How To Check If mysqli_fetch_array($result,MYSQLI_NUM) Was Successfully Fetched Or Not?

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I am beginner in php learning procedural style. Learning prepared Statements and Mysqli. I am trying to learn how to check if fetching rows was successful or problematic. Note: I am not trying to check if there are matching rows or not after querying my mysql database for a keyword. No. To search for search results you usually use the mysqli_stmt_num_rows(). But I use the COUNT function. Like so:

$query = "SELECT COUNT(id) From links WHERE keyword = ?";
$stmt = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt,$query))
{
    mysqli_stmt_bind_param($stmt,'s',$keyword);
    mysqli_stmt_execute($stmt);
    mysqli_stmt_bind_result($stmt,$row_count);
    if($result = mysqli_stmt_fetch($stmt))
    {
        echo 'Row Count: ' .$row_count; echo '<br>';    
    }
    else
    {
        echo 'Record fetching failed!';
        echo 'Error: ' .mysqli_stmt_error($conn);
        echo 'Error: ' .mysqli_stmt_errno($conn);
    }
    mysqli_stmt_close($stmt);
}
else
{
    echo 'Search Preparation Failed!';
}
mysqli_close($conn);

I just need to check if the rows were fetched or not without any issue. need to know whether there were any trouble or not when trying fetching them. And if there are any troubles, then what the trouble is. I need php to tell me this. Here is my attempt:

<?php
//5.
//MULTIPLE RECORDS FETCHING - mysqli_stmt_get_result(): mysqli_fetch_array().

//ERROR REPORTING.
ini_set('display_errors','1');
ini_set('display_startup_errors','1');
error_reporting(E_ALL);

//MYSQLI CONNECTION.
//BAREBONE TEMPLATE.
mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT);

$server = 'localhost';
$user = 'root';
$password = '';
$database = 'bruteforce';

if(!$conn = mysqli_connect("$server","$user","$password","$database"))
{
    echo 'Mysqli Connection Error' .mysqli_connect_error($conn);
    echo 'Mysqli Connection Error Number' .mysqli_connect_errno($conn);
}

mysqli_set_charset($conn,'utf8mb4');

//QUERY DATABASE.
$keyword = 'keywords';

$query = "SELECT id,date_and_time,domain,domain_email,ip,url,anchor,title,description,keyword,keyphrase From links WHERE keyword = ?";
$stmt = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt,$query))
{
    mysqli_stmt_bind_param($stmt,'s',$keyword);
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_get_result($stmt);
    while($col = mysqli_fetch_array($result,MYSQLI_NUM))
    {
        $id = $col['0'];
        $url = $col['5'];
        $title = $col['7'];
        $description = $col['8'];
        $keyword = $col['9'];
        
    echo 'Id: ' .$id; echo '<br>';
    echo 'Url: ' .$url; echo '<br>';
    echo 'Title: ' .$title; echo '<br>';
    echo 'Description: ' .$description; echo '<br>';
    echo 'Keyword: ' .$keyword; echo '<br>';
    }
//WHICH ONE OF THESE 15 IFs BELOW ARE VALID ?
//if($col == FALSE) //Output: Record fetching failed!Error: Error: 0
    //if($col === FALSE) //Output: Blank
    //if(!$col) //Output: Record fetching failed!Error: Error: 0
    //if(!$col = mysqli_fetch_array($result)) //Output: Record fetching failed!Error: Error: 0
    //if(!$col == mysqli_fetch_array($result)) //Output: Blank
    //if(!$col === mysqli_fetch_array($result)) //Output: Blank
    //if($col != mysqli_fetch_array($result)) //Output: Blank
    //if($col !== mysqli_fetch_array($result)) //Output: Blank
    //if($col !=== mysqli_fetch_array($result)) //Output: Parse error: syntax error, unexpected '=' in ...
    //if(mysqli_fetch_array($result) == FALSE)//Output: Record fetching failed!Error: Error: 0
    //if(mysqli_fetch_array($result) === FALSE) //Output: Blank
    
    if(!mysqli_fetch_array($result))//Output: Record fetching failed!Error: Error: 0
    //if(mysqli_fetch_array($result) != FALSE) //Output: Blank
    //if(mysqli_fetch_array($result) !== FALSE) //Output: Record fetching failed!Error: Error: 0---
    //if(mysqli_fetch_array($result) !=== FALSE) //Output: Parse error: syntax error, unexpected '=' in ...
    
{
    echo 'Record fetching failed!';
    echo 'Error: ' .mysqli_stmt_error($stmt);
    echo 'Error: ' .mysqli_stmt_errno($stmt);
}
mysqli_stmt_close($stmt);
}
mysqli_close($conn);
?>

I know there are more than one way of checking if fetching data was successful or not using mysqli. And I was experimenting with 15 different ways to see if "mysqli_fetch_array($result,MYSQLI_NUM)" worked or not. I wrote down my experiment results on the comments of each of the 15 lines that I tested. Kindly note the 15 IFs near the bottom of my code and point-out to me all the valid ones out of those 15. That way, I learn more than one way of coding. And as a bonus, do not mind showing me any better ways of checking than my 15 IFs, using mysqli and prepared statements.

