PHP Javascript Mysql Insert - javascript

I am trying to use java-script prompt to insert values into MySQL. I have a html file and a php file.
The html files saved as Test.html:
<button onclick="myFunction()">Create Project</button>
<script>
function myFunction() {
var project = prompt("Please enter project name");
if (project != null && project !="") {
$.post("conn.php", { project : project });
}
}
</script>
The conn.php:
<?php
$servername = "localhost";
$username = "root";
$password = "Password";
$dbname = "db1";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$proj = $_POST['project'];
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
The conn.php works perfectly and inserts values if I remove the post parameter.
For example, instead of this:
$proj = $_POST['project'];
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";
if i use:
$proj = "NewProject";
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";
NewProject gets inserted into the database.
I am not sure if I am missing something in my index.html which is not posting the value in the prompt to php script. I have tried echo $_POST['project']; instead of inserting into MySQL. The echo is missing.

I have run your given code, it runs only i have added the jquery link above script code
Please check this correction,
<button onclick="myFunction()">Create Project</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
function myFunction() {
var project = prompt("Please enter project name");
if (project != null && project !="") {
$.post("conn.php", { project : project },function(response){
console.log(response);
});
}
}
</script>
and also i have added isset() function with $_POST param in conn.php file
<?php
$servername = "localhost";
$username = "root";
$password = "Password";
$dbname = "db1";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$proj = isset($_POST['project'])?$_POST['project']:'';
$sql = "INSERT INTO projects (project_name) VALUES ('$proj')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>

Related

How can I make a simple admin page to change the content inside a p element

I just want to change the text inside a p element dynamically instead of manually changing the content by opening the html file itself
You could try querying the text from a MySQL database. Here's a PoC:
Inside index.php you should have the code to query the DB and get your data.
index.php
<?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 htmldata WHERE element='pobject'"; #SQL SELECT query
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$pdata = $row['data'];
}
}
$conn->close();
?>
<html>
<body>
<p><?php echo $pdata; ?></p>
</body>
</html>
Now here's the PHP code for admin.php.
Note: This is only the code that will receive a POST parameter and insert it in the database. You should include session checking so unauthorized users cannot edit the p data, and also make a frontend that will send the POST request.
admin.php
<?php
if(!isset($_POST['pdata'])){
echo "No data passed";
}else{
$pdata = $_POST['pdata'];
$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);
}
$sql = "UPDATE htmldata SET data='$pdata' WHERE element='pobject'";
$result = $conn->query($sql);
$conn->close();
}
?>

PHP Coding to Connect MYSQL Issues

I am creating a database to make my 'PHP' website but I couldn't do this. My website is cruzapp that is related to rideshare companies and changing it in to php to get details about our users. But I can't connect MYSQL by using the following PHP code:
?php
$username = "name";
$password = "password";
$hostname = "host";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Not connected to MySQL");
echo "Connected to MySQL<br>";
//select a database to work with
$selected = mysql_select_db("examples",$dbhandle)
or die("Could not select examples");
//execute the SQL query and return records
$result = mysql_query("SELECT id, model,year FROM cars");
//fetch tha data from the database
while ($row = mysql_fetch_array($result)) {
echo "ID:"$row{'id'}." Name:".$row{'model'}."Year: ". //display the results
$row{'year'}.<br>";
}
//close the connection
mysql_close($dbhandle)
?>
Can anyone help me to debug this code?
I will be very thankful to you.
Try this one out. It uses MySQLi with error echoing.
<?php
$username = "name";
$password = "password";
$hostname = "host";
$database = "examples";
$con = mysqli_connect($hostname, $username, $password, $database);
if (!$con) {
exit("Connection failed: " . mysqli_connect_error());
}
$result = mysqli_query($con, "SELECT id, model,year FROM cars");
if (mysqli_error($con)) {
exit("Error: " . mysqli_error($con));
}
while ($row = mysqli_fetch_array($result)) {
echo "ID:" . $row['id'] . " Name:" . $row['model'] . "Year: " . $row['year'] . "<br>";
}
mysqli_close($con);
First of all you should not use mysql because with PHP 7 mysql extension does not work anymore. so you must consider to change it to mysqli or PDO. PDO is recommended. Any how for a quick fix $selected = mysql_select_db($dbhandle,"examples") do this and also check all your values like hostname database name table name and make sure there are no mistakes.

