I have simple php script that shoud output json, but it dont work
<?php
require 'connect.php';
$sql = "SELECT * FROM horizont";
$result = $con->query($sql);
$rows = array();
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$rows[] = $row;
}
}
echo '<pre>';
var_dump($rows);
echo '</pre>';
echo json_encode($rows);
?>
i get result on var dump , but not json code.
var dump result:
connect.php
<?php
$con = mysqli_connect("localhost","root","","horizont");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
Fixed with :
$con->set_charset("utf8");
For encoding problem try this:
$con->set_charset("utf8");
Related
<?php
$con = mysqli_connect("localhost", "alsmwsk3", "skrktkzl12", "alsmwsk3");
$result = mysqli_query($con, "select * from NOTICE ORDER BY noticeDate DESC");
$response = array();
$num = 0;
while($row = mysqli_fetch_array($result))
{
array_push($response, array("noticeContent"=>$row[0],"noticeName"=>$row[1],"noticeDate"=>$row[2]));
echo $num;
}
//php를 읽었을떄 json타입이 아닐 readed php not but json?
//print (json_encode($response));
// echo json_encode(array("response"));
echo json_encode(array("response"=>$response));
mysqli_close($con);
?>
when I open this phpsource on browser there are no data on browser I think am i wrong some kind of miss?
and it is mysql setting
using php database connection i want to display data in json format which data are fatched from database(MySql),but i can't displaying in json format. http://takeyourtime.16mb.com/fatchData.php
$con = mysqli_connect($host, $username, $pwd, $db) or die('Unable to connect');
if (mysqli_connect_error($con))
{
echo "Failed to Connect to Database ".mysqli_connect_error();
}
$name = $_POST['Query'];
$sql = "SELECT * FROM playerstb";
$query = mysqli_query($con,$sql);
if ($query)
{
$rows = array();
while ($r = mysql_fetch_assoc($query)) {
$rows['root_name'] = $r;
}
}
echo json_encode($rows);
mysqli_close($con);
Just use json_encode. BTW, your script has an syntax error in the ending if block:
if($query){
$rows = array();
while($r = mysql_fetch_assoc($query)) {
$rows['root_name'][] = $r; // probably must be an array
}
echo json_encode($rows);
}else{
/*
This will show up when you have a query error
nothing to do with the results found.
I would consider changing the message below
*/
echo('Not Found');
}
inside your while loop you dont save all results you each one writen over the before one
you have to store it in array like note this ([])*
while($r = mysql_fetch_assoc($query)) {
$root_names[] = $r;
}
echo json_encode(['root_name'=>$root_names]);
You have to store first your result in array then after that create an array name your desire key ($array["name"])
$con=mysqli_connect($host,$username,$pwd,$db) or die('Unable to connect');
if(mysqli_connect_error($con))
{
echo "Failed to Connect to Database ".mysqli_connect_error();
}
$name=$_POST['Query'];
$sql="SELECT * FROM playerstb";
$query=mysqli_query($con,$sql);
if($query)
{
$rows = array();
while($r = mysql_fetch_assoc($query)) {
$rows[] = $r;
}
$data["data"]=$rows;
echo json_encode($data);
}
}else
{
echo('Not Found ');
}
mysqli_close($con);
?>
I have the following query that is returning:
"Code1""Code2""Code3"
What I need is for the result to be in CSV format like this:
'Code1','Code2','Code3'
How do I modify my code to accomplish this?
<?php
$conn = new mysqli("127.0.0.1", "user", "password", "table", "3306");
if(mysqli_connect_errno($conn)) {
echo "Unable to connect to database server";
}
$result = mysqli_query($conn,"SELECT code FROM sourcecodes");
while($row = $result->fetch_assoc()) {
$mydata = ($row['code']);
echo json_encode($mydata);
}
$result->free();
$conn->close();
?>
Per every iteration you should be pushing to an array. Once the array is built, you can turn it into json and echo it.
$mydata = array();
while ($row = $result->fetch_assoc()) {
$mydata[] = $row['code'];
}
$json = json_encode($mydata);
$string = str_replace(array('[', ']'), '', $json);
echo $string;
Try this
<?php
$conn = new mysqli("127.0.0.1", "user", "password", "table", "3306");
if(mysqli_connect_errno($conn)) {
echo "Unable to connect to database server";
}
$result = mysqli_query($conn,"SELECT code FROM sourcecodes");
while($row = $result->fetch_assoc()) {
$mydata[] = $row['code'];
}
$result->free();
$conn->close();
echo "'".implode("','",$mydata)."'";
?>
Or you need json_encode?
i am trying to get this output from a database of countries. i had use the mysqli_fetch_object() function but it does not work with me
"countries":[{"countryname":"India","flag":"http:\/\/wptrafficanalyzer.in\/p\/demo1\/india.png","language":"Hindi","capital":"New Delhi","currency":{"code":"INR","currencyname":"Rupee"}},{"countryname":"Pakistan","flag":"http:\/\/wptrafficanalyzer.in\/p\/demo1\/pakistan.png","language":"Urdu","capital":"Islamabad","currency":{"code":"PKR","currencyname":"Pakistani Rupee"}}]}
and i am use this php script
<?php
require 'config.php';
$con=mysqli_connect($servername,$username,$password,$db);
if(!$con)
{
die ("Erro in connection" . mysqli_connect_error);
}
else{ $encode = array();
$sql="select * from country ";
$res=mysqli_query($con,$sql);
if(mysqli_num_rows($res)>0)
{
$temp_array=array();
while($row=mysqli_fetch_object($res))
{
//$temp_array[]=$row;
$encode=$row;
}
//echo json_encode($temp_array);
echo json_encode($encode);
}
else
{
echo " 0 Rows";
}
}
?>
if anybody can help me ?
