I want to get information from a database, and I want to put it in selectbox as option.
I tried to do it but I could not not put it what is my mistake?(db can connect I just delete server name )
I am not sure how I can put db rows in selectbox as option.
therefore, I think my code has a problem.
p.php
<?php
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM test" ;
$result = mysqli_query($conn, $sql) or die("Query: ($sql) [problem]");
$row = mysqli_fetch_assoc($result);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_row($result)) {
display("<option value=$row[seat_id]>",$row[seatnumber]."\n");
}
display ("</select>", "\n");
} else {
echo "0 results";
}
mysqli_close($conn);
function display($tag , $value) {
echo $tag . $value ;
}
?>
p.html
<html>
<head>
<meta charset="utf-8">
<link href="" rel="stylesheet" type="text/css" />
</head>
<script type="text/javascript">
function transfer(){
var pix = document.getElementById('pix').value;
document.abc.test.value =pix;
}
</script>
<script
src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js">
</script>
<script>
function ajaxWay() {
// syntax: $.post(URL,data,callback);
$.get("p.php", function(dataFromtheServer) {
$("#result").html(dataFromtheServer);
});
}
</script>
<body>
<div id="a" style="text-align:center;">
<form name="abc" method="get" action="p.php">
<select id='pix' onchange='ajaxWay()'>
<input type="button" value="click" onclick="transfer();">
<input type="text" name="test" id="test">
</form>
</div>
</body>
</html>
If your main problem is that you are not able to embed options into your HTML, try something like this:
<html>
<head>
<meta charset="utf-8">
<link href="" rel="stylesheet" type="text/css" />
</head>
<script type="text/javascript">
function transfer(){
var pix = document.getElementById('pix').value;
document.abc.test.value =pix;
}
</script>
<script
src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js">
</script>
<script>
function ajaxWay() {
// syntax: $.post(URL,data,callback);
$.get("prefinal.php", function(dataFromtheServer) {
$("#result").html(dataFromtheServer);
});
}
</script>
<body>
<div id="a" style="text-align:center;">
<form name="abc" method="get" action="p.php">
<?php
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM test" ;
$result = mysqli_query($conn, $sql) or die("Query: ($sql) [problem]");
?>
<select id='pix' onchange='ajaxWay()'>
<?php
$row = mysqli_fetch_assoc($result);
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_row($result))
{
echo '<option value="' . $row[0] . ">' . $row[0] . '</option>';
}
}
mysqli_close($conn);
?>
</select>
<input type="button" value="click" onclick="transfer();">
<input type="text" name="test" id="test">
</form>
</div>
</body>
</html>
Note that in the above I am using your code from p.php to generate the actual content for the select options. The form should NOT submit to p.php, but to some other script that will process the form and perform the required actions. I would help more if I knew more about what you were trying to achieve!
Related
I want to make a program what will count clicks after click button.
I have this code but it don't work. I use mysqli to connect to database and I use query to insert value to database and query to select from database.
<html>
<head>
<meta charset="UTF-8">
<title>Click</title>
</head>
<body>
<form action="#" method="post">
<input type="submit" name="click" value="Klikni mě">
<br>
<?php
if(isset($_POST["click"])){
$connection=new mysqli("hidden","hidden","hidden","hidden");
if($connection == false){
die("Sorry jako");
}
$query="INSERT INTO klik (klikcount) VALUES ('$klik')";
if($connection->query($query) == false){
die("Promiň");
}
$sql="SELECT klikcount FROM klik";
$result=$connection->query($sql);
if($result->num_rows>0){
while($row=$result->fetch_assoc()){
echo $row["klikcount"];
}
}
$klik=$klik+1;
}
?>
</form>
</body>
</html>
thanks.
I try solve your code and I made some changes.
Change position of "$klik = $klik+1;"
Add another SELECT
My new code:
<html>
<head>
<meta charset="UTF-8">
<title>Click</title>
</head>
<body>
<form action="#" method="post">
<input type="submit" name="click" value="Klikni mě">
<br>
<?php
if(isset($_POST["click"])) {
$connection = new mysqli("hidden","hidden","hidden","hidden");
if($connection == false) {
die("Sorry jako");
}
$sql="SELECT klikcount FROM klik";
$result=$connection->query($sql);
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()){
$klik = $row["klikcount"];
}
}
$klik = $klik+1;
$query = "INSERT INTO klik (klikcount) VALUES ('$klik')";
if($connection->query($query) == false) {
die("Promiň");
}
$sql = "SELECT klikcount FROM klik";
$result = $connection->query($sql);
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo $row["klikcount"];
}
}
}
?>
</form>
</body>
</html>
I am trying to realize a most simple way to query a postgreSQL db via a web site form. My current approach is the following, which results in a broken looking page (even the print command is ignored) and not reacting to submitting the values, i.e., not producing any output.
