[giggly kisses]

Right now I'm working on an admin page that allows the user to look at all the users in the database and edit their fields (except the USER_ID which is the primary key). My problem is when I go to update the database I get an error message saying "Catchable fatal error: Object of class mysqli_result could not be converted to string in E:\wamp\www\adminpage.php on line 22". I'm getting this error because in my database USER_ID is an auto number of type int. To fix this problem I tried using the settype() method on $value in my foreach look so it inputs a value of type int but I'm still getting the error. Any ideas on how to fix this? Here is my code:

<?php
include("dbconnect.php");
if(isset($_REQUEST['select'])) {
	$checkboxes = ($_REQUEST['select']);
}


if(isset($_REQUEST['lName'])) {
	$lName = ($_REQUEST['lName']);
	
	foreach($lName as $key => &$value) {
		if($value == null || $value == "") {
			$value = mysqli_query($db, "SELECT USER_LNAME FROM $tablename WHERE USER_LNAME='$lName[$key]';");

		}
	}
}

if(isset($_REQUEST['submit'])) {
	foreach($checkboxes as $key => &$value) {
		if(isset($value)) {
			$query2 = "UPDATE $tablename SET USER_LNAME='$lName[$key]' WHERE USER_ID='settype($value, integer)';"; //Where I'm getting the error.

			mysqli_query($db, $query2) or die("Could not update database");
		}
	}
}

if(isset($_REQUEST['delete'])) {
	echo"B";
	foreach($checkboxes as $value) {
		if(isset($value)) {
			echo"A<br />";
		}
	}
}

$query1 = "SELECT * FROM $tablename;";

$result = mysqli_query($db, $query1) or die("Could not connect to database");
$row = mysqli_fetch_row($result);
?>

[nkasei28]

The $value is assigned the first mysqli select query which is an object. Then you go ahead to use that $value variable as a parameter to the second query. That is not right. First, try to get the result from the first query into a different variable and then pass that one to the second query.

[giggly kisses]

Are talking about here:

if(isset($_REQUEST['lName'])) {
	$lName = ($_REQUEST['lName']);
	
	foreach($lName as $key => &$value) {
		if($value == null || $value == "") {
			$value = mysqli_query($db, "SELECT USER_LNAME FROM $tablename WHERE USER_LNAME='$lName[$key]';"); //here?

		}
	}
}

if(isset($_REQUEST['submit'])) {
	foreach($checkboxes as $key => &$value) {
		if(isset($value)) {
			$query2 = "UPDATE $tablename SET USER_LNAME='$lName[$key]' WHERE USER_ID='settype($value, integer)';"; //Where I'm getting the error.

			mysqli_query($db, $query2) or die("Could not update database");
		}
	}
}

I thought the first $value and second $value are different variables as they are in different scopes like an 'i' variable would be in a for loop.

This post has been edited by giggly kisses: 19 March 2011 - 03:57 PM

[nkasei28]

Why not try it like so:

   
if(isset($_REQUEST['submit'])) {  

     foreach($checkboxes as $key => &$thevalue) {  

         if(isset($thevalue)) {  
             $thevalue=settype($thevalue, integer); //change the type here before using it in query
             $query2 = "UPDATE $tablename SET USER_LNAME='$lName[$key]' WHERE USER_ID='$thevalue'"; 

    

             mysqli_query($db, $query2) or die("Could not update database");  

         }  

     }  

 }  

This post has been edited by nkasei28: 19 March 2011 - 04:06 PM

[giggly kisses]

When I do that I get a Notice saying "Notice: Use of undefined constant integer - assumed 'integer' in E:\wamp\www\adminpage.php on line 22".

[nkasei28]

Put the 'integer' in the settype in quotes. sorry i didn't notice that earlier

$thevalue=settype($thevalue, "integer");

[giggly kisses]

Still a no go.

[nkasei28]

What is the error you get now? Try echo-ing the '$thevalue' variable to see what it holds at runtime. Comment out the $query2 before you do that.

[giggly kisses]

Well it turns out the error was with in this part of the code although it was pointing at the second query:

if(isset($_REQUEST['lName'])) {
	$lName = ($_REQUEST['lName']);
	
	foreach($lName as $key => &$value) {
		if($value == null || $value == "") {
			$value = mysqli_query($db, "SELECT USER_LNAME FROM $tablename WHERE USER_LNAME='$lName[$key]';");

		}
	}
}

I'm going to have to find another way to check if the text fields are blank.

This post has been edited by giggly kisses: 19 March 2011 - 05:05 PM

[JackOfAllTrades]

empty()