Sorry for my English. I'm trying to output the data to a database format json. It seems to do everything right, but it is not true outputs. Here is my link which is obtained: http://ksupulse.tk/get_all.php if I did check the validity of the site http://jsonlint.com/ or http://jsonformatter.curiousconcept.com/ get an error.
get_all.php
<?php
header('Content-Type: application/json; charset=utf-8');
?>
<?php
$response = array();
require 'db_connect.php';
$db = new DB_CONNECT();
$result = mysql_query("SELECT * FROM demo") or die(mysql_error());
if (mysql_num_rows($result) > 0) {
$response["demo"] = array();
while ($row = mysql_fetch_array($result)) {
$product = array();
$product["id"] = $row["id"];
$product["name"] = $row["name"];
$product["detaly"] = $row["detaly"];
array_push($response["demo"], $product);
}
$response["success"] = 1;
echo json_encode($response);
} else {
$response["success"] = 0;
$response["message"] = "No products found";
echo json_encode($response);
}
?>
I've honestly spent so much time searching for this problem, but the answer is not found. All the same, why do I have prints to the database is not json? DB encoded utf_unicode_ci, and the table in utf8_general_ci
There is an extraneous carriage return in front of your { } sequence (which is valid, in itself).
You should not close ?> and then reopen <?php your script after the header instruction.
It outputs garbage to the browser. You really want your stream to begin with the { first character.
In other words (for #KnightRider) the lines 5-7 of the script should be removed!
05 ?>
06
07 <?php
Sorry for answering this is because I could not attempt but have to write my views
Hi, I have checked this which you posted in the comment and JSONlint verifies it as a valid JSON
{"demo":[{"id":"3","name":"123123","detaly":"123123123"},{"id":"4","name":"4444","detaly":"555555"}],"success":1}
What else do you need?
Related
I created an API for the Java desktop application because I want to get the data from an online database. The beginning was fine. But some parts were not shown as required. Below is how the data is in the database.
patient_id patient_name patient_nic patient_dob patient_note
PTT00001 Rebecca J Burns 988249675V 1998-12-17 Had previously taken medicine for...
PTT00002 Erica L Prom 926715648V 1992-06-21 To show up a second time for...
The PHP code I used to get this as JSON is as follows and it doesn't show any output(A blank page appeared)
PHP Code :
<?php
$con = mysqli_connect("localhost", "root", "", "on_dam_sys");
$response = array();
if($con){
$sql = "select * from patient";
$result = mysqli_query($con,$sql);
if($result){
header("Content-Type: JSON");
$i = 0;
while($row = mysqli_fetch_assoc($result)){
$response[$i]['patient_id'] = $row ['patient_id'];
$response[$i]['patient_name'] = $row ['patient_name'];
$response[$i]['patient_nic'] = $row ['patient_nic'];
$response[$i]['patient_dob'] = $row ['patient_dob'];
$response[$i]['patient_note'] = $row ['patient_note'];
$i++;
}
echo json_encode($response, JSON_PRETTY_PRINT);
}
}
?>
But when the patient_name is removed using the same code, everything except it appears as below. What is the reason for that?
PHP code 2 :
<?php
$con = mysqli_connect("localhost", "root", "", "on_dam_sys");
$response = array();
if($con){
$sql = "select * from patient";
$result = mysqli_query($con,$sql);
if($result){
header("Content-Type: JSON");
$i = 0;
while($row = mysqli_fetch_assoc($result)){
$response[$i]['patient_id'] = $row ['patient_id'];
$response[$i]['patient_nic'] = $row ['patient_nic'];
$response[$i]['patient_dob'] = $row ['patient_dob'];
$response[$i]['patient_note'] = $row ['patient_note'];
$i++;
}
echo json_encode($response, JSON_PRETTY_PRINT);
}
}
?>
Output for PHP code 02 :
[
{
"patient_id": "PTT00001",
"patient_nic": "988249675V",
"patient_dob": "1998-12-17",
"patient_note": "Had previously taken medicine for fever and still not cured. The body is lifeless."
},
{
"patient_id": "PTT00002",
"patient_nic": "926715648V",
"patient_dob": "1992-06-21",
"patient_note": "To show up a second time for heart disease. She is ready for surgery"
}
]
I also need to get the patient_name
Probably in one of the patient "name" there is some invalid char, in this case json_encode() simply return false, add JSON_THROW_ON_ERROR so the execution stop throwing an error.
echo json_encode($response, JSON_PRETTY_PRINT|JSON_THROW_ON_ERROR);
Probably, adding also JSON_INVALID_UTF8_IGNORE will solve the problem.
Anyway, it is worth to find the offending row.
I am creating an Android Application that requires information to be retrieved from a MySQL database on a MAMP server. I have written PHP code to try retrieve the information from the database however there is no information being retrieved from the database. I have checked the code using PHP code checker and there is no issues found. Can anyone find any issues or provide any links to help. Thank you in advance.
