i am using this tutorial and i dont know what is error is this
<?php
header('Content-type: application/json'); // this is the magic that sets responseJSON
// Connecting, selecting database
$link = mysql_connect($dbhost, $dbuser, $dbpass)
or die('Could not connect: ' . mysql_error());
mysql_select_db($dbname) or die('Could not select database');
switch($_POST['op']) {
case 'getAllRecords': {
$table = $_POST['table'];
$query = sprintf("SELECT * FROM %s", mysql_real_escape_string($table));
// Performing SQL query
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
$all_recs = array();
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
$all_recs[] = $line;
}
break;
}
}
echo json_encode($all_recs);
// Free resultset
mysql_free_result($result);
// Closing connection
mysql_close($link);
?>
error is :- null
Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /home/ajay/public_html/mapleleafrealities.com/test.php on line 26
This is Example
My Error
if you have any simple example of getting json from external db please give me link or code
The op POST parameter is not "getAllRecords", so you're trying to encode something that doesn't exist, and free a result that was never taken. Try putting them inside.
Related
I've used the below code to generate JSON from a MySQL table. It works great when I'm only generating 2 arrays, but for some reason when I try and generate 3 or 4 arrays, I get the "white screen of death". I thought it was happening because I needed to bump up my PHP Memory Limit, but I've done that and still get the same problem. See below:
The code that works is:
<?php
//Create Database connection
$db = mysql_connect("localhost","user","dbpassword");
if (!$db) {
die('Could not connect to db: ' . mysql_error());
}
//Select the Database
mysql_select_db("dbname",$db);
//Replace * in the query with the column names.
$result = mysql_query("select * from customer", $db);
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$row_array['customerfname'] = $row['customerfname'];
$row_array['customerlname'] = $row['customerlname'];
//push the values in the array
array_push($json_response,$row_array);
}
echo json_encode($json_response);
//Close the database connection
fclose($db);
?>
But when I attempt to call more values, this is where I get the blank screen:
<?php
//Create Database connection
$db = mysql_connect("localhost","user","dbpassword");
if (!$db) {
die('Could not connect to db: ' . mysql_error());
}
//Select the Database
mysql_select_db("dbname",$db);
//Replace * in the query with the column names.
$result = mysql_query("select * from customer", $db);
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$row_array['customerfname'] = $row['customerfname'];
$row_array['customerlname'] = $row['customerlname'];
$row_array['customeremail'] = $row['customeremail'];
$row_array['customertel'] = $row['customertel'];
//push the values in the array
array_push($json_response,$row_array);
}
var_dump($json_response);
echo json_encode($json_response);
?>
Update**
The error appears to be with my last line (echo json_encode ($json_response);
The following output is generated when I add print_r($row) before that last line.
Array ( [id] => 1 [customerfname] => First [customerlname] => Last [customeremail] => test#test.com [customerphone] => 000-000-0000
Update 2
Adding var_dump($json_response); before the echo json_encode line results in the blank screen.
For future reference: If someone else is visiting this question due to having a similar problem, the error with the blank page occurred since the query returned too many rows and json_encode() did not seem to be able to handle it.
Try the following, and share with us your result,
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
//Create Database connection
$db = mysql_connect("localhost","user","dbpassword");
if (!$db)
die('Could not connect to db: ' . mysql_error());
mysql_select_db("dbname",$db);
mysql_query('SET CHARACTER SET utf8');
$result = mysql_query("select `customerfname`, `customerlname`, `customeremail`, `customertel` from customer", $db) or die("Error: ".mysql_error());
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$json_response[] = $row;
}
var_dump($json_response);
/* If the var_dump works, you can try and uncomment this section
if (function_exists('json_encode'))
{
echo json_encode($json_response);
echo "JSON Error: ".json_last_error();
}
else { echo "json_encode() is not supported"; }
*/
?>
When you use flcose($db) that is going to cause some trouble. remove that line.
flcose is used to close ressources such as writing to text files.
to close connection use mysql_close():
mysql_close($db);
and also Your loop can be reduce to this:
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
//push the values in the array
array_push($json_response,$row);
}
//Close the database connection
mysql_close($db);
echo json_encode($json_response);
and I would probably echo the json after closing the connection but thats no biggie.
