Php - Mysqli_Fetch_Array() Expects Parameter 1 to Be Mysqli_Result, String Given

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

You have:

 mysqli_fetch_array($query)

Should be:

mysqli_fetch_array($result) 

Also in line 3 you have:

if($query === FALSE) { die(mysql_error()); } 

Should be rather:

if ($result === FALSE) { die(mysql_error()); } 

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).

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result when I have 2 parameters

Use $row=mysqli_fetch_array($result); in place of $row=mysqli_fetch_array($query);

As mysqli_fetch_array() function fetches a result row as an associative array and you were passing $query which is of type string. So, its throwing error.

How to solve mysqli_fetch_array() expects parameter 1 to be mysqli_result,string in php

In this code you did not execute the queries so you cannot fetch.

Try this approach:

<?php
    if(isset($_GET['id'], $_GET['twd'])){
        /*Importing our db connection script*/
        require_once('dbConnect.php');
        /* check connection */
        if (mysqli_connect_errno()) {
            printf("Connect failed: %s\n", mysqli_connect_error());
            exit();
        }

        $sql = sprintf("SELECT MAX(id) as MaxId FROM work_details WHERE twd = '%s'", 
                       mysqli_real_escape_string($_GET['twd']));

        if ($result = mysqli_query($con, $sql)) {
            /* fetch associative array */
            if ($row = mysqli_fetch_row($result)) {
                if($row[0] === $_GET['id']){
                    $sql = sprintf("DELETE FROM work_details WHERE id='%s';", 
                                  mysqli_real_escape_string($_GET['id']));
                    if ($result = mysqli_query($con, $sql)) {
                        echo 'success';
                    }else{
                        echo 'failed';
                    }
                }

            }

            /* free result set */
            mysqli_free_result($result);
        }
        /* close connection */
        mysqli_close($con);
    }
?>

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
}

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

remove apostrophe around $result. This mysqli_fetch_array('$result') should be mysqli_fetch_array($result)

Note the difference:

'$result' is a string (single apostrophe)

$result is a resource returned by mysqli_query

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