I am trying to store data using html form into mysql database.
It is somehow not working when I click submit it should work but it seems so, But the data is not stored in my database.
Here is my php code;
<?php
$db_host = "localhost";
$db_username = "root";
$db_pass = "";
$db_name = "test_database";
#mysql_connect ("$db_host","$db_username","$db_pass") or die ("Could not connect to mysql!");
#mysql_select_db ("$db_name") or die ("No Database");
$value = $_POST['input1'];
$value2 = $_POST['input2'];
$sql = "INSERT INTO 'users' ('username','pid') VALUES ('$value')('$value2')";
echo "Hello World!";
mysql_close();
?>
and here is my other php file which includes the form
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form action = "mysql_connect.php" method="post" />
<p>Input 1: <input type="text" name="input1" /> </p>
<p>Input 1: <input type="number" name="input2" /> </p>
<input type="submit" value="Submit" />
</form>
</body>
</html>
You have not executed the SQL.
$sql = "INSERT INTO 'users' ('username','pid') VALUES ('$value')('$value2')";
mysql_query($sql) or die ("Failed Executing");
echo "Hello World!";
Please use following in place of existing SQL Query your executing
$sql = mysql_query("INSERT INTO 'users' ('username','pid') VALUES ('$value', '$value2')") or die (mysql_error());
Try to replace the SQL-line with:
$sql = mysql_query("INSERT INTO 'users' ('username','pid') VALUES ('$value', '$value2')");
Your code should be:
<?php
$db_host = "localhost";
$db_username = "root";
$db_pass = "";
$db_name = "test_database";
$conn=mysql_connect($db_host,$db_username,$db_pass) or die ("Could not connect to mysql!");
mysql_select_db ($db_name,$conn) or die ("No Database");
$value = $_POST['input1'];
$value2 = $_POST['input2'];
$sql = "INSERT INTO users (username,pid) VALUES ('".mysql_real_escape_string($value)."','".mysql_real_escape_string($value2)."')";
$q=mysql_query($sql);
if($q){
echo "Inserted!";
}
else
{
echo "Not inserted";
}
mysql_close();
?>
Buddy, before ask something , please research deeply.
$sql = "INSERT INTO 'users' ('username','pid') VALUES ('$value'), ('$value2')";
This will prepare insert query string.
But you need to call mysql api to insert data.
You are no executing your sql statement. Execute your sql statement, this will work for you.
if(mysql_query($sql)){ echo "Data inserted successfully."; }else{ echo "Unable to execute query." };
Related
when i hit the submit button, nothing happens. perhaps the database is not connected. i am trying to make a form using php and html. i am using xampp, i wrote the code in notepad++ and i saved form.php in htdocs. i don't know what is wrong. maybe the names i used for the variables.
this is the html code:
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<form method="post" action="C:\xampp\htdocs\form.php">
Nume de utilizator : <input type="text" name="nume_de_utilizator" placeholder="Enter Your Name" >
Email : <input type="text" name="email" placeholder="Enter Your Email">
Parola: <input type="password" name="parola">
<input type="submit" value="submit" >
</form>
</body>
</html>
this is form.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "autentificare";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$nume_de_utilizator = mysqli_real_escape_string($conn, $_POST['nume_de_utilizator']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$parola = mysqli_real_escape_string($conn, $_POST['parola']);
// Attempt insert query execution
$sql = "INSERT INTO utilizatori (nume_de_utilizator, email, parola) VALUES ('$nume_de_utilizator', '$email', '$parola')";
if(mysqli_query($conn, $sql))
printf("%d Row inserted.\n", mysqli_affected_rows($con));
else
{ echo "ERROR: Could not able to execute $sql. " . mysqli_error($conn);}
// Close connection
mysqli_close($conn);
?>
and this is the "autentificare", the database
my database
Try to see the "online link"
eg: "http://localhost:8080/form.php"
Do a simple echo msg - file to check and after replace
action="C:\xampp\htdocs\form.php" with action=http_link
PHP only adding Numbers to MySQL in column of VARCHAR instead of texts
when using query directly in MySQL it works...but if I use $_POST from HTML, IT fails
I don't know the reason how it is getting failed. what is the problem here ?