Thank You.

EDIT: After reading responses to my questions, I realise, especially from Dharman's response, that it is not necessary to check if anything went wrong with mysqli_fetch_array() or not. And so, when I asked you people which of the following IFs are valid, I now realise none of them are valid:

//if($col == FALSE) //Output: Record fetching failed!Error: Error: 0
    //if($col === FALSE) //Output: Blank
    //if(!$col) //Output: Record fetching failed!Error: Error: 0
    //if(!$col = mysqli_fetch_array($result)) //Output: Record fetching failed!Error: Error: 0
    //if(!$col == mysqli_fetch_array($result)) //Output: Blank
    //if(!$col === mysqli_fetch_array($result)) //Output: Blank
    //if($col != mysqli_fetch_array($result)) //Output: Blank
    //if($col !== mysqli_fetch_array($result)) //Output: Blank
    //if($col !=== mysqli_fetch_array($result)) //Output: Parse error: syntax error, unexpected '=' in ...
    //if(mysqli_fetch_array($result) == FALSE)//Output: Record fetching failed!Error: Error: 0
    //if(mysqli_fetch_array($result) === FALSE) //Output: Blank
    
    if(!mysqli_fetch_array($result))//Output: Record fetching failed!Error: Error: 0
    //if(mysqli_fetch_array($result) != FALSE) //Output: Blank
    //if(mysqli_fetch_array($result) !== FALSE) //Output: Record fetching failed!Error: Error: 0---
    //if(mysqli_fetch_array($result) !=== FALSE) //Output: Parse error: syntax error, unexpected '=' in ...

I have checked many tutorials. They usually show something like this (which Dharman too demonstrated):

if(mysqli_stmt_prepare($stmt,$query))
{
    mysqli_stmt_bind_param($stmt,'s',$keyword);
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_get_result($stmt))
    while($col = mysqli_fetch_array($result,MYSQLI_NUM))
    {
        $id = $col['0'];
        $url = $col['5'];
        $description = $col['8'];
        $keyword = $col['9'];
            
        echo 'Id: ' .$id; echo '<br>';
        echo 'Url: ' .$url; echo '<br>';
        echo 'Description: ' .$description; echo '<br>';
        echo 'Keyword: ' .$keyword; echo '<br>';
    }
    mysqli_stmt_close($stmt);
}
mysqli_close($conn);

Note that on tutorials, they do not have anything like the following to check if mysqli_stmt_get_result() or mysqli_fetch_array() failed or not.

1.

if(!$result = mysqli_stmt_get_result($stmt))

if(!mysqli_fetch_array($result))

But I still wanted to check if anything goes wrong on these two lines or not. Since you people have told me that, nothing is likely to go wrong on the first one. I now still have remaining to question you about the second one. Hence, I ask: Is there anything likely to go wrong when trying to get result with mysqli_stmt_get_result() ? If so, then how to check this ? In short, which of the following new IFs are valid, if any ?

//if($result == FALSE) //Output: Blank
    //if($result) === FALSE //Output: Parse error: syntax error, unexpected '===' (T_IS_IDENTICAL) in ..
    //if(!$result) //Output: Blank
    //if(!$result = mysqli_stmt_get_result($stmt)) //Output: Record fetching failed!Error: Error: 0
    //if(!$result == mysqli_stmt_get_result($stmt)) //Output: Record fetching failed!Error: Error: 0
    //if(!$result === mysqli_stmt_get_result($stmt)) //Output: Record fetching failed!Error: Error: 0
    //if($result != mysqli_stmt_get_result($stmt)) //Output: Blank
    //if($result !== mysqli_stmt_get_result($stmt)) //Output: Blank
    //if($result !=== mysqli_stmt_get_result($stmt)) //Output: Parse error: syntax error, unexpected '=' in ...
    //if(mysqli_stmt_get_result($stmt) == FALSE)//Output: Record fetching failed!Error: Error: 0
    //if(mysqli_stmt_get_result($stmt) === FALSE) //Output: Record fetching failed!Error: Error: 0

Here is the new context of the script:

<?php
//5.
//MULTIPLE RECORDS FETCHING - mysqli_stmt_get_result(): mysqli_fetch_array().