Your code should be something like below;
<?PHP
require 'config.php';
$con=mysqli_connect($servername,$username,$password,$db);
if(!$con)
die ("Error in connection" . mysqli_connect_error);
else{
$encode = array();
$sql = "select * from country ";
$res = mysqli_query($con,$sql);
if(mysqli_num_rows($res)>0)
{
$temp_array=array();
while($row=mysqli_fetch_object($res))
{
$encode[]=$row;
}
}
echo json_encode($encode);
}
?>
You do not need to write down 0 rows because you are returning a json object which has an array so when you check result.length it tells you the row count.
As an advice you may use something like below;
<?php
require 'config.php';
$con = mysqli_connect($servername,$username,$password,$db);
$resultArray = array();
$resultArray["error"] = true; //That will tell your javascript client if any error exits.
$resultArray["errorMessage"] = ""; //We will set this value if any error exits;
if(!$con)
{
$resultArray["error"] = true;
$resultArray["errorMessage"] = "Error in connection" . mysqli_connect_error();
}
else{
$resultArray["error"] = false;
$itemCollection = array();
$sql = "select * from country ";
$res = mysqli_query($con,$sql);
if(mysqli_num_rows($res)>0)
while($row = mysqli_fetch_object($res))
$itemCollection[]=$row;
$resultArray["itemCollection"] = $itemCollection;
}
echo json_encode($resultArray);
?>
In my sample you are going to get a json result something like below;
{"error":true,"errorMessage":"ErrorMessageIfExists","itemCollection":[yourObject,yourObject]}
Hope this helps you.
I'm currently stuck with some PHP code. I want to access a table in my database and retrieve the data in a JSON format. Therefore, I tried the following code :
<?php
$con = mysqli_connect("......","username","pwd","DBName");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT * FROM users";
if ($result = mysql_query($con, $sql))
{
$resultArray = array();
$tempArray = array();
while($row = $result->fetch_object())
{
$tempArray = $row;
array_push($resultArray, $tempArray);
}
echo json_encode($resultArray);
}
mysqli_close($con);
?>
However, it's getting me an empty page. It worked once but only with a special number of row in the table, so not very efficient as you might guess.
Does anybody have an idea why i'm getting those weird results?
EDIT 1 :
I Just tried to add this to my code :
echo json_encode($resultArray);
echo json_last_error();
And it's returning me 5. It seems to be an error from the data encoding in my table. Therefore I added that code :
$tempArray = array_map('utf8_encode', $row)
array_push($resultArray, $tempArray);
And I got the following output : [null,null,null]0 (The zero comes from the echo json_last_error();)
So here I am, can anybody help me with this ?
I would start by changing if ($result = mysql_query($con, $sql)) to if ($result = mysqli_query($con, $sql)) because they are different database extensions
Another thing would be to change while($row = $result->fetch_object()) to while ($row = mysqli_fetch_object($result)) { (Procedural style vs. Object oriented style)
If you still see blank screen, try adding error_reporting(E_ALL); at the top of your script, and you'll be able to know exactly where the bug is
<?php
$con = mysqli_connect("......","username","pwd","DBName")
or die("Failed to connect to MySQL: " . mysqli_connect_error());
$sql = "SELECT * FROM users";
$query = mysqli_query($con, $sql) or die ("Failed to execute query")
if ($result = $query)
{
$resultArray = array();
while($row = $result->fetch_object())
{
array_push($resultArray, $row);
}
$result->close()
echo json_encode($resultArray);
}
mysqli_close($con);
?>
This code works for me, try it out:
<?php
$con = mysqli_connect("......","username","pwd","DBName");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT * FROM users";
if ($result = mysqli_query($con, $sql))
{
while($row = $result->fetch_object())
{
$resultArray[] = $row;
}
echo json_encode($resultArray);
}
mysqli_close($con);
?>
EDIT 1:
As a test replace this code:
while($row = $result->fetch_object())
{
$resultArray[] = $row;
}
echo json_encode($resultArray);
with this code:
while($row = $result->fetch_assoc())
{
print_r($row);
}
What output do you get?
I finally found a solution ! That was indeed an encoding problem, the json_encode() function accepts only strings encoded in utf8. I changed the interclassement of my table to utf8_general_ci and I modified my code as follows :
<?php
//Create Database connection
$db = mysql_connect(".....","username","pwd");
if (!$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 users", $db);
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$row_array['id'] = $row['id'];
$row_array['name'] = utf8_encode($row['name']);
$row_array['lastName'] = utf8_encode($row['lastName']);
//push the values in the array
array_push($json_response,$row_array);
}
echo json_encode($json_response);
//Close the database connection
fclose($db);
?>
And I got the expected output.