Cannot pass data from my JavaScript to my PHP file

When I click on "subcribeButton" I want it to save the value and pass that value to my PHP file. I want that PHP file to then make a post to the SQL database.
This is my js:
$(".subscribeButton").on( "click", function() {
var email = $(".subscribeInput").val();
var isValid = isEmailValid(email);
if(isValid){
$(".errorMsg").css( "display", "none" );
modal.style.display = "block";
$.post( "save.php", { email: email });
$(".subscribeInput").val("");
} else {
$(".errorMsg").css( "display", "initial" );
};
});
$(".subscribeInput").on( "click", function() {
$(".subscribeInput").val("");
});
This is my php code, I would like my php code to accept data from my js file and then post the data to my sql database:
<?php
$servername = "localhost";
$username = "user";
$password = "pw";
$dbname = "db_name";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$text = $_POST['email'];
echo $text
$sql = "INSERT INTO MyGuests (email)
VALUES ($text)";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
The result is I'm getting the following error:
POST http://flockto.it/home/save.php 500 (Internal Server Error)
send # jquery.js:4
ajax # jquery.js:4
m.(anonymous function) # jquery.js:4
(anonymous function) # email.js:13
dispatch # jquery.js:3
r.handle # jquery.js:3
so why not adding a callback in your ajax request so you can debug it in console or with an alert see this it may help you
$.post( "save.php", { "email" : email } , function(result){
console.log(result);//here the result will display what you will echo after getting the post in your php file
});
so in your php file you can add this
if (isset($_POST['email'])){
$text = $_POST['email'];
echo $text;
}
//so if you check the console you will get the email
//value from your php file so then you can insert it at the DB as you want
You have a syntax error.
Your code is missing the ; at the end of this line:
echo $text
It should be:
echo $text;
The solution:
In the js file url was wrong and quotation marks around email were missing
$(".subscribeButton").on( "click", function() {
var email = $(".subscribeInput").val();
var isValid = isEmailValid(email);
if(isValid){
$(".errorMsg").css( "display", "none" );
modal.style.display = "block";
$.post( "data/save.php", { "email": email });
$(".subscribeInput").val("");
} else {
$(".errorMsg").css( "display", "initial" );
};
});
$(".subscribeInput").on( "click", function() {
$(".subscribeInput").val("");
});
In the php file used the wrong variable
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$text = $_POST['email'];
if (isset($_POST['email'])){
$text = $_POST['email'];
echo $text;
}
$sql = "INSERT INTO MyGuests (email)
VALUES ('$text')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
A million thanks to everyone who replied to this thread and helped me solve the issue.