//ERROR REPORTING.
ini_set('display_errors','1');
ini_set('display_startup_errors','1');
error_reporting(E_ALL);

//MYSQLI CONNECTION.
//BAREBONE TEMPLATE.
mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT);

$server = 'localhost';
$user = 'root';
$password = '';
$database = 'brute';

if(!$conn = mysqli_connect("$server","$user","$password","$database"))
{
    echo 'Mysqli Connection Error' .mysqli_connect_error($conn);
    echo 'Mysqli Connection Error Number' .mysqli_connect_errno($conn);
}

mysqli_set_charset($conn,'utf8mb4');

//QUERY DATABASE.
$keyword = 'keywordsd';

$query = "SELECT id,date_and_time,domain,domain_email,ip,url,anchor,title,description,keyword,keyphrase From links WHERE keyword = ?";
$stmt = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt,$query))
{
    mysqli_stmt_bind_param($stmt,'s',$keyword);
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_get_result($stmt);
    //WHICH OF THESE FOLLOWING IFs ARE VALID ?
    //if($result == FALSE) //Output: Blank
    //if($result) === FALSE //Output: Parse error: syntax error, unexpected '===' (T_IS_IDENTICAL) in ..
    //if(!$result) //Output: Blank
    //if(!$result = mysqli_stmt_get_result($stmt)) //Output: Record fetching failed!Error: Error: 0
    //if(!$result == mysqli_stmt_get_result($stmt)) //Output: Record fetching failed!Error: Error: 0
    //if(!$result === mysqli_stmt_get_result($stmt)) //Output: Record fetching failed!Error: Error: 0
    //if($result != mysqli_stmt_get_result($stmt)) //Output: Blank
    //if($result !== mysqli_stmt_get_result($stmt)) //Output: Blank
    //if($result !=== mysqli_stmt_get_result($stmt)) //Output: Parse error: syntax error, unexpected '=' in ...
    //if(mysqli_stmt_get_result($stmt) == FALSE)//Output: Record fetching failed!Error: Error: 0
    //if(mysqli_stmt_get_result($stmt) === FALSE) //Output: Record fetching failed!Error: Error: 0
    {
        die('mysqli_stmt_get_result() Failed!);
    }
    else
    {
        while($col = mysqli_fetch_array($result,MYSQLI_NUM))
        {
            $id = $col['0'];
            $url = $col['5'];
            $description = $col['8'];
            $keyword = $col['9'];
            
            echo 'Id: ' .$id; echo '<br>';
            echo 'Url: ' .$url; echo '<br>';
            echo 'Description: ' .$description; echo '<br>';
            echo 'Keyword: ' .$keyword; echo '<br>';
        }
    mysqli_stmt_close($stmt);
    }
}
mysqli_close($conn);
echo '<b>'; echo __LINE__; echo '</b>'; echo '<br>';
?>
prepared-statement mysqli php
1 Answers

Your question is not as stupid as one might think. However, it's extremely unlikely that fetching the results would fail. I have tried to crash this function myself, but I was never able to achieve it.

This is so unlikely that I would advise to stop worrying about it. As long as you write the whole thing properly, the code should not fail. If you want to be thorough then you can call mysqli_error() right after calling mysqli_fetch_array to see if there's an error. Automatic error reporting doesn't even cover this scenario, but just in case there was a problem PHP will display a warning. If you ever see that warning "Error while reading a row" then it's likely you found a yet undiscovered bug in PHP rather than anything else.

So, stop worrying about this function.

P.S. Your code is full of cargo cult practices. As a rule of thumb, if you have enabled mysqli error reporting, then you don't need to check for any function's return value. PHP will automatically trigger an error if it happens. There's still a remote chance that a function call will fail without an error, but it's so unlikely that you should not worry about it in your code.

Here it is the same thing, but cleaned up without any redundant code:

<?php

//5.
//MULTIPLE RECORDS FETCHING - mysqli_stmt_get_result(): mysqli_fetch_array().

//ERROR REPORTING.
ini_set('display_errors', '1');
ini_set('display_startup_errors', '1');
error_reporting(E_ALL);

//MYSQLI CONNECTION.
//BAREBONE TEMPLATE.
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

$server = 'localhost';
$user = 'root';
$password = '';
$database = 'bruteforce';

$conn = mysqli_connect($server, $user, $password, $database);

mysqli_set_charset($conn, 'utf8mb4');

//QUERY DATABASE.
$keyword = 'keywords';

$query = "SELECT id,date_and_time,domain,domain_email,ip,url,anchor,title,description,keyword,keyphrase From links WHERE keyword = ?";

$stmt = mysqli_prepare($conn, $query);
mysqli_stmt_bind_param($stmt, 's', $keyword);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
while ($col = mysqli_fetch_row($result)) {
    $id = $col[0];
    $url = $col[5];
    $title = $col[7];
    $description = $col[8];
    $keyword = $col[9];

    echo 'Id: ' .$id;
    echo '<br>';
    echo 'Url: ' .$url;
    echo '<br>';
    echo 'Title: ' .$title;
    echo '<br>';
    echo 'Description: ' .$description;
    echo '<br>';
    echo 'Keyword: ' .$keyword;
    echo '<br>';
}
user_18394390
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