Actually I guess the question is duplicate, but I didn't understand whats the WRONG with my SQL code. My error is
Notice: Undefined index: CI in
C:\wamp\www\LOCATIONVIEWER\exampleDB.php on line 30.
I need to convert all decimal value of column into hexadecimal value. Help me to do it... :)
Here is my code:
<html>
<head>
<title>test</title>
</head>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "locationviewer";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//CI is a column name of the table
$sql = "SELECT CONV(CI,10,16) FROM locationdata";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "CI: " . $row["CI"]."<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
You should change following code:
$sql = "SELECT CONV(CI,10,16) FROM locationdata";
To
$sql = "SELECT CONV(CI,10,16) AS `CI` FROM locationdata";
this issue happened because php can not see CI index and current index is somthing like CONV(CI,10,16)
Related
I got this work for me, but I'm sure there's a better way to get this done. But, I've searched many hours without finding the exact answer to what I'm looking to do. Basically getting the variable usrID from the URL, I need to search MySQL for the corresponding information to this user. Later I want to use the different fields on my page (better website) to personalize the experience.
<?php
$servername = "localhost";
$username = "authorized-user";
$password = "secret";
$dbname = "agentDB";
$usrID = "001";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM agentInfo WHERE usrID = '$usrID'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$Lname = $row["Lname"];
$Fname = $row["Fname"];
$tl = $row["tl"];
}
}
mysqli_close($conn);
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Load MySQL Data into Corresponding PHP Variables</title>
</head>
<body>
here is the body<br>
My name is: <?php echo $Fname; ?> <?php echo $Lname; ?><?php echo $tl; ?>
</body>
</html>
You could create a variable to store a full name and then "tl" on it like this:
$user_info = $Lname . ", " . $Fname . ": " . $tl;
Then:
<?php echo $user_info; ?>
Wherever you need that information.
If you want to minimize the amount of variables being assigned you could wrap it in a function and return the desired data field:
function fetchUserData(userData) {
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM agentInfo WHERE usrID = '$usrID'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$userData = $row[userData];
}
}
return $userData;
}
mysqli_close($conn);
You can the get the specified data like this:
<?php echo fetchUserData("Fname"); ?>
After I add a column to my database, I want to retrieve it but not expected.
In PHP, I try reopening apache and mysql still not work.
Does anyone know how to resolve it? Thanks!
your question is not fully explanatory but with what I could try to understand you want to retrieve data or records from your database
you could try the code below and tweak it to work your way
<?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 databaseName";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data
while($row = $result->fetch_assoc()) {
print $row"<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
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
I have a fairly simple PHP page which displays some fields from a database.
For some reason I get:
Notice: Undefined index: entries.id in /var/www/html/originalprices.php on line 24
I can't see whats wrong, could anyone help? Thanks
<?php
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "zxdb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT entries.id, entries.title, entrytypes.text,entries.original_price, entries.budget_price,labels.name FROM entries,publishers, labels, entrytypes
where entries.entrytype_id = entrytypes.id
and publishers.entry_id = entries.id
and publishers.label_id = labels.id
and labels.id = '1371'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["entries.id"]. " - Name: " . $row["entries.title"]. " " . $row["entrytypes.text"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The Name of the returned field will be "id" not entries.id.
You can check that by using var_dump()
I have just set-up LAMP and am trying to query a mysql database and write the results on screen. The issue I have is that instead of writing the results of my query onscreen, my syntax is written on screen instead?
What is the proper way of querying a mysql database with php and writing the results on screen?
This is my syntax that does not work?
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "Select state from PHPTests.state";
$result = $conn->query($sql);
if ($result->num_rows > 0)
{
while($row = $result->fetch_assoc())
{
echo $row['state'];
}
}
else
{
echo "0 results";
}
$conn->close();
?>
<html>
<body>
<h4>Hello - This is test site with php</h4>
</body>
</html>