And I find white screen of death very funny :D
I have very strange problem with PHP which I am starting to learn .. I have created tables in MySQL database with some data, and now I want to show them in webpage.
This is my source where I have this problem:
<?php
// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $con);
// I check the connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else {
// It always goes here
echo "Connected to database!";
}
// I am testing very simple SQL query.. there should be no problem
$result = mysql_query("SELECT * FROM cathegories", $con, $db);
if (!$result) {
// but it always dies
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
mysql_close($con);
?>
What is wrong?
Thanks in advance!
You are mixing mysql and mysqli.
Try something like:
<?php
$con= new mysqli("localhost","user","passwd","database");
if ($con->connect_errno){
echo "could not connect";
}
$select = "SELECT * FROM tablename";
if($result = $con->query($select)){
while($row = $result->fetch_object()){
echo $row->rowname."<br>";
}
}
else { echo 'no result'; }
$con->close();
?>
// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $connection);
change to
// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $con);
mysql_query only takes two parameters - the actual SQL and then the link identifier (I assume in your case that's stored in $con; therefore remove $db from the third parameter).
You don't even need the second $con parameter really.
Where's the actual logic to connect to the database initially? Just because mysqli_connect_errno() doesn't return an error it doesn't mean the connection actually exists and that $con is available in the current scope.
I'd var_dump($con) before the mysql query to make sure it's a valid connection.
I have been trying for the past while to echo out the result of a a MySQL query in PHP. I can't seem to get it to work so I am a bit lost. I know for a fact that the query works as I have done it in PHPMYADMIN and it is working fine, however whenever I load the webpage, nothing is outputted. For the purposes of this question I have generalized the things in the query as I obviously don't want anyone accessing my database. This is what I have tried:
<?php
$connect = mysql_connect('host', 'user', 'password', 'dbname');
if (!$connect) {
die('Could not connect: ' . mysql_error());
}
if (!mysql_select_db('dbname')) {
die('Could not select database: ' . mysql_error());
}
$result = mysql_query('SELECT SUM(row name) FROM `table`');
if (!$result) {
die('Could not query:' . mysql_error());
}
echo mysql_result($result);
mysql_close($connect);
?>
I look forward to your replies.
mysql_result need the second parameter (The row number from the result that's being retrieved).
echo mysql_result($result, 0);
you have to change
echo mysql_result($result);
by
echo mysql_result($result,0);
or you can use
print_r(mysql_fetch_assoc($result));
Try this
echo mysql_result($result,0);
For more information, please give a look on http://www.php.net/mysql_result
or you can use mysql_fetch_row instead of mysql_result.
$row = mysql_fetch_row($result);
echo $row[0];
You may use this following format:
Give your Sql command like this:
$result = mysql_query('SELECT SUM(row name) as new_column_name FROM `table`');
then use this loop
while($fetched_values = mysql_fetch_array($result)){
echo $fetched_values['new_column'];
}
In your mysql_connect you have provided 4 parameters. it accepts 3 params
host, username and password. dbname is given inside mysql_select_db
as you can see from official docs. So remove the 4th param from mysql_connect function.
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
http://php.net/function.mysql-connect
$db_selected = mysql_select_db('foo', $link);
http://www.php.net/manual/en/function.mysql-select-db.php
but these mysql_* function will be deprecated use mysqli instead.
pass a row number as second parameter in mysql_result
echo mysql_result($result, 1);
or
use print_r(mysql_fetch_array($result));
where $result is your result resource
if you have any doubt you may refer my following example program.
<?php
$connect = mysql_connect('localhost', 'user_name', 'password', 'db_name');
if (!$connect)
{
die('Could not connect: ' . mysql_error());
}
if (!mysql_select_db('db_name'))
{
die('Could not select database: ' . mysql_error());
}
$row = mysql_query('SELECT (Tamil+English+Maths) as total FROM table_name');
if (!$row)
{
die('Could not query:' . mysql_error());
}
while($fetched_row = mysql_fetch_array($row)){
echo $fetched_row['total'];
}
mysql_close($connect);
?>
here Tamil,English and Maths are the column values of my table
I created a script on a windows platform which connects to the mysql database and returns the results of a table. A very basic script which I wrote to simply test my connection worked. The script works fine on my windows machine but not on my new mac. On the mac it simply does not display any records at all.