Could you give me a clue what is wrong with the code?
<!DOCTYPE html>
<head>
<title>Insert data to PostgreSQL with php - creating a simple web application</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<style>
li {listt-style: none;}
</style>
</head>
<body>
<h2>Enter Query</h2>
<ul>
<form name="insert" action="pg.html" method="POST" >
<li>RA:</li><li><input type="text" name="ra" /></li>
<li>Dec:</li><li><input type="text" name="dec" /></li>
<li>Radius:</li><li><input type="text" name="radius" /></li>
<li><input type="submit" /></li>
</form>
</ul>
<?php
$servername = "localhost";
$username = "foo";
$password = "bar";
$dbname = "foobar";
print "bla $dbname";
// Create connection
$conn = pg_connect("host=$servername user=$username, password=$password dbname=$dbname");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM foobar WHERE q3c_radial_query(RAJ2000, DECJ2000, ra, dec, radius)";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "RA: " . $row["RAJ2000"]. ", DEC: " . $row["DECJ2000"]. ", r" . $row["rmag"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
This is my PHP code.
<html><head>
<title>login.php</title>
<link rel = "stylesheet" href="login-style.css">
</head>
<body>
<div class = "container">
<div class = "message">
<?php
define('DB_NAME','mydb');
define('DB_USER','root');
define('DB_PASSWORD','');
define('DB_HOST','127.0.0.1');
$link = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD);
if($link){
die('could not connect:'. mysql_error());
}
$db_selected = mysql_select_db(DB_NAME,$link);
if(!$db_selected){
die('can\'t use'.DB_NAME . ': ' . mysql_error());
}
$value1 = $_POST['name'];
$value2 = $_POST['password'];
$sql = "INSERT INTO Account (username,password) VALUES ('$value1','$value2')";
if(!mysql_query($sql))
{die('ERROR'.mysql_error());
}
mysql_close();
?>
<h1>Thank you for logging in </h1>
<form action = "form.html">
<p class ="submit">
<button type ="submit" >
GO TO FORM
</button>
</p>
</div>
</div>
</body>
</html>
I want to store the data in MySQL but when I run the html code (which i have connected to this php code using action =" name of this file"), no entry is done in database I created, can you tell me what is done wrong and help me to correct it?
<DOCTYPE! html>
<html lang = "en-US">
<head>
<meta charset = "UTF-8">
<title>Sign-In</title>
<link rel = "stylesheet" href ="style-sign.css">
<script type = "text/javascript">
function validateForm(){
var x = document.forms["forma"]["login"].value;
if(x == null || x == "")
{ alert("fill the field ");
return false;
}
var y = document.forms["forma"]["password"].value;
if( y == null || y == "")
{
alert("fill the field");
return false;
}
}
</script>
</head>
<body >
<div class="container">
<div class="login">
<h1>Login</h1>
<form name = "forma" method="post" onsubmit="return validateForm()" action = "login.php">
<p>
<input type="text" name="name" value="" placeholder="Username or Email">
</p>
<p>
<input type="password" name="password" value="" placeholder="Password">
</p>
<p class="submit">
<input type="submit" name="commit" value="Login" >
</p>
</form>
</div>
</div>
</body>
</html>
this is my html code which i have connected to this php file ,I am using xampp and using phpmyadmin , I am not getting any error and apache and mysql both are running
help me with this problem
Change the following:
$sql = "INSERT INTO Account (username,password) VALUES ('$value1','$value2')";
To:
$sql = "INSERT INTO Account (username,password) VALUES ($value1,$value2)";
EDIT:
You do not have input fields setup for the username and password. So POST will not work.
First you From needs an action and a method.
Best woould be to rename the file to form.php
<form action = "form.php" method="POST">
in the result file ...
Connect to the database
$mysqli= #new mysqli('localhost', 'fake_user', 'my_password', 'my_db');
if ($mysqli->connect_errno) {
die('Connect Error: ' . $mysqli->connect_errno);
}
escape your value strings
$query= "INSERT INTO Account (username,password) VALUES ('".$mysqli->real_escape_string($value1)."','".$mysqli->real_escape_string($value2)."')"
Execute query
$result=$mysqli->query($query);
Verify results if any or check if an Error occured
if(!$result) {
$ErrMessage = $mysqli->error . "\n";
$mysqli->close();
die( $ErrMessage) ;
}
my name is Luis, well Im trying to make a file in html, css and php which gets information from the database and outputs it. But it has to be in the same page as the html so I can edit with CSS... thank you all!