<?php
/*
* Following code will list all the products
*/
// array for JSON response
$response = array();
// include db connect class
require_once('connect.php');
// connecting to db
//$db = new db_name();
// get all products from products table
$result = mysqli_query("SELECT * FROM tbl_book");
// check for empty result
if (mysqli_num_rows($result) > 0) {
// looping through all results
// book node
$response["books"] = array();
while ($row = mysqli_fetch_array($result)) {
// temp user array
$book = array();
$book["id"] = $row["id"];
$book["title"] = $row["title"];
$book["description"] = $row["description"];
$book["bookID"] = $row["bookID"];
// push single book into final response array
array_push($response["books"], $book);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No book found";
// echo no users JSON
echo json_encode($response);
}
?>
I want to send database records with a PHPH file via json to my app I am making with IntelXDK. Because I can't use PHP code with the Intel XDK, I needed to use JSON. I want to show the two records 'quote' and 'author' from my 'quotes' table on my screen. Someone helped me to this code but it just returns [null,null]instead of the two records I need.. I tried debugging but I am new to PHP so I can'get it to work.. Anyone who can help or sees an error in this code? Thanks!
PS: Yes I now there are already multiple questions asked on this subject by other people. I have read them all but none of them solves my question..
<?php
if(isset($_GET["get_rows"]))
{
//checks the format client wants
if($_GET["get_rows"] == "json")
{
$link = mysqli_connect("localhost", "xxxxx", "xxxxx", "xxxx");
/* check connection */
if (mysqli_connect_errno()) {
echo mysqli_connect_error();
header("HTTP/1.0 500 Internal Server Error");
exit();
}
$query = "SELECT quote, author FROM quotes WHERE id = " . date('d');
$jsonData = array();
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
$row = $result->fetch_assoc($result);
// Create a new array and assign the column values to it
// You can either turn an associative array or basic array
$ret= array();
$ret[] = $row['quote'];
$ret[] = $row['author'];
//encode to JSON format
echo json_encode($ret);
}
else {
echo json_encode($ret);
}
/* close connection */
mysqli_close($link);
}
else
{
header("HTTP/1.0 404 Not Found");
}
}
else
{
header("HTTP/1.0 404 Not Found");
}
?>
You have a bug in fetch_assoc() function call - remove $result parameter. If you had error reporting enabling, you should see:
Warning: mysqli_result::fetch_assoc() expects exactly 0 parameters, 1 given
Just change it to:
$row = $result->fetch_assoc();
In javascript to parse this response, just do this:
var obj = JSON.parse(xmlhttp.responseText);
document.getElementById("quote").innerHTML = obj[0];
document.getElementById("author").innerHTML = obj[1];
I think your problem is with fetch_assoc()
Try to use that :
$row = mysqli_fetch_assoc($result);
instead of
$row = $result->fetch_assoc($result);
It's works for me with your example
change this
$row = $result->fetch_assoc($result);
to
$row = $result->fetch_assoc();
Just change it to:
$row = $result->fetch_assoc();
Updated:
response = JSON.parse(xmlhttp.responseText);
you can now access them independently as:
reponse.quote and response.author
I am unable to parse data in PHP from MySQL.
Here's my code
<?php
header('Content-Type: text/html; charset=utf-8');
// array for JSON response
$response = array();
echo 'भगवान';
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all products from products table
$result = mysql_query("SELECT *FROM create_event") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// products node
$response["create_event"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$product = array();
$product["id"] = $row["id"];
$product["desc"] = $row["desc"];
$text;
array_push($response["create_event"], $product);
}
// success
$response["success"] = 1;
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No products found";
// echo no users JSON
echo json_encode($response);
}
?>
It's my database, I have set all the collation in utf-8 ci format but still not working:
I have tried all possible solutions and help online and have gone through popular posts and answers and also have set browsers settings to support Hindi lang but still displaying ? marks in output. Here is the output format
भगवान{"id":"1","desc":"???? ??????"}
mysql_query('SET character_set_results=utf8');
Use this after connecting to database or before using any select statement.
I'm new to PHP and I don't understand why there is an extra word ARRAY infront of the JSON string.
Heres the output of JSON String:
Array{"Users":[{"UserID":"1","FirstName":"lalawee","Email":"12345","Password":null},{"UserID":"2","FirstName":"shadowblade721","Email":"12345","Password":null},{"UserID":"3","FirstName":"dingdang","Email":"12345","Password":null},{"UserID":"4","FirstName":"solidsnake0328","Email":"12345","Password":null}],"success":1}
This is the PHP file:
<?php
/*
* Following code will list all the Users
*/
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all Users from Users table
$result = mysql_query("SELECT * FROM Users") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// Users node
$response["Users"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$user[] = array();
$user["UserID"] = $row["UserID"];
$user["FirstName"] = $row["FirstName"];
$user["Email"] = $row["Email"];
$user["Password"] = $row["Password"];
// push single User into final response array
array_push($response["Users"], $user);
}
// success
$response["success"] = 1;
echo $response;
// echoing JSON response
echo json_encode($response);
}
else {
// no Users found
$response["success"] = 0;
$response["message"] = "No Users found";
// echo no users JSON
echo json_encode($response);
}
?>
Remove
echo $response;
which is printing the word Array. If you try to echo the array, it will display the word 'Array' rather than printing the content of an array itself. Use print_r() function to display the content of an array.
print_r($response);
With the exception of classes that use the __to_string magic method, echo and print will only output the string interpretation of a variable's value. Number types (integers and floats), strings, and (I think) booleans have a straightforward string representation. Any other variable (arrays, objects, resources) will either output nothing, their variable type, or throw a fatal error.
For arrays and objects, print_r() will go through each member/property and attempt to convert it to a string (print_r being shorthand for print_recursive). So print_r($response) will give you the full array.
Bear in mind that generally the full array is only useful to output for debugging. Passing a php string version of an array to javascript is likely to be useless.