How to write PHP to Mysql page - php

Currently I just done HTML and I trying writing PHP Script on connection Mysql, But all web site on google is confuse, because now I need specify write connection by used PHP Only,
Please could you help explain step by step on code from HTML to PHP and write PHP to MySql, Thank you very much.

There is many docs about this, but you can start from here:
http://www.w3schools.com/php/php_mysql_select.asp
And here is the example code, what i suggest you:
<?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);
?>

Related

Always getting blank page while using php and mysql connections [duplicate]

This question already exists:
PHP's white screen of death [duplicate]
Closed 5 years ago.
I am always getting blank page when using Php and mysql connections
this is my php file
this is my html file
Try this code instead:
<?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();
?>

Need help to display data from mysql database

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

Getting User Data Based on Their Information

This first field is where a web visitor will enter in the 'cardname' hit submit and be directed to another page (dashboard2.php) where only his or her content will appear.
Enter your cardname to access your content<br>
<form action='dashboard2.php'>
<input type='text' name='cardname'/><input type='submit' value='retrieve card'/>
</form>
</body>
The page below is the page that is directed after the user enters in the 'cardname' from the first input field. However, I only want this second page to show the information based on the cardname that was entered. Right now, it shows every single cardname, questionone, answerone from that table.
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "flashcards";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT cardname, questionone, answerone FROM cards";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> ". $row["cardname"]. " ". $row["questionone"]. " " . $row["answerone"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
You have to modify the query to accept a WHERE clause. For instance, WHERE cardname = mysqli_real_escape_string($conn, $_GET['cardname']) (The default method for any form is GET unless you specify method="post".).
You should learn about prepared statements for MySQLi and perhaps consider using PDO, it's really not hard.
It seems that you want to perform a search and not a display all the records.
Usually a search returns records that match a certain field, unless a specific ID or unique value was entered in the search. I'm not sure this is the case.
I put this together a little quick but hopefully it helps...
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "flashcards";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// escape the string to avoid SQL injections
$searchEscaped = $conn->real_escape_string($_POST['cardname']);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT cardname, questionone, answerone FROM cards WHERE cardname = '$searchEscaped' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
if($result->num_rows == 1){
// only one result found, show just that
$row = $result->fetch_assoc()
echo "<br> ". $row["cardname"]. " ". $row["questionone"]. " " . $row["answerone"] . "<br>";
}else{
// multiple rows found, show them all
while($row = $result->fetch_assoc()) {
echo "<br> ". $row["cardname"]. " ". $row["questionone"]. " " . $row["answerone"] . "<br>";
}
}
} else {
echo "0 results";
}
$conn->close();
?>

Mysql output displaying into XCODE? With webview?

I have a PHP script that outputs my events from my ICAgenda calendar on the website.
It outputs "name" "date" and "location" / event.
How can I import this into XCODE to show the events?
I thought styling the php script, and then add it as a webview in XCODE ? Or I'll guess there is another workaround to get the data in XCODE ?
This is the code that gathers the events:
<?php
$servername = "myservername";
$username = "username";
$password = "password";
$dbname = "dbname";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT next, title, city FROM jos_icagenda_events ORDER BY next";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "Datum & tijd: " . $row["next"]. "<br>";
echo "Plaats: " . $row["title"]. "<br>";
echo "Locatie: " . $row["city"]. "<br><br>";
}
} else {
echo "0 results";
}
$conn->close();
?>

Connecting PHP to mySQL and retrieving data

i am new to PHP, and i am attempting to create a simple connection from html to mySQL using php. I met with some problems when running my codes.
this is my code:
<?php
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT username FROM users";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["userid"]. ;
}
} else {
echo "0 results";
}
$conn->close();
?>
after running on a browser, this is displayed:
connect_error) { die("Connection failed: " . $conn->connect_error); } $sql = "SELECT username FROM users"; $result = $conn->query($sql); if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "
id: ". $row["userid"]. ; } } else { echo "0 results"; } $conn->close(); ?>
Do you have mysql running on your localhost machine? You must verify that it is working first before you can connect via php. Also, make sure you have TCP/IP sockets open in mysql and to make sure it isn't just listening via unix sockets.
echo "<br> id: ". $row["userid"]. ;
this is a syntax error, no need to end it with a full stop.also the correct syntax to connect to sql server is
mysqli_connect("localhost","my_user","my_password","my_db") or die("");
Debug the code before posting it here.
maybe try testing it with a try catch statement. Its what I've done. this way you can display your error messages a little more nicely. As for the cause, PressingOnAlways is probably right
If you are using wampserver or any other server you need to put your php files into c:\wamp\www folder and run them in browser by typing localhost/nameofyourfile.php (change c:\wapm\www for your server installation path or type)
you have the syntax error in blow line
echo "<br> id: ". $row["userid"]. ;
. (dot) should not be there.
second you fetch usernamestrong text by following query
$sql = "SELECT username FROM users";
but you try to get userid
echo "<br> id: ". $row["userid"]. ;
try below code.
<?php
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "databasename";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT usersname FROM users";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> id: ". $row["usersname"] ;
}
} else {
echo "0 results";
}
$conn->close();
?>

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