MYSQLi PHP function fetch array - php

I am trying to retrieve a zone by running a PHP function based off of a place that has already been submitted.
Using FORM method GET, after submission, the variable that I am retrieving is:
$place = mysqli_real_escape_string($_GET['place]);
The variable immediately after is zone:
$zone = getZone($pol); // here is the PHP function call
Above both of these variables is the function getZone, which looks like this:
function getZone($place)
{
$searchZone = "SELECT ZONE FROM zones WHERE PLACE = '".$place."'";
$result = mysqli_query($dbc, $searchZone);
$row = mysqli_fetch_array($result);
return $row['ZONE'];
}
I can run the query in the database, and it returns the ZONE.
Now, the mysqli_fetch_array, which normally works for me, is failing to produce the result from the query.
Does anyone see what is wrong?

You've forgotten about PHP's variable scope rules:
$result = mysqli_query($dbc, $searchZone);
^^^^---- undefined
Since $dbc is undefined in the function, you're using a local null handle, which is invalid. If you'd had ANY kind of error handling in your code, you'd have been told about the problem.
Try
global $dbc;
$result = mysqli_query(...) or die(mysqli_error($dbc));
instead. Never assume success. Always assume failure, check for that failure, and treat success as a pleasant surprise.

This might help
//Assuming $dbc as connection variable
function getZone($dbc,$place)
{
$searchZone = "SELECT ZONE FROM zones WHERE PLACE = '".$place."'";
$result = mysqli_query($dbc, $searchZone);
$row = mysqli_fetch_array($result);
return $row['ZONE'];
}
include 'path/to/connectionfile';//Only if you haven't already done that
$zone = getZone($dbc,$pol);

Ok... I figured it out, thanks to the assistance of Marc B. I took into account that I was not providing my connection string, so I added it to the file. Problem is, I needed to add it to the actual function, like so:
function getZone($place)
{
include ("../include/database.php");
// then the rest of the code
After I included the database connection, I am now able to retrieve the zone.
Thank you.

Related

issue with getting result object of mysql query into a function

I have a function to process info from a database. This is called multiple times in a page. And I don't want to query the database every time. So I put the query outside. If I do that, the function doesn't work. I know this can be done because, there was a similar question somewhere in SO. But that addressed a different situation. I don't know what is wrong here. Any help will be greatly appreciated.
If I put all this code into a separate test file including the conn file and query, it works. But in my main page, where I have the functions.php included first, then conn.php and then the query and then the display code called by js fadein event, the $result refuses to work inside the function
EDIT : This code has been cleaned up as per comments received (globals replaced with variables passed to the function and variable names rationalised)
function total($item,$result,$val){
global $totRate;
while($getRates=$result->fetch_assoc()){
$gotItem= strtolower(preg_replace('/[^(\x20-\x7F)]*/',"",$getRates['item']));
$gotItem=str_replace(array("_"," ","/"),"",$gotItem);
if($item==$gotItem){
$rate= $getRates['rate'];
$totRate=$val*$rate;
return $totRate;
}
}
}
The Result Call PHP file
$query = "SELECT * FROM rates ORDER BY item";
$result = $orderdb->query($query)
if (isset($_POST[$itemname]) && !empty($_POST[$itemname])) {
$val=$_POST[$itemname];
total($itemname);
echo $totprate;
} else {
echo "0";
}
I am writing this with the assumption that your SQL is working but are having problems displaying what you want - this may help. The code below saves your $result variable from your query and then passes it into the total function as a second parameter. Previously you were returning $totprate from total but you were not saving it anywhere - it is now saved to the $totprate variable.
Note: I cannot see $orderdb anywhere in your code, I'm assuming you have that in your file and that it is working.
function total($item, $result){
global $val;
global $pid;
global $pitem;
global $prate;
global $totprate;
global $gotitem;
global $getratess;
// global variable for $result removed so it doesn't overwrite variable passed to function
while($getratess=$result->fetch_assoc()){
$gotitem= strtolower(preg_replace('/[^(\x20-\x7F)]*/',"",$getratess['item']));
$gotitem=str_replace(array("_"," ","/"),"",$gotitem);
if ($item==$gotitem) {
$pid=$getratess['id'];
$pitem= $getratess['item'];
$prate= $getratess['rate'];
$totprate=$val*$prate;
return $totprate;
}
}
}
$query = "SELECT * FROM rates ORDER BY item";
$result = $orderdb->query($query);
if (isset($_POST[$itemname]) && !empty($_POST[$itemname])) {
$val=$_POST[$itemname];
$totprate = total($val, $result); // pass itemname as first parameter and result array as second parameter and save it to the $totprate variable
echo $totprate;
} else {
echo "0";
}
Let me know if this helps.