<?php
$link=mysqli_connect("localhost","root","","home_ac");
if(mysqli_connect_error()) {
die("error in database");
}
$name =$_POST["name"];
$query = "INSERT INTO `test`(`number`, `name`) VALUES (NULL,$name)";
if(mysqli_query($link, $query)){
echo "done";
}
else {
echo "failed";
}
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<form method="post">
<input type="text" placeholder="enter a name" name="name">
<input type="submit" value="add">
</form>
</body>
</html>
You need quotes around text
$query = "INSERT INTO `test`(`number`, `name`) VALUES (NULL,'$name')";
Please, think about prepared query. It solve quotes problem and protect from SQL injection.
You have to use PHP Prepared Statements or PHP Data Objects (PDO).
For example, using PDO:
<html>
<head>
<meta charset="utf-8">
<title> Example PDO Insert </title>
</head>
<body>
<form method="post" action="" name="myForm" id="myForm">
<input type="text" placeholder="Enter Your Name" name="name" required="required">
<input type="submit" name="submit" value="add">
</form>
</body>
</html>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "home_ac";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
if ( isset($_POST['submit']) && !empty($_POST['name']) ) {
# code...
$sql = "INSERT INTO test (number,name) VALUES (NULL,'$name')";
// use exec() because no results are returned
$conn->exec($sql);
echo "New record created successfully";
}
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
?>
For the past couple days, I have been trying to learn how to search an mysql database. So far I have the code below. for some reason it isn't searching and giving me results back. my database is named score and the table is all scores. Someone please help me with this.
It should be searching my database but it's coming up with no results. I have made sure everything is correct.
This file is searching.php
<?php
if (isset($_POST['search'])) {
$id = $_POST['id'];
$connect = mysqli_connect("localhost", "root", "root", "score");
$query = "SELECT `name` FROM `all_scores` WHERE `id` = $id LIMIT 1";
$result = mysqli_query($connect, $query);
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
$name = $row['name'];
}
} else {
echo "Undifined ID";
$gameid = "";
}
mysqli_free_result($result);
mysqli_close($connect);
} else {
$gameid = "";
}
this is search.php
<!DOCTYPE html>
<html>
<head>
<title> PHP FIND DATA </title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<form action="searching.php" method="post">
Id:<input type="text" name="id"><br><br>
<input type="submit" name="search" value="Find">
</form>
</body>
</html>
To get the form values inside the php file you need to use $_POST. Here's an example using PDO. You're only retrieving one row so you don't need the while loop.
searching.php
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$q = $conn->prepare("SELECT `name` FROM `all_scores` WHERE `id` = :id LIMIT 1");
$q->bindValue(':id', $_POST['id'], PDO::PARAM_STR, 50);
$q->execute();
if ($q->rowCount() > 0) {
$check = $q->fetch(PDO::FETCH_ASSOC);
$row_id = $check['id'];
// do something
}
Html:
<form action="searching.php" method="post">
Id:<input type="text" name="id"><br><br>
<input type="submit" name="search" value="Find">
</form>
Take some time look at several other examples
I am trying to create a pastebin type of thing.
I have got my php code
<?php
printf("uniqid(): %s\r\n", uniqid());
?>
<!DOCTYPE html>
<meta charset="UTF-8">
<head>
<link rel="stylesheet" type="text/css" href="style.css">
<title>Xdev - CodeShare</title>
</head>
<form method="post" action="cslink.php">
<textarea cols="100" rows="20" name="textbox"></textarea>
<br />
<input type="submit" value="Submit now" />
</form>
and my php file to store it in my mysql database
<?php
require 'connect.php';
// Connecting to the MySQL database
$conn = mysql_connect($host, $user, $pass);
mysql_select_db($db);
$username = $_POST['textbox'];
$sql = "INSERT INTO `text` (`id`, `text` ) VALUES (NULL, '$username' );";
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
?>
and my view file (also in php)
<?php
require 'connect.php';
$sql = 'SELECT * FROM text WHERE id = 3';
$result = mysql_query($sql, $conn);
while ($row = mysql_fetch_assoc($result)){
$id = $row['id'];
$text = $row['text'];
echo "id: " . $id . "<br/>";
echo "text: " . $text . "<br/>";
echo $text;
}
?>
So if I type in:
hello
world
testing
(each on a separate line)
It will show up as this: hello world testing - all in a single line
I want to know if I can and how to output it as separate lines
As #itachi mentioned, nl2br should solve this problem.