<html>
<head>
</head>
<body>
<script>
</script>
<form action="<?php include 'file.php';?>" method="POST">
<b>Busque Jรก</b>
<input type="text" name="prestador" placeholder="prestador">
<input type="text" name="cidade" placeholder="cidade">
<input type="submit">
</form>
</body>
</html>
File in PHP:
<?php
$servername = "localhost";
$username = "username";
$password = "";
$dbname = "test";
$namer = $_POST['prestador'];
$cities = $_POST['cidade'];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT name,address,city FROM tb_prestadores WHERE name = '$namer' AND city = '$cities'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "name: " . $row['name']. " Endereço - : " . $row['address']." Cidade :".$row['city']."<br>";
}
} else {
echo "It is going to else";
}
$conn->close();
?>
You are including / inserting the file wrongly.
What you should do:
1) Copy your html code and paste it at the bottom of your php code i.e. after the ending '?>' tag. Your code should look like this:
<?php
$servername = "localhost";
$username = "username";
$password = "";
$dbname = "test";
$namer = $_POST['prestador'];
$cities = $_POST['cidade'];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT name,address,city FROM tb_prestadores WHERE name = '$namer' AND city = '$cities'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "name: " . $row['name']. " Endereço - : " . $row['address']." Cidade :".$row['city']."<br>";
}
} else {
echo "It is going to else";
}
$conn->close();
?>
<html>
<head>
<!-- include your css here-->
</head>
<body>
<script>
</script>
<form action="" method="POST">
<b>Busque Jรก</b>
<input type="text" name="prestador" placeholder="prestador">
<input type="text" name="cidade" placeholder="cidade">
<input type="submit">
</form>
</body>
</html>
Btw, this is not the 'optimal' way to code applications. Please look into Smarty or some other templating language as well as MVC once you get the hang of php and html basics. Good luck!
To begin with, your <?php include 'file.php';?> is miss-placed and should not be in the action tag of your form rather it should be right before it, like so:
<html>
<head></head>
<body>
<?php include 'file.php';?>
<form action="" method="POST">
<b>Busque Jรก</b>
<input type="text" name="prestador" placeholder="prestador">
<input type="text" name="cidade" placeholder="cidade">
<input type="submit">
</form>
</body>
</html>
I want to select query in specific page. I created table in database:
First, I added phpexec on plugin for using PHP on page. Then, I tested Select query and it was ok. Finally, I want to create form for checking Serial Number. Here is my code:
<html>
<head>
<title></title>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"><br><br>
Serial Number: <input type="Text" Name="Num1"><p>
<input type="Submit" value="Calculate">
</form>
<phpcode>
<?php
if (count($_POST) > 0 && isset($_POST["Num1"])
{
$servername = "localhost";
$username = "******";
$password = "*******";
$dbname = "******";
$serialNum = $_POST["Num1"];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Install FROM SN WHERE serial = $serialNum";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "id: " . $row["Install"];
}
} else {
echo "no result";
}
$conn->close();
}
?>
</phpcode>
</body>
</html>
Actual output is:
How to solve this issue?
Your form is broken... Try it this way
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<p>Serial Number: <input type="Text" Name="Num1" value=""></p>
<input type="Submit" value="Calculate">
</form>
hi maybe you can try with this
is a little example and check your values and id, names for values
and sintaxis for html and php
<html>
<head>
<title></title>
</head>
<body>
<form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
<br><br>
Serial Number: <input type="text" name="Num1" id="Num1">
<br><br>
<input type="submit" name="submit" id="submit" value="Calculate">
<br><br>
</form>
<phpcode>
<?php
if(isset($_POST["submit"]))
{
function Conect()
{
if (!($link=mysql_connect("localhost","username","password")))
{
echo "error to conect to database.";
exit();
}
if (!mysql_select_db("databasename",$link))
{
echo "Error to select database.";
exit();
}
return $link;
}//end function Conect
$serialNum = mysql_real_escape_string($_POST["Num1"]);
$query="SELECT Install FROM SN WHERE serial = $serialNum";
$action=mysql_query($query,$link) or die("Error: ".mysql_error());
if(mysql_num_rows($action) > 0)
{
?>
<table border="1">
<tr COLSPAN=2 BGCOLOR="#6D8FFF">
<td>INSTALL</td>
</tr>
<?php
while($row=mysql_fetch_array($action))
{
echo "<tr>".
"<td>".$row["Install"]."</td>".
"</tr>";
}//end while
}
else
{
echo "don't exist recordsfor list ";
}//end if
mysql_close($link);
}//end if
?>
</phpcode>
</body>
</html>
good luck ..!!