php - mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given [duplicate]

Question

I'm trying to write a PHP code that inserts and updates values and rows in a mySQL database with phpmyadmin, the update on an existing row works fine, but insert does not work. To give you some context, deviceID is the primary key, if deviceId does not already exist in the database, an INSERT should be done.

I think the problem lies in the comparison of whether the if-statement returns the empty set or a result. Help would be appreciated, thanks in advance!

Warning:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in C:\wamp\www\android_connect\update.php on line 43

line 43 is: $arr = mysqli_fetch_array($query);

Here is the php code:

if ($_SERVER ["REQUEST_METHOD"]=="POST"){
require'connectiontest.php';
createStudent();
}

$response = array();

function createstudent()
{

// check for required fields
if (isset($_POST['deviceId']) && isset($_POST['buildingId']) && 
isset($_POST['levelId']) && isset($_POST['floorplanId']) && 
isset($_POST['latitude']) && isset($_POST['longitude'])
&& isset($_POST['x']) && isset($_POST['y']) && isset($_POST['i']) && i 
sset($_POST['j']) && isset($_POST['heading']) 
&& isset($_POST['probability']) && isset($_POST['roundtrip'])) {
// extract data from POST into variables
global $connect;

<?php


$deviceId = $_POST['deviceId'];
$buildingId = $_POST['buildingId'];
$levelId = $_POST['levelId'];
$floorplanId = $_POST['floorplanId'];
$latitude = $_POST['latitude'];
$longitude = $_POST['longitude'];
$x = $_POST['x'];
$y = $_POST['y'];
$i = $_POST['i'];
$j = $_POST['j'];  
$heading = $_POST['heading'];
$probability = $_POST['probability'];
$roundtrip = $_POST['roundtrip'];
$query = "SELECT * FROM devicelocations WHERE deviceId = '$deviceId';";
$arr = mysqli_fetch_array($query);

if (sizeof($arr) == 1) {
    error_log("Database is empty, doing an INSERT.", 0);
$query = "INSERT INTO devicelocations (deviceId, buildingId, levelId, floorplanId, latitude, longitude, x, y, i, j, heading, probability, roundtrip) VALUES ('$deviceId', '$buildingId', '$levelId', '$floorplanId', '$latitude', '$longitude', '$x', '$y', '$i', '$j', '$heading', '$probability', '$roundtrip');";}

else {
    error_log("Database already has that deviceId, doing an UPDATE.", 0);
    $query = "UPDATE devicelocations SET buildingId = '$buildingId', levelId = '$levelId', floorplanId = '$floorplanId', latitude = '$latitude', longitude = '$longitude', x = '$x', y = '$y', i = '$i', j = '$j', heading = '$heading', probability = '$probability', roundtrip = '$roundtrip' WHERE deviceId = '$deviceId';";}

mysqli_query($connect, $query) or die (mysqli_error($connect));
mysqli_close($connect);



// check if row inserted or not
if ($query) {
    // successfully inserted into database
    $response["success"] = 1;
    $response["message"] = "Entry successfully inserted or updated.";

    // echoing JSON response
    echo json_encode($response);
} else {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "Entry was not successfully inserted or updated.";

    // echoing JSON response
    echo json_encode($response);
}
}  else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";

// echoing JSON response
echo json_encode($response);
}

}

?>

@Dharman this is a different problem actually. However silly it is, any error message deserves an answer on Stack Overflow. — Your Common Sense
– Your Common Sense, Commented Nov 1, 2019 at 7:19

jasinth premkumar · Accepted Answer · 2018-05-15 15:48:02Z

1

you didn't even qxecute the query before that connection with database must be made.

$query = "SELECT * FROM devicelocations WHERE deviceId = '$deviceId';";
$result=mysqli_query($con,$query);//executing query


$arr = mysqli_fetch_array($result);

$con=mysqli_connect(ip,"username","password",database)

is connection to database

answered May 15, 2018 at 15:48

jasinth premkumar

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3 Comments

Jed Over a year ago

The connection to the databse works, I am able to update values in rows that I inserted with another code. Now that I executed the query as you said, i dont get any warnings, but it is still not possible to insert a new row.

Jed Over a year ago

I made an edit; "require'connectiontest.php';" is where the connection is done. By insert a new row i mean that the "INSERT INTO" statement does not work.

jasinth premkumar Over a year ago

sixe arr return s size of result which will be greater than one so use if(mysqli_num_rows==1)

Jed · Accepted Answer · 2018-05-15 18:36:54Z

0

This solved the problem:

$result=mysqli_query($connect,$query);
$count=mysqli_num_rows($result);
if($count>0) {
update query } else{
insert query }

answered May 15, 2018 at 18:36

Jed

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Comments

Ginxxx · Accepted Answer · 2018-05-25 04:01:39Z

0

Use

$count=mysqli_num_rows($your_query);  
if($count>0){    
   update query
}
else
{    
   insert query
}

edited May 25, 2018 at 4:01

Ginxxx

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answered May 15, 2018 at 15:42

Sachin Tripathi

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