Warning Problem: Expects Parameter 1 to Be MySQLi_Result

warning problem: expects parameter 1 to be mysqli_result

mysqli_query() returns FALSE if there was an error in the query. So you should test for it...

/* Select queries return a resultset */
if ($result = mysqli_query($dbc, "SELECT Name FROM City LIMIT 10")) {
    printf("Select returned %d rows.\n", $result->num_rows);

    /* free result set */
    $result->close();
}

See this link for the mysqli_query reference
http://php.net/manual/en/mysqli.query.php

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in

mysqli_fetch_array()'s 1st parameter must be a result of a query. What you are doing is you are passing the connection (which doesn't make sense) and the query command itself.

Read the doc here: http://php.net/manual/en/mysqli-result.fetch-array.php

To fix this, execute the query first, then store the result to a variable then later fetch that variable.

$sql = "select * from privinsi";
$result = mysqli_query($connection,$sql);
while($r = mysqli_fetch_array($result))
{
    // your code here
}

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in on line 21

The mysqli_fetch_array() function fetches a result row as an associative array, a numeric array, or both.

Basically replaces this loop here:

$query = "SELECT Lastname,Age FROM Persons ORDER BY Lastname";
$res = $con->query($query);

while ($row = $res->fetch_assoc()) {
      //do something
}

With this:

<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");

$sql="SELECT Lastname,Age FROM Persons ORDER BY Lastname";
$result=mysqli_query($con,$sql);

 // Numeric array
 $row=mysqli_fetch_array($result,MYSQLI_NUM);
 printf ("%s (%s)\n",$row[0],$row[1]);

 // Associative array
 $row=mysqli_fetch_array($result,MYSQLI_ASSOC);
 printf ("%s (%s)\n",$row["Lastname"],$row["Age"]);

 // Free result set
 mysqli_free_result($result);

 mysqli_close($con);
?>

So in your case you just have to take the result of the query and (since I can see you have specified you want it as associative array choose that option) it would look like this:

 while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
 //do something
 }

error mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given

Problem

You are missing how to pass the argument to mysqli_fetch_array().

Solution

Therefore, this line:

if(mysqli_query($cons, $result)) {

should be

if($res = mysqli_query($cons, $result)) { // assign the return value of mysqli_query to $res

^{(FWIW, I'd go with $res = mysqli_query($cons, $result); then do if($res) {.)}

And then do

while($row = mysqli_fetch_array($res)) // pass $res to mysqli_fetch_array instead of the query itself

Why?

You were giving to mysqli_fetch_array() - as an argument - the string that contains your query. That's not how it works. You should pass the return value of mysqli_query() instead. Therefore, you could also write: while($row = mysqli_fetch_array(mysqli_query($cons, $result))) {} (but it's not adviced, it is just to show you how it works).

getting warning mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, array given

Try changing your code to:

...
if (!empty($result)) { // check if result is empty or not..
  if (mysqli_num_rows($result)>0) {
    while ($r = mysqli_fetch_assoc($result)) {
      // changing $result to $output to avoid confusions in using same array for printing.
      $output = array(
         'email' => $r['email']
      );
      array_push($all, $output);
    }
  }
  else {
       echo '{"success":0}';
  }
else {
  print "No results: " . $sql_query; // check whats the issue with query
}
...

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