Php MySQL into json - trying to put query into function

Question

I'm trying to build a very basic API, I've got a query that pulls data from a MySQL view, I can echo the query out as json no problem but I want to put the query in a function in order to be able to call it from the API script...I'm just having some trouble putting the code into a function.

Here's the ode that works.....

<?php
$servername = "database.com";
$username = "username";
$password = "password";
$dbname = "db1";

// Create connection
$conn = new mysqli($servername, $username, $password,          $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * from DB_Available_Dates";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo $row["year"]. " - " . $row["Month"]. " " . $row["the_days"]. "<br>";
}
} else {
echo "0 results";
}


$conn->close();
?>

...and this is my attempt to put it into a function (that doesn't work!)...

//Connection as above
function available_dates() {
$sql = "SELECT * from DB_Available_Dates";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row

$encodeArray = array();

while($row = $result->fetch_assoc()) {
    $encodeArray[] = $row;
}
} else {
echo "0 results";
}
$dates = array();
$dates = json_encode($encodeArray);
return $dates;
}

available_dates();

$conn->close();
?>

I've just started with functions so I expect my errors to be quite comical! ...do I need to use echo to call the function?

Sumit Kumar · Accepted Answer · 2016-04-08 12:22:24Z

1

you are not returning anything,you should assign return value to some thing.You should call your function like this.

function available_dates() {
$sql = "SELECT * from DB_Available_Dates";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row

$encodeArray = array();

while($row = $result->fetch_assoc()) {
    $encodeArray[] = $row;
}
} else {
echo "0 results";
}
$dates = array();
$dates = json_encode($encodeArray);
return $dates;
}

$result=available_dates();

answered Apr 8, 2016 at 12:22

Sumit Kumar

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Comments

M4R1KU · Accepted Answer · 2016-04-08 12:37:34Z

1

You are doing nothing with the return value of your function.

You can echo it or put it in a variable.

echo available_dates();

or

print_r(available_dates());

or

$results = available_dates();

edited Apr 8, 2016 at 12:37

answered Apr 8, 2016 at 12:19

M4R1KU

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1 Comment

monkee Over a year ago

print_r(available_dates());, missed a bracket.

starex · Accepted Answer · 2016-04-08 12:54:02Z

0

You're returning an array in your code:

$dates = array();
$dates = json_encode($encodeArray);
return $dates;

You should simply echo the json_encode like this:

echo json_encode($encodeArray);

And then take care of the formatting/displaying (alternatively, you could do it like in your first example and echo back the output ready for displaying):

while($row = $result->fetch_assoc()) {
    $encodeArray[] = $row["year"]. " - " . $row["Month"]. " " . row["the_days"].<br>";
}

and then

echo json_encode($encodeArray);

answered Apr 8, 2016 at 12:54

starex

161 bronze badge

4 Comments

dazzathedrummer Over a year ago

so, would the return value be 'return json_encode($encodeArray);' ?

starex Over a year ago

If you're calling directly for the result (like in your first example that works) it is echo. But if you're calling the function (in php), then you should return $encodeArray And then when you're returning to jQuery, echo json_encode(available_dates())

dazzathedrummer Over a year ago

I'm also getting an error that refers to the '$result = $conn' line - it says "Call to a member function query() on a non-object"

dazzathedrummer Over a year ago

...fixed the error, it was a scope issue - so I put the connection inside the function - working properly now!!

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