The connection to the database is successful (it's on a server, not local), the query works fine (I echo the name, street works fine), BUT it never echoes the json string back to me.
I tried several tutorials from Google how to convert mysql to json but I never get a result out of it. No errors either.
<?php
include "dbconnect.php";
$sql = "SELECT * FROM Test ";
$result = mysqli_query($conn, $sql);
$json_array = array();
while($row = mysqli_fetch_assoc($result))
{
$json_array[] = $row;
echo "<br>";
echo $row["name"];
echo $row["street"];
echo $row["farbe"];
}
echo json_encode($json_array);
?>
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 have a php script which should echo a bunch of datasets in a json encoded string. Though the page is blank.
<?php
$con = mysqli_connect("SERVER", "USER", "PASSWORD", "DATABASE");
if(!$sql = "SELECT news.title, news.content, login.username, news.id, news.date, news.timestamp, news.importance, news.version FROM news INNER JOIN login ON news.id = login.id ORDER BY timestamp DESC") {
echo "FAIL";
}
mysqli_query($con,$sql);
$res = mysqli_query($con,$sql);
$result = array();
while($row = $res->fetch_array())
{
array_push($result,
array('title'=>$row[0],
'content'=>$row[1],
'author'=>$row[2],
'id'=>$row[3],
'date'=>$row[4],
'timestamp'=>$row[5],
'importance'=>$row[6],
'version'=>$row[7]
));
}
$oldjson = json_encode(["result"=>$result]);
echo json_encode(array("result"=>$result));
mysqli_close($con);
?>
What is the problem here? I tried some error detection with if(!..) but it did not help. I think the problem may be the array creation and/or echo, though I cannot figure out how to fix that.
You should check the result of json_encode, since you are encoding data that comes from the database you might have some stuff that requires escaping.
$json = json_encode(...);
if ($json === false) {
echo "Error = " . json_last_error() . " " . json_last_error_msg();
// You may want to var_dump the $result var here to figure out what the problem is
} else {
echo $json; // Should be ok
}
It's not a duplicate question about UTF-8 Unicode.
I am new to php and I am trying to create a json response.
I added data into my database properly as follows.
Then I tried to connect to DB and i did it successfully .
but after that when I tried to create a json response by using the following code, it doesn't shows any json response .
My PHP code is :
<?php
define('HOST','localhost');
define('USER','root');
define('PASS','qwerty');
define('DB','carol');
$con = mysqli_connect(HOST,USER,PASS,DB);
if (!$con)
{
echo "Please try later..";
}
else
{
echo "DB Connected..";
}
$sql = "SELECT * from songs";
$res = mysqli_query($con,$sql);
if (!$res)
{
echo "query failed..";
}
else
{
echo "Query success..";
echo (mysqli_num_rows($res));
}
$result = array();
while($row = mysqli_fetch_array($res)){
array_push($result,
array('title'=>$row[0]),
array('url'=>$row[1]),
array('lyrics'=>$row[2])
);
}
echo json_encode(array("result"=>$result));
mysqli_close($con);
?>
I'm getting only echo of DB Connected , Query Success and 14 (no of rows)
I'm trying PHP for the first time by using some online tutuorials.
if I did any mistake in my code,please help me to find my mistake.
Thank you in advance.
After I added echo var_dump($res);
I got
Helo evry1
i want to display data from phpmyadmin table to my php page but the problem is tht it not displaying here is my code
<?php
include('config.php');
$res = mysql_query("SELECT * FROM accord");
while($row = mysql_fetch_assoc($res))
{
$id=$row['id'];
$rupes=$row['rs'];
$mob=$row['mobilen'];
$carmodel=$row['modelcar'];
}
?>
PKR:<?php echo $rupes;?>
here is my code plz help
There might be two problems
1) You have issue with config.php
2) Your query has some issue try putting die like this
mysql_query("SELECT * FROM accord") or die(mysql_error());
You must echo your variables:
<?php
include('config.php');
$res = mysql_query("SELECT * FROM accord");
while($row = mysql_fetch_array($res))
{
echo $id=$row['id'];
echo $rupes=$row['rs'];
echo $mob=$row['mobilen'];
echo $carmodel=$row['modelcar'];
}
?>
Why is this not working?
<?php
$select = "select * from messages where user='$u'";
$query = mysqli_query($connect,$select) or die(mysqli_error($connect));
$row = mysqli_num_rows($query);
$result = mysqli_fetch_assoc($query);
$title = mysqli_real_escape_string($connect,trim($result['title']));
$message = mysqli_real_escape_string($connect,trim($result['message']));
while(($result = mysqli_fetch_assoc($query))){
echo $title;
echo '<br/>';
echo '<br/>';
echo $message;
}
?>
where as this works -
<?php
echo $title;
?>
SORRY TO SAY, BUT NONE OF THE ANSWERS WORK. ANY OTHER IDEAS?
If your mysqli query is returning zero rows then you will never see anything printed in your while loop. If $title and $message are not set (because you would want reference them by $result['title'] & $result['message'] if that are the field names in the database) then you will only see two <br /> tags in your pages source code.
If the while loop conditional is not true then the contents of the while loop will never execute.
So if there is nothing to fetch from the query, then you won't see any output.
Does you code display anything, or skip the output entirely?
If it skips entirely, then your query has returned 0 rows.
If it outputs the <br /> s, then you need to check your variables. I could be wrong, not knowing te entire code, but generally in this case you would have something like
echo $result['title'] instead of echo $title
If $title and $message come from your mysql query then you have to access them through the $result array returned by mysqli_fetch_assoc.
echo $result['title'];
echo $result['message'];
Also if your using mysqli you'd be doing something like this:
$mysqli = new mysqli("localhost", "user", "password", "db");
if ($result = $mysqli->query($query)) {
while ($row = $result->fetch_assoc()) {
print $row['title'];
}
$result->close();
}
Try this:
<?php
$result = mysql_query($query);
while($row = mysqli_fetch_assoc($result)){
echo $title.'<br/><br/>'.$message;
}
?>
Does this work;
<?php
$select = "select * from messages where user='$u'";
$query = mysqli_query($connect,$select) or die(mysqli_error($connect));
$row = mysqli_num_rows($query);
while(($result = mysqli_fetch_assoc($query))){
echo $result['title'];
echo '<br/>';
echo '<br/>';
echo $result['message'];
}
?>
Basically I've made sure that it's not picking the first result from the query & then relying on there being more results to loop through in order to print the same message repeatedly.