I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine.
The Mac has mysql (with data) and apache running for php.
Please could someone help as I have no idea what to do now?
Script below:
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'root';
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
$dbname = 'test';
mysql_select_db($dbname);
mysql_select_db("test", $conn);
$result = mysql_query("SELECT * FROM new_table");
while($row = mysql_fetch_array($result))
{
echo $row['test1'] . " " . $row['test2'] . " " . $row['test3'];
echo "<br />";
}
mysql_close($con);
There are various things you could do to debug this.
Show all PHP errors.
ini_set('display_errors','On');
ini_set('error_reporting',E_ALL);
Catch all possible MySQL errors, not only the ones concerning whether you connected successfully.
mysql_select_db("test", $conn) or die(mysql_error());
mysql_select_db($dbname) or die(mysql_error());
$result = mysql_query("SELECT * FROM new_table") or die(mysql_error());
Side note: There's no reason to select which database you wish to use twice.
Its very difficult to see what is wrong ... so add some basic error checking, like changing this
$result = mysql_query("SELECT * FROM new_table");
to
$result = mysql_query("SELECT * FROM new_table") or die(mysql_error());
This will show you the error you are getting from your query (if there is one) .. you ill see in the documentation for mysql_query that it returns a boolean if there was an error
Also note that you have a mistake in the variable name for closing the MySQL Connection :
mysql_close($con);
should be
mysql_close($conn);
Check to see if the SELECT query was successful or not before fetching the rows.
<?php
$result = mysql_query("SELECT * FROM new_table");
if(!$result)
die('SQL query failed: ' . mysql_error());
The only thing i can think of is that Mac file system is case sensitive while windows isn't so it might be that you mispelled the name of the table. In any cas you should do
$result = mysql_query("SELECT * FROM new_table") or die("error:".mysql_error());
to view the error
I think you should use the improved PHP mysql driver
try
{
$db = new mysqli("localhost","root","root","test");
if ($db->connect_errno) {
throw new Exception($db->connect_error);
}
if ($result = $db->query("SELECT * FROM new_table")) {
printf("Select returned %d rows.\n", $result->num_rows);
while($row = $result->fetchAssoc())
{
echo $row['test1'] . " " . $row['test2'] . " " . $row['test3'];
echo "<br />";
}
/* free result set */
$result->close();
}
$db->close();
}
catch(Exception $e)
{
printf("Database Error: %s\n", $e->getMessage());
exit();
}
I'm using jqueryui and its Autocomplete plugin. It use a json to extract items.
I want to modify it so that items will be extracted from my db.
Here is how items should be :
$items = array(
"Great <em>Bittern</em>"=>"Botaurus stellaris",
"Great2 <em>Bittern</em>"=>"Botaurus stellaris 2"
);
How to make an sql query that extract data from a table and write it like the code above into the php file ?
Table : customer
id_customer | name_customer | country_customer
I want that array produce id_customer => name_customer
The query is just:
SELECT id_customer, name_customer FROM customer
and you can generate the array like so (assuming you are using MySQL):
$items = array();
$result = mysql_query($sql);
while(($row = mysql_fetch_assoc($result))) {
$items[$row['id_customer']] = $row['name_customer'];
}
References: MySQL SELECT syntax, mysql_query(), mysql_fetch_assoc()
<?php
//Use mysql_connect for connect to a Db
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
// Select a DB
$db_selected = mysql_select_db('db_name', $link);
if (!$db_selected) {
die ('Can\'t use dbame_n : ' . mysql_error());
}
//Build a query
$sql = "SELECT id_customer, name_customer FROM customer";
//Send de query to db
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
// Initialize Array
$arr_customers = array();
while(($row = mysql_fetch_assoc($result))) {
$arr_customers[$row['id_customer']] = $row['name_customer'];
}
// convert to JSON
$json = json_encode($arr_customers);
// Send to JqueryUI
echo $json;
exit();
?>