What kind of object can I send to this view in order to get this PHP code to work?

I am terrible at PHP and I need to retrieve data from a database and give it to an index.php view. The view is pre-made and has this code:
//This is simplified - it has error handling that is not shown
$results = getAll($tableName);
//This is the line where it is failing
//Undefined Offset
$columns = empty($results) ? array() : array_keys($results[0]);
$idColumn = $columns[0];
There is all the rest of it but I just need to know what on earth it is that this bit of code is expecting. I have not even got the first clue what is supposed to be sent to this thing. I just need to get it to work.
This is what I have tried so far:
function getAll($tablename)
{
$mysqlConnection = getDbConnection();//Just the normal PDO db connection
$sql = "SELECT * FROM ".$tablename;
$sth = $mysqlConnection->prepare($sql);
$sth->execute();
$resultSet = $sth->fetch(PDO::FETCH_ASSOC);
return $resultSet;
}
I have tried various different PDO::FETCH_... types but nothing is working. There is no information about what it is that I am supposed to send that part of the view.
If you want all the rows from fetch(), you will need to loop through the result set because it will return a single row. In the loop you can place them in an array.
You can use fetchAll() instead. It will return all the results as an array.

Unreachable Statement in PHP

require_once 'C:/wamp/www/FirstWebsite/CommonFunctions.php';
function SelectRowByIncrementFunc(){
$dbhose = DB_Connect();
$SelectRowByIncrementQuery = "SELECT * FROM trialtable2 ORDER BY ID ASC LIMIT 1";
$result = mysqli_query($dbhose, $SelectRowByIncrementQuery);
$SelectedRow = mysqli_fetch_assoc($result);
return $SelectRowByIncrementQuery;
return $SelectedRow; //HERE is the error <-------------------------------
return $result;
}
$row = $SelectedRow;
echo $row;
if ($row['Id'] === max(mysqli_fetch_assoc($Id))){
$row['Id']=$row['Id'] === min(mysqli_fetch_assoc($Id));#TODO check === operator
}
else if($row['Id']=== min(mysqli_fetch_assoc($Id))){
$row['Id']=max(mysqli_fetch_assoc($Id));#TODO check === operator //This logic is important. DONT use = 1!
Ok, I am trying to write a program for the server end of my website using PHP. Using Netbeans as my IDE of choice I have encountered an error while attempting to write a function which will store a single row in an associative array.
The issue arises when I try to return the variable $SelectedRow. It causes an 'Unreachable Statment' warning. This results in the program falling flat on its face.
I can get this code to work without being contained in a function. However, I don't really feel that that is the way to go about solving my issues while I learn to write programs.
Side Notes:
This is the first question I have posted on SO, so constructive criticism and tips are much appreciated. I am happy to post any specifications that would help an answer or anything else of the sort.
I do not believe this is a so-called 'replica' question because I have failed to find another SO question addressing the same issue in PHP as of yet.
If anybody has any suggestions about my code, in general, I'd be stoked to hear, as I have only just started this whole CS thing.
You can only return one time. Everything after the first return is unreachable.
It's not entirely clear to me what you want to return from that function, but you can only return one value.
The return command cancels the rest of the function, as once you use it, it has served its purpose.
The key to this is to put all of your information in to an array and return it at the end of the function, that way you can access all of the information.
So try changing your code to this:
require_once 'C:/wamp/www/FirstWebsite/CommonFunctions.php';
function SelectRowByIncrementFunc(){
$dbhose = DB_Connect();
$SelectRowByIncrementQuery = "SELECT * FROM trialtable2 ORDER BY ID ASC LIMIT 1";
$result = mysqli_query($dbhose, $SelectRowByIncrementQuery);
$SelectedRow = mysqli_fetch_assoc($result);
$returnArray = array();
$returnArray["SelectRowByIncrementQuery"] = $SelectRowByIncrementQuery;
$returnArray["SelectedRow"] = $SelectedRow;
$returnArray["result"] = $result;
return $returnArray;
}
And then you can access the information like so:
$selectedArray = SelectRowByIncrementFunc();
$row = $selectedArray["SelectedRow"]
And so forth...

