I've been at this for hours trying to figure out why this will not work. I am trying to read data from my mysql database using php and jquery. In my php file it loops through a table and should echo the results of that table. I am using the jquery .get method to try and retrieve it but i get a 404 error file not found. The 404 error goes away and everything works fine if i change my php file to something simple like:
<?php echo 'Hello'; ?>
I believe the problem is in my php file because when i create a new file with just echoing a string and nothing else it returns everything fine.
$(document).ready(function() {
$.get("php/listtasks.php", function(data){
alert("Data: " + data);
})
});
My php file where i think there is a problem contains:
<?php // sqltest.php
require_once 'session.php';
require_once 'login.php';
$conn = new mysqli($db_hostname, $db_username, $db_password, $db_database);
if($conn->connect_error) die($conn->connect_error);
$query = "SELECT * FROM list";
$result = $conn->query($query);
if (!$result) die ("Database access failed: " . $conn->error);
$rows = $result->num_rows;
for ($j = 0 ; $j < $rows ; ++$j)
{
$result->data_seek($j);
$row = $result->fetch_array(MYSQLI_NUM);
echo $row[2];
}
$result->close();
$conn->close();
?>
Related
this is my code. i've done this before in other computer and it's okay, but now when try it in my laptop,it can't be done. idk what is the problem, it will show blank in phpmyadmin. i'm using xampp v3.2.2, is that will be the problem?
<html><head><title>Your Data</title></head>
<body>
<?php
$n = $_POST["n"];
$c = $_POST["contact"];
$e = $_POST["email"];
$cm = $_POST["campus"];
$m1 = $_POST["member1"];
$m2 = $_POST["member2"];
$m3 = $_POST["member3"];
$connect = mysqli_connect("localhost","root","") or die("Unable to connect MySQL".mysqli_error());
$db = mysqli_select_db($connect,"multimedia_db") or die("Unable to select database");
$query1 = "INSERT INTO teams(advisor_name,advisor_contact,advisor_email,advisor_campus,member1,member2,member3) VALUES ('$n','$c','$e','$cm','$m1','$m2','$m3')";
$data1 = mysqli_query($connect,$query1) or die("SQL statement failed"); //records are assigned to variable data
echo "You've succesfully register";
?>
</body>
</html>
I don't use MySQLi very often. So I'll explain how to use PDO. Just so you know PDO means PHP Data Objects. The reason I'm explaining, PDO is because, if done properly, it makes SQL injection almost impossible.
Connection
connecting to your database is generally done in a separate file. Here is an example:
con.php
<?php
$hostname = '';
$username = '';
$password = '';
$dbname = '';
try {
$dbh = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password, array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo 'Connection failed: ' . $e->getMessage();
}
?>
This is just connecting to the database, so we don't have to keep connecting to other pages, we just refer to this page with an include, like this:
<?php include 'con.php'; ?>
We can put this on any page and it'll include the connection to the database. For example, if you want to select from a database:
<?php
include 'con.php';
$load_data = $dbh->prepare("SELECT * FROM user_table");
if ($load_data->execute()) {
$load_data->setFetchMode(PDO::FETCH_ASSOC);
}
while ($row = $load_data->fetch()) {
$name = $row['name'];
echo $name;
}
?>
This would simply SELECT everything from the user_table from the column name and would display all the matching records.
If you're trying to do an INSERT instead:
<?php
include 'con.php';
$post_name = $_POST['post_name'];
$stmt = $dbh->prepare("INSERT INTO user_table (name) VALUES (:user_name)");
$stmt->bindParam(':user_name', $post_name, PDO::PARAM_STR);
if ($stmt->execute()) {
echo "Success";
} else {
echo "Failed";
}
?>
So the $post_name would be the name you give your input on a form in this case name="post_name" that would be inserted into the user_table.
Hope this helps and FYI here is a very good tutorial on how to do INSERT, UPDATE and DELETE using PDO.
i've found the solution for my question. It's just that i forgot to put localhost in front of the 'url'. no wonder it showed blank.
like 'localhost/sem5/saveRegistration.php'.
i'm sorry for the inconvenience. still a beginner using this hehe
I have a simple AJAX function bound to a button that should execute a PostgreSQL query. However, when I click the button that I bound the ajax query to, all I get is the confirmation that the database connection was successful. Nothing seems to happen withe the ajax result (should be printing to console in the handleAjax() function. What am I doing wrong?
