["middle_name"]."</td><td>".$row["subject_code"]."</td><td>" .$row["subject_desc"]."</td><td>".$row["trans_final"]."</td><td>".$row["remarks"]."</td><td>".$row["subject_prof"]."</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
so i want to separate the distinct statements so that the other sql statement does not distinct like the first one. can you help me? so new for this
Related
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "job_database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['Get']))
{
$jtitle = $_POST['jobtitle'];
$location = $_POST['location'];
$category = $_POST['categories'];
//Query specified database for value
$sql = "SELECT * FROM addjob where jtitle ='$jtitle' ∧ location ='$location' ∧ category ='$category' " ;
$result = $conn->query($sql);
}
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "jtitle: " . $row["jtitle"]. "location:" . $row["location"]. "category" . $row["category"]."<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Change your sql query as follows,
$sql = "SELECT * FROM addjob WHERE jtitle =".$jtitle." AND location =".$location." AND category =".$category." ;
since ∧ is wrong, what it does is, it ends the SQL query because there is ; in the end. So change ∧ into and and check.
And read more about form validation and SQL injection from these links.
Good luck! =)
What I wan't to do is create buttons that are automatically generated from the database. So when I add a new record in the database the button is created Is this possible with a loop? So yes how do I create the button.
This is what I have so far:
<?php
$servername = "localhost";
$username = "root";
$password = "Iamthebest1009";
$dbname = "dktp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM theme";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "". $row["theme_name"]. "<br>";
}
} else {
echo "no results";
}
$conn->close();
?>
Yes it is possible. you need to echo html
<?php
$servername = "localhost";
$username = "root";
$password = "Iamthebest1009";
$dbname = "dktp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM theme";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$your_url ="https://www.google.com";
echo "". $row["theme_name"]. "<br>";
echo '<input type="button" name="' . $row["theme_name"]. '" value="'. $row["theme_name"].'">';
}
} else {
echo "no results";
}
$conn->close();
?>
I would need some help with showing data that I have on my database but I can't seen to be able to.`
$servername = "servername";
$username = "username";
$password = "password";
$dbname = "dbname";
$connect = mysqli_connect($servername, $username, $password, $dbname) or die ("connection failed");
//Query
$query = "SELECT * FROM 'Students'";
mysqli_master_query($dbname, $query) or die ("Error while Query");
$result = mysqli_master_query($dbname, $query);
$row = mysql_fetch_array($result);
while ($row = mysql_fetch_array($result)) {
echo "<p>".$row['Name']."</p>";
};
mysql_close($connect);
?>`
I am pretty new to this so I could have missed something simple. Any help appreciated.
Below is a sample code of the normal procedure to connect to a database and to select data from it. Please follow this type of coding since MySQL is now deprecated and MySQLi is used.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
For further reference check out http://php.net/manual/en/book.mysqli.php and also https://www.w3schools.com/php/php_mysql_insert.asp
This question already has answers here:
Reference - What does this error mean in PHP?
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Closed 6 years ago.
<?php
$servername = "localhost";
$username = "user";
$password = "pass";
$dbname = "test";
$tablename = "mapcoords";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
{
echo "Failure";
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql = "SELECT (lat, lng) FROM mapcoords";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
{
echo "ok";
}
$conn->close();
?>
Here is the code. So like I said, it can connect successfully, but the code won't successfully query. What's weird is that if I copy and paste the same code, which seems to be EXACTLY the same, it works. It makes no sense. I can't find a single difference between their code and my code besides the way they space things and the way I space things. Here is their code:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT (lat, lng) FROM mapcoords";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row["lat"]. " " . $row["lng"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The problem is that this query:
SELECT (lat, lng) FROM mapcoords
returns the folloowing error:
[21000][1241] Operand should contain 1 column(s)
You have to change the query to
SELECT lat, lng FROM mapcoords
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
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Using the code below I am trying to import users table into my php page and only want to show the last ten entries in a table, what else do I need to add to my code to achieve this
<?php
$servername = "localhost";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM dispenses";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "Amount: " . $row["amount"];
}
} else {
echo "0 results";
}
$conn->close();
?>
$query = "SELECT * FROM 'users'";
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM users";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"];
}
} else {
echo "0 results";
}
$conn->close();
?>
This should help:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
http://www.w3schools.com/php/php_mysql_select.asp