Script not posting information from dropdown to textbox

I am trying to get data from a dropdown and post it to a textbox. But by some reason I dont get any response also the Error message that needs to be shown in the textbox.
First of all, this is my dropdown:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM products";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<select class='form-control' id='product1' name='product1' onChange='getProduct1(this.value)' style='width: 100%;'>";
echo "<option selected disabled hidden value=''></option>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row["id"]. "'>" . $row["article_id"]. " | " . $row["name"]. "</option>";
}
echo "</select>";
} else {
echo "0 results";
}
$conn->close();
?>
After selecting a item in the dropdown the scripts needs to paste . $row["name"]. into the following textbox:
<input type="text" class="form-control" id="product11" name="product11">
The jquery script that I am using to paste the name is the following script:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function getProduct1(selectedItem) { // Do an Ajax request to retrieve the information
console.log("getProduct1 before ajax", jQuery('#product1').val());
jQuery.ajax({
url: 'get.php',
method: 'POST',
data: {'product1' : jQuery('#product1').val()},
success: function(response){
// and put the price in text field
console.log("getProduct1 after ajax", jQuery('#product1').val());
jQuery('#product11').val(response);
},
error: function (request, status, error) {
alert(request.responseText);
},
});
}
</script>
The script uses the following PHP script that connects with the database and retrieves the relevant information:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname) ;
// Check connection
if ($conn->connect_error) {
die('Connection failed: ' . $conn->connect_error) ;
}else {
$product1 = isset($_POST['produc1t'])?$_POST['product1']:'';
$product11 = isset($_POST['product11'])?$_POST['product11']:'';
$query = 'SELECT * FROM products WHERE id="' . mysqli_real_escape_string($conn, $product1) . '"';
$res = mysqli_query($conn, $query) ;
if (mysqli_num_rows($res) > 0) {
$result = mysqli_fetch_assoc($res) ;
echo $result['product11'];
}else{
$result = mysqli_fetch_assoc($res) ;
echo "Error";
}
}
?>
When I run the script by selecting an option in the dropdown, nothing is happening. Does anyone know what is wrong with my script?
I am not sure you should query the database again for a value you already retrieved. Something like this should work:
jQuery( document ).ready(function() {
jQuery( "#product1" ).change(function(){
var name = jQuery( "#product1 option:selected" ).text().split('|')[1];
jQuery("#product11").val(name);
});
});
You don't need the javascript/jQuery command in the HTML

JavaScript PHP $_Get issue

so I have a JavaScript function which calls to a PHP file using an asyncronous method. This is my code
JavaScript
var xmlHttp = new XMLHttpRequest();
xmlHttp.onreadystatechange = function() {
if (xmlHttp.readyState == 4 && xmlHttp.status == 200)
callback (xmlHttp.responseText);
}
xmlHttp.open("GET", "http://127.0.0.1/formulario/insertReporte.php?"+'nombreAlumno='+nombreAlumno+'&noCta='+noCta+'&semestre='+semestre, true);
xmlHttp.send(null);
And here is my PHP File
<?php
$servername = "myServerName";
$username = "myUserName";
$password = "myPassWord";
$dbname = "myDb";
$nombreAlumno = $_GET['nombreAlumno'];
$noCta = intval($_GET['noCta']);
$semestre = intval($_GET['semestre']);
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO infoalumno (nombreAlumno,noCta,noSemestre)
VALUES ($nombreAlumno,$noCta,$semestre)";
if ($conn->query($sql) === TRUE) {
$last_id = $conn->insert_id;
echo $last_id;
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
My issue comes on this line $nombreAlumno = $_GET['nombreAlumno']; as I get this error
On my database my nombreAlumno field is declared as a varchar.
I know my connection works because if I change that line into $nombreAlumno = intval($_GET['nombreAlumno']); it inserts 0 into my database.
Any ideas what am I doing wrong?
You need to add quotes around your values. Also use mysqli_real_escape_string before insert into database to prevent sql injection
$nombreAlumno=mysqli_real_escape_string($conn, $nombreAlumno);
$noCta=mysqli_real_escape_string($conn, $noCta);
$semestre=mysqli_real_escape_string($conn, $semestre);
$sql = "INSERT INTO infoalumno (nombreAlumno,noCta,noSemestre)
VALUES ('".$nombreAlumno."','".$noCta."','".$semestre."')";
Try this
<?php
$servername = "myServerName";
$username = "myUserName";
$password = "myPassWord";
$dbname = "myDb";
$nombreAlumno = $_GET['nombreAlumno'];
$noCta = intval($_GET['noCta']);
$semestre = intval($_GET['semestre']);
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO infoalumno (nombreAlumno,noCta,noSemestre)
VALUES ('".$nombreAlumno."','".$noCta."','".$semestre."')";
if ($conn->query($sql) === TRUE) {
$last_id = $conn->insert_id;
echo $last_id;
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>

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