Getting information from MySQL

I'm having trouble getting info from my MySQL database.
Here is my code :
/********************
* Database Info
********************/
$host = "localhost";
$user = "admin";
$pass = "admin#";
$database = "db_admin";
/********************
* Database connection
********************/
$con = mysqli_connect( $host, $user, $pass, $database );
if (mysqli_connect_errno ()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error ();
}
$result = array();
if (isset($_POST['ID'])) {
$id = $_POST['ID'];
$query = "SELECT * FROM Servers WHERE PID='" .$id. "'";
$result = mysqli_query($con, $query);
}
print("<pre>".print_r($result,true)."</pre>");
My first question, Did I use "isset" function currectly?
Because it doesnt seem like it is actually going though the if statement.
The Url I am using is : #..com/view.php?ID=1
My second question, Did I use the $query correctly?
Because I echo $id and that echoed out a "MySQL Object()"
Finally, the print printed out "Array()"
I'm just starting on PHP, Thanks for the help :)
A few things:
If you're passing the variable in the query string, use $_GET instead of $_POST to retrieve the values.
$result will return an pointer to the recordset, not the rows themselves. You will have to use mysqli_fetch_array() to fetch the rows.
ADD:
If you are sure that you will only have 1 record returning, you can use:
$row = mysqli_fetch_assoc($query);
echo $row['field_name'];
More then 1 record?
while($row = mysqli_fetch_assoc($query)){
echo $row['field_name'];
}
# your first question: if you have a input field with the name="ID", then its good.
Please also post your HTML :)
$var = 'Hello world';
if(isset($var)){ //If the var $var has been set (in this case it is)
echo $var;
} else {
//If $var is not set, then we get in the else
echo 'The var $var is not set';
}
The best thing is debugging the code with a debugger, you may use XDebug, or at least use var_dump(); to see what happens
var_dump($_REQUEST, $result);
Answer to your first question: It's hard to say if you've used it correctly when you haven't said what you're trying to do. I'm presuming that you only want run the code enclosed in the if-statement if the POST variable 'ID' has been received. If so, yes you've done it correctly.
Answer to your second question: I'm presuming on this line you're trying to build a string with a valid MySQL query. You've done that correctly, assuming $_POST['ID'] is a string (or can be converted to a string, see http://www.php.net/manual/en/language.types.string.php#language.types.string.casting).
If you're echoing $id and it's returning an object, however, you'll have a problem. You can't combine a string and an object like that. You'd need to iterate the object with a foreach, for example, and extract the id from that. The rest of the code won't work until that part is resolved.
The thing to investigate now is why $_POST['ID'] is returning an object. You'll need to provide the form code at the very least.

Passing PHP MySQL Result Object to Function

I'm trying to take a MySQL result row and pass it to a function for processing but the row isn't getting passed. I'm assuming this is because the actual row comes back as a object and objects can't get passed to function?
E.G
function ProcessResult($TestID,$Row){
global $ResultArray;
$ResultArray["Sub" . $TestID] = $Row["Foo"] - $Row["Bar"];
$ResultArray["Add" . $TestID] = $Row["Foo"] + $Row["Bar"];
}
$SQL = "SELECT TestID,Foo,Bar FROM TestResults WHERE TestDate !='0000-00-00 00:00:00'";
$Result= mysql_query($SQL$con);
if(!$Result){
// SQL Failed
echo "Couldn't find how many tests to get";
}else{
$nRows = mysql_num_rows($Result);
for ($i=0;$i<$nRows;$i++)
{
$Row = mysql_fetch_assoc($Result);
$TestID = $Row[TestID];
ProcessResult($TestID,$Row);
}
}
What I need is $ResultArray populated with a load of data from the MySQL query. This isn't my actual application (I know there's no need to do this for what's shown) but the principle of passing the result to a function is the same.
Is this actually possible to do some how?
Dan
mysql_query($SQL$con); should be mysql_query($SQL,$con); The first is a syntax error. Not sure if this affects your program or if it was just a typo on here.
I would recommend putting quotes around your array keys. $row[TestID] should be $row["TestID"]
The rest looks like it should work, although there are some strange ideas going on here.
Also you can do this to make your code a little cleaner.
if(!$Result){
// SQL Failed
echo "Couldn't find how many tests to get";
}else{
while($Row = mysql_fetch_assoc($Result))
{
$TestID = $Row['TestID'];
ProcessResult($TestID,$Row);
}
}
mysql_fetch_assoc() returns an associative array - see more
If you need an object, try mysql_fetch_object() function - see more
Both array and object can be passed to a function. Thus, your code seems to be correct, except for one line. It should be:
$Result= mysql_query($SQL, $con);
or just:
$Result= mysql_query($SQL);

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