Trouble converting mysql statement to a mysqli prepared statement

[darklighterz]

Im in the proces of learning and converting some mysql statements into mysqli prepared statements  However i am stuck with this one  anybody able to help?

 

The error i get is

 

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

 

My understanding is that this error is due to my while loop failing due to it not obtaining some needed criterea....however i have no idea why as it works perfectly if i dont use a prepared statement, and i just cant figure out whats going on.

 

Any help would be greatly appreciated.

 

The code is...

 

<?php 
$dbc = mysqli_connect("localhost", "user", "pass", "databasename") or die("Problem connecting: ".mysqli_error());

$Output = "";
if (!isset($_SESSION["cart_array"]) || count($_SESSION["cart_array"]) < 1) { 
$Output = "<h2 align='center'>No data to render page</h2>";
} 
else {
$i = 0;
foreach ($_SESSION["cart_array"] as $each_item){
	$unique_id = $each_item["unique_id"];
	$query = "SELECT firstname, lastname, location FROM people WHERE id=?";
	$stmt = mysqli_prepare($dbc, $query) or die(mysqli_error($dbc));
	mysqli_bind_param($stmt, "i", $unique_id);
	mysqli_stmt_execute($stmt);
/* do i need to store the result??? is there any down side in doing so */
	$result = mysqli_stmt_store_result($stmt);
/* ive tried using $row = mysqli_fetch_array($result) below aswel to no avail */
		while($row = mysqli_fetch_array($stmt)) {
		$fname = $row["firstname"];
		$lname = $row["lastname"];
		$uname = $row["location"];
	}
$Output .= 'Name: ' .$fname. ' ' .$lname. '<br />' .$uname. '<br /><br />';
}
mysqli_free_result($result);
mysqli_stmt_close($stmt);
mysqli_close($dbc);
}
?>

[PFMaBiSmAd]

That's not how mysqli_stmt_store_result is used.

 

I'll get back with an answer shortly...

 

Edit: When you use mysqli_stmt_store_result, you will still use mysqli_stmt_fetch($stmt) to fetch each row.

 

Perhaps you were trying to use mysqli_stmt_get_result? Which would allow you to use mysqli_fetch_xxxxx statements to fetch each row?

[darklighterz]

That's not how mysqli_stmt_store_result is used.

 

I'll get back with an answer shortly...

 

Edit: When you use mysqli_stmt_store_result, you will still use mysqli_stmt_fetch($stmt) to fetch each row.

 

Perhaps you were trying to use mysqli_stmt_get_result? Which would allow you to use mysqli_fetch_xxxxx statements to fetch each row?

 

Thank you, I had been stumped for two days. your reply led me down a new avenue and i ended up with this....

 

Is there any other way for me to not have an empty while loop??? the $Output below is for testing purposes i just want the variables $fname, $lname and $uname to have the appropriate db information for each row id, so i can use that info elsewhere. If theres a better way of doing this please let me know.

 

 

<?php
$query = "SELECT firstname, lastname, location FROM people WHERE id=?";
$stmt = mysqli_prepare($dbc, $query) or die(mysqli_error($dbc));
mysqli_bind_param($stmt, "i", $item_id);
    /* execute query */
		mysqli_stmt_execute($stmt);
	mysqli_stmt_bind_result($stmt, $fname, $lname, $uname);

	while(mysqli_stmt_fetch($stmt)) {
	}

                        $Output .= 'Name: ' .$fname. ' ' .$lname. '<br />' .$uname. '<br /><br />';
}
mysqli_free_result($result);
mysqli_stmt_close($stmt);
mysqli_close($dbc);
}
?>

[PFMaBiSmAd]

There's no stored result, so - mysqli_free_result($result); isn't needed (it would be mysqli_free_result($stmt); if it was being used.)

[darklighterz]

There's no stored result, so - mysqli_free_result($result); isn't needed (it would be mysqli_free_result($stmt); if it was being used.)

 

Oh yea, forgot about that in my excitement.lol. Im going to be using it anyway as i realised a better way rather than needlessly using a while loop (I think this is what i was originally going for before troubleshooting and frustration mutated my code.lol).

 

Again thank you very much for your help! You made me re-read my code with a fresh pair of eyes so to speak and as always the solution seemed so obvious.

 

<?php 
$dbc = mysqli_connect("localhost", "user", "pass", "databasename") or die("Problem connecting: ".mysqli_error());

$Output = "";
if (!isset($_SESSION["cart_array"]) || count($_SESSION["cart_array"]) < 1) { 
$Output = "<h2 align='center'>No data to render page</h2>";
} 
else {
$i = 0;
foreach ($_SESSION["cart_array"] as $each_item){
	$unique_id = $each_item["unique_id"];
	$query = "SELECT firstname, lastname, location FROM people WHERE id=?";
	$stmt = mysqli_prepare($dbc, $query) or die(mysqli_error($dbc));
	mysqli_bind_param($stmt, "i", $unique_id);
		mysqli_stmt_execute($stmt);
	mysqli_stmt_bind_result($stmt, $fname, $lname, $uname);
	mysqli_stmt_store_result($stmt);
	mysqli_fetch($stmt);
        $Output .= 'Name: ' .$fname. ' ' .$lname. '<br />' .$uname. '<br /><br />';
}
mysqli_free_result($stmt);
mysqli_stmt_close($stmt);
mysqli_close($dbc);
}
?>