This is the javascript code (with jquery):
$(document).ready(function() {
function sendAjax() {
$.ajax({
url: "db/database.php",
success: function (result) {
handleAjax(result);
}
});
}
function handleAjax(result) {
console.log(result);
}
$("#submit-button").on("click", sendAjax);
});
And this it the contents of database.php:
<?php
function dbconn(){
ini_set('display_errors', 1); // Displays errors
//database login info
$host = 'localhost';
$port = 5432;
$dbname = 'sms';
$user = 'postgres';
$password = 'postgres';
// establish connection
$conn = pg_connect("host=$host port=$port dbname=$dbname user=$user password=$password");
if (!$conn) {
echo "Not connected : " . pg_error();
exit;
} else {
echo "Connected.";
}
}
$conn = dbconn();
$sql = "SELECT * FROM numbers;";
$result = pg_query( $sql ) or die('Query Failed: ' .pg_last_error());
$count = 0;
$text = 'error';
while( $row = pg_fetch_array( $result, null, PGSQL_ASSOC ) ) {
$text = $row['message'];
//echo $text;
}
pg_free_result( $result );
?>
The problem is in the database.php file, all you get is "Connected." because you don't print your result at the end. Ajax only receive the output of the php file.
So at the end of your php file you should add :
echo $text;
And you also should remove the echo "Connected.";
AJAX is not a magic wand that in magic way reads PHP code. Let's say AJAX is a user. So what does user do.
Open web page
Wait until PHP execute code and display data
Tells you what he sees
If you don't display anything, ajax can't tell you what he saw.
In thi's place is worth to say that the best way to communicate between PHP and AJAX is using JSON format.
Your code generally is good. All you have to do is to display your data. All your data is in your $text var. So let's convert your array ($text) to JSON.
header('Content-Type: application/json');
echo json_encode($text);
First you set content-type to json, so ajax knows that he reads json. Then you encode (convert) your PHP array to js-friendly format (JSON). Also delete unnecessary echoes like 'Conntected' because as I said, AJAX reads everything what he sees.
You should return $conn from dbconn()
if (!$conn) {
echo "Not connected : " . pg_error();
exit;
} else {
echo "Connected.";
return $conn;
}
I having written a php script which makes an SQL query and fetches a list of unique names from the database.
I am making an AJAX GET request using jQuery to the php script. When I check resources in the console I see that the php script is being called, and when I check the response it contains a list of unique names.
However, the jquery GET request is failing, and is displaying an error message in the console.
It may be easier and clearer to look at my code, as I have no idea what is the issue here. Please see code below.
php
<?php
header('Content-Type: application/json');
$servername = "****";
$username = "****";
$password = "****";
$dbname = "****";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT DISTINCT(name) FROM customer";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo json_encode(array('customer' => $row["name"]));
}
} else {
echo "0 results";
}
$conn->close();
?>
JS
$.ajax({
type: 'GET',
url: 'getcustomers.php',
success: function(data){
console.log(data);
},
error: function() {
console.log('error');
}
});
In the console it simply says error, meaning it has executed the error function.
When I load the php file in the browser it displays the following.
{"name":"Peter"}{"name":"Alan"}{"name":"Mike"}
Your JSON response is not a valid one. You are printing each data row on each iteration. So replace the while statement with this one,
if ($result->num_rows > 0) {
$return = array();
while($row = $result->fetch_assoc()) {
$return[] = array('customer' => $row["name"]);
}
echo json_encode($return);
} else {
echo "0 results";
}
Considering your script returns any result (I hope you've tried running it in broswer) then you can use something like this:
$.get('path/to/file/filename.php').done(function(response) {
$('#exampleDiv').html(response);
});
Although, common errors because you must specify the directory path if the php file you're requesting is outside the current working directory.
change your error handler function header to the following:
error: function (jqXHR, textStatus, errorThrown) {
then print that and see what the error is
you are echoing json_encode string in side while loop, instead of that you will have to push row in an array and at the end you can echo json string only once.