I have written this code to create a search form and get results from mysql table:
<?
$db_hostname = 'localhost';
$db_username = 'root';
$db_password = '';
$db_database = 'jatc_university_j32';
// Database Connection String
$con = mysql_connect($db_hostname, $db_username, $db_password);
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db_database, $con);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title></title>
</head>
<body>
<form action="" method="post">
Search: <input type="text" name="FieldValue" /><br />
<input type="submit" value="Submit" />
</form>
<?
if (!empty($_REQUEST['FieldValue'])) {
$FieldValue = mysql_real_escape_string($_REQUEST['FieldValue']);
$result = mysqli_query($con, "SELECT Fieldvalue from #__rsform_submission_values WHERE FieldName = 'candidatname'");
while ($row = mysqli_fetch_array($result)) {
echo $row;
echo "<br>";
}
}
?>
</body>
</html>
In my database table I have FieldName: candidatname, candidatsurname and FieldValue: John, Wayne etc.
I want to search entering a name and return the other details for this candidate
Anyway when I run code nothing happens
Can you please check if I am doing wrong something because I get the same result in a lot of trials
1.Nothing happens beacuse you are using <? ?> for php, but you should use <?php ?>
2.Your form method is post then you should check variblae like this on form submit:
if (!isset($_POST['FieldValue']))
3. "SELECT Fieldvalue from #__rsform_submission_values WHERE FieldName = 'candidatname'"
instead of candidatename, give the value that you got from the form:
"SELECT Fieldvalue from #__rsform_submission_values WHERE FieldName = '".$FieldValue."'" <BR>
4. <input type="submit" value="Submit" /> add a name property to this like:
<input type="submit" value="Submit" name="Submit"/>
and it will work, i tested after applying these changes!
Try this:
<?php
$formpartone = <<<EODformpartone
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title></title>
</head>
<body>
<form action="{$_SERVER['PHP_SELF']}" method="get">
Search: <input type="text" name="FieldValue" value=""/><br />
<input type="submit" value="Submit" />
</form>
EODformpartone;
$formparttwo = <<<EODformparttwo
</body>
</html>
EODformparttwo;
if (!isset($_GET["FieldValue"]))
{
echo $formpartone;
echo $formparttwo;
}
ELSE {
function search_candidate(){
$host = "localhost";
$user = "root";
$password = "";
$database = "jatc_university_j32";
$searchstring = $_GET["FieldValue"];
$link = mysqli_connect($host, $user, $password, $database);
IF(!$link){
echo ('unable to connect to database');
}
ELSE {
$query = "SELECT candidatename, candidatesurname
FROM (
SELECT c.SubmissionID,
max(CASE WHEN c.FieldName='candidatename' THEN c.Fieldvalue ELSE 0 END) AS 'candidatename',
max(CASE WHEN c.FieldName='candidatesurname' THEN c.Fieldvalue ELSE 0 END) AS 'candidatesurname'
FROM mf2sn_rsform_submission_values as c
GROUP BY c.SubmissionID
) a
WHERE a.candidatename LIKE '".$searchstring."' OR a.candidatesurname LIKE '".$searchstring."'";
$result = mysqli_query($link, $query);
echo "<table>";
while($row = mysqli_fetch_array($result, MYSQLI_BOTH)){
echo "<tr><td>".$row['candidatename']."</td><td>".$row['candidatesurname']."</td></tr>";
} echo "</table>";
}
mysqli_close($link);
}
echo $formpartone;
echo search_candidate();
echo $formparttwo;
}
?>
Sample table mysql
CREATE TABLE candidates
(
id int auto_increment primary key,
SubmissionID int,
FieldName varchar(20),
Fieldvalue varchar(30)
);
INSERT INTO candidates
(SubmissionID, FieldName, Fieldvalue)
VALUES
(1,'candidatename','Bob'),
(1,'candidatesurname', 'Smith'),
(2,'candidatename','Jack'),
(2,'candidatesurname', 'Doe');
SQLFiddle demo of the sample data
You should look into the validation and sanitation of the search string to avoid SQL injection. And what to do when no candidates are found with the name you inserted. However, I think for now it is important to test if everything works.