$outputArr = array();
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
array_push(outputArr ,array('customer' => $row["name"]));
}
} else {
echo "0 results";
}
echo json_encode($outputArr);
I have the following PHP code that is capturing and encoding JSON data ready to be used by an Ext.js file:
<?php
$Query = "SELECT `Department`,`DepartmentHeadID` FROM `Department`";
$Result = mysql_query($Query) or die("Error 01: " . mysql_error());
while($r = mysql_fetch_array($Result))
{
// Create JSON Data:
$rows[] = $r;
echo $r[0];
echo "<br />";
}
$Data = json_encode($r);
echo "<hr />";
echo $Data;
?>
$Data returns "false" when I echo it out by accessing the file directly.
I am then trying to capture and use this data with Ext.js and until I can resolve this "false" issue I'm a bit stuck.
No PDO being used due to the server PHP version as this is not a production environment and it's running on an internal server.
Any help greatly appreciated.
you are encoding $r in JSON.
$r is false when your loop has finished.
Encode $rows instead ;-)
i am trying to modify a session variable in each ajax call (so as to later retreive only the new content from db ) but one session variable $_SESSION['s'] is getting unset in every alt. ajax call ! however the unmodified session variable $_SESSION["iuser"] is working fine . I am testing this on localhost / wamp and coding in npp .
<?php
session_start();
//if(isset($_SESSION["s"]))
$in=$_SESSION["s"];
//echo $_SESSION['sid'];
$link=mysql_connect("localhost","root","");
mysql_select_db("secure_scrapbook") or die("Sorry :( Connection Error");
$query="Select data,user,id from public where public=1 && id>'$in' ORDER BY id DESC";
$result=mysql_query($query);
$i=mysql_fetch_array($result);
$_SESSION["s"]=$i["id"];
//echo $_SESSION['sid'];
if(!($_SESSION["s"]==$in))
{echo 0;
echo $in;
echo $_SESSION["iuser"];
echo $_SESSION["s"];
echo "check<br>";
}
?>
--jquery /ajax -----
<script>
setInterval(function() {
$.ajax({
url: "gets.php",
type : "get" ,
dataType : 'html',
success : function(data){
//$("#public").html(data);}
$("#public").html(data+$("#public").html()+'-');}
});}, 5000); //5 seconds
</script>
May I suggest you debug your calls like this:
<?php
session_start();
if (empty($_SESSION["s"])){
echo 'EMPTY SESSION ';
die;
}
$in = $_SESSION["s"];
$link=mysql_connect("localhost","root","");
mysql_select_db("secure_scrapbook") or die("Sorry :( Connection Error");
$query="Select data,user,id from public where public=1 && id>'$in' ORDER BY id DESC";
$result=mysql_query($query);
$i=mysql_fetch_array($result);
if (empty($i["id"])){
echo 'EMPTY $i';
die;
}
$_SESSION["s"]=$i["id"];
//echo $_SESSION['sid'];
if(!($_SESSION["s"]==$in))
{echo 0;
echo $in;
echo $_SESSION["iuser"];
echo $_SESSION["s"];
echo "check<br>";
}
?>
Each time you get the message EMPTY SESSION or EMPTY $i check:
the ajax headers and try to step-by-step track your problem.
the element in the database does have an ID
the $in is within the boundries of possible ID's
the $in is a number
<?php
session_start();
//if(isset($_SESSION["s"]))
$in=$_SESSION["s"];
//echo $_SESSION['sid'];
$link=mysql_connect("localhost","root","");
mysql_select_db("secure_scrapbook") or die("Sorry :( Connection Error");
$query="Select data,user,id from public where public=1 && id>'$in' ORDER BY id DESC";
$result=mysql_query($query);
if($i=mysql_fetch_array($result))
{
$_SESSION["s"]=$i["id"];
//echo $_SESSION['sid'];
if(!($_SESSION["s"]==$in))
{
echo 0;
echo $in;
echo $_SESSION["iuser"];
echo $_SESSION["s"];
echo "check<br>";
}
}
else
echo "No more Data in table";
?>
Try this and if "No more Data in table" is echoed, it means that your db doesn't have any data satisfying the conditions you set.
One more thing, seeing your code it looks like $_SESSION["s"] will never be equal to $in so i guess you don't